Nuclear physics

Class 10 Physics, unit 21, book pages 309–350

Nuclear physics studies the atomic nucleus, which is made of protons and neutrons called nucleons. This unit follows how the idea of the atom changed from the plum pudding model to Rutherford's nucleus, how nuclei are written and how isotopes differ, and how unstable nuclei decay by alpha, beta and gamma emission with a half-life that is used to date ancient objects. It then covers ionizing radiation, its effects and safe use, its applications and background sources, nuclear fission and fusion with the mass-energy relation $E = m c^{2}$ , the nuclear reactions that power the Sun, and the ideas of dark matter and falsifiability.

Learning outcomes (25)

Describe the structure of the atom.
Justify the findings of the alpha-particle scattering experiments.
Define the terms proton number (atomic number) $Z$ and nucleon number (mass number) $A$ , and calculate the number of neutrons in a nucleus.
Recall the term nuclide and use the nuclide notation $_{Z}^{A} X$ .
Explain what is meant by an isotope and state that an element may have more than one isotope.
Explain what is meant by background radiation.
State the sources that make a significant contribution to background radiation.
Describe the emission of radiation from a nucleus as spontaneous and random.
Describe alpha particles, beta particles and gamma radiation.
Justify qualitatively the order of strength of alpha particles, beta particles and gamma radiation in terms of (a) their relative ionizing effects and (b) their relative penetrating powers.
Describe the deflection of alpha particles, beta particles and gamma radiation in electric and magnetic fields.
Explain that radioactive decay is a change in an unstable nucleus that can result, most commonly, in the emission of alpha particles, beta particles or gamma radiation.
Use decay equations, using nuclide notation, to show the emission of alpha particles, beta particles and gamma radiation.
Describe nuclear reactions (fission and fusion) with examples.
Recognise that matter can be converted to energy and vice versa (in this way the law of conservation of energy still holds).
Apply the equation $E = m c^{2}$ to calculate the energy released in the process of nuclear reactions.
Describe the activity of a radioactive material in terms of counts per unit time.
Define and infer the half-life of materials.
Explain and apply the concept of carbon dating to solve problems.
Explain how the type of radiation emitted and the half-life of the isotope determine which isotope is used for applications.
State the effects of ionizing nuclear radiations on living things, including cell death, mutations and cancer.
Explain how radioactive materials are moved, used and stored in a safe way.
Explain the nature of the Sun.
Describe that it is hypothesized that most of the matter in the universe is made up of dark matter.
Explain, with examples in physics, falsifiability as the idea that a theory is scientific only if it makes assertions that can be disproved.

21.1Discovery of the atomic nucleus

British physicist J.J. Thomson made a groundbreaking discovery on 30 April 1897: atoms are made up of smaller components. In 1898 Thomson suggested that an atom is a positively charged chunk of matter with negatively charged electrons scattered throughout it, like raisins in a fruitcake. This idea is known as the plum pudding model.

Figure 21.1: The Thomson plum pudding model of the atom. Electrons are embedded in a uniform sphere of positive charge.

By the end of the nineteenth century, scientists had found that electrons make up only a tiny part of an atom's total mass. Since atoms are neutral, the rest of the atom must be positive charge, and it must carry most of the mass. The question was: how are the mass and the positive charge spread out in the atom?

Rutherford's alpha-particle scattering experiment

Thomson's plum pudding model was disproved in 1911 by Ernest Rutherford's alpha-particle scattering experiment. Rutherford had studied the radioactive decay of uranium and found that it emits two types of radiation. One type, called alpha ( $α$ ) rays, is less penetrating than the other type, which he named beta ( $β$ ) rays. Both are studied in detail later in this unit.

Rutherford and his research assistant Hans Geiger built a special device (Figure 21.2) to see how alpha particles interact with a thin gold foil. They directed alpha rays at a piece of gold foil and used a screen coated with zinc sulfide to detect the scattered alpha particles. Whenever an alpha particle hit the zinc sulfide, it gave out a flash of light that could be seen through a detector. Rutherford moved the detector to different positions around the foil to find out whether the alpha rays were deflected.

Figure 21.2: Rutherford's experiment aimed a beam of alpha rays from a polonium source at a very thin sheet of gold foil. The detector can be moved round the foil.

According to Thomson's model, the alpha particles should have passed straight through the foil without much deflection. At first most of the alpha particles did pass straight through, as expected. However, after a few days Geiger excitedly told Rutherford that some alpha particles were scattering at angles greater than $9 0^{\circ}$ .

Rutherford was amazed. He compared it to the unlikely event of a 15-inch shell bouncing back after hitting a piece of tissue paper. The observations showed consistently that about 1 in every 20,000 alpha particles was deflected through more than $9 0^{\circ}$ .

Rutherford's nuclear model of the atom

Rutherford decided that the alpha particles could only be repelled by an immensely powerful electrostatic field. For such a field to exist, all the positive charge had to be confined within a minuscule space at the core of the atom. He therefore put forward the nuclear model of the atom: the majority of an atom's positive charge and mass is concentrated in a tiny region at the centre, which he called the nucleus, while the negatively charged electrons orbit around it, like planets revolving around the Sun (Figure 21.3).

Figure 21.3: Rutherford's atomic model. Light electrons orbit a very small, heavy, positively charged nucleus.

In interpreting his results, Rutherford built a model of atomic structure with the following features.

A very small nucleus surrounded by mostly empty space. This accounts for the observation that most alpha particles passed through this great volume of empty space and were undeflected.
A nucleus containing most of the mass of the atom. Only a few alpha particles were redirected at wide angles, and some even rebounded back almost completely. This can only happen if there is a dense, concentrated mass at the centre of the atom.
A nucleus that is positively charged. The force of electric repulsion from this small, positive nucleus deflected a positive alpha particle only when it came very close to the nucleus. Since very few particles were deflected by any appreciable angle, the nucleus must be so small that most alpha particles did not approach it closely enough to be deflected.
A number of very light, negatively charged electrons that move in orbits around the nucleus. As the alpha particles travelled straight through the empty space in the atom, every so often one of them would come across an electron, attract it by electrostatic force and continue on its path, hardly affected by the small mass of the electron it picked up.

What Rutherford observed and what he concluded
Observation	What it shows
Most alpha particles pass straight through the foil	The atom is mostly empty space and the nucleus is very small
A few are deflected through wide angles; about 1 in 20,000 by more than $9 0^{\circ}$ , some almost straight back	The mass is packed into a dense nucleus at the centre
Deflection happens only when an alpha particle comes very close to the nucleus	The nucleus carries all the positive charge and repels the positive alpha particle
An alpha particle passing an electron is hardly affected	Electrons are very light and move round the nucleus

How a positive alpha particle is repelled by the small positive nucleus: it is deflected only when it passes close to it, and an almost head-on approach throws it back.

The gold atom, now missing an electron, would become a positive ion. Rutherford proposed that these electrons orbit the positive nucleus, kept in place by electric attraction, much as the planets orbit the Sun because of gravity. The model accepted today is quite different from the Rutherford model.

The atom today

The simple Rutherford picture has been replaced by a more accurate and advanced quantum mechanical description of the atom. In this model the nucleus is made up of protons and neutrons, which are tightly packed together in a very small volume. Protons have a positive charge, while neutrons have no charge. The number of protons in the nucleus determines the element of the atom. An atom is mostly empty space, with the nucleus at the centre.

The nucleus is typically about 10,000 times smaller than the atom as a whole. The size difference is like a ping pong ball in the centre of a 2 km circle. The electrons are much smaller and lighter than the protons and neutrons. They are constantly in motion, buzzing around the nucleus at incredibly high speeds. However, their exact location and movement cannot be determined with certainty. Instead, we can only describe their probable location, using the probabilities given by quantum mechanics.

Electron cloud of the hydrogen atom

r_{max}

highest

radial distance

probability

Figure 21.4: The probability of finding the electron of a hydrogen atom in its ground state, against the radial distance from the nucleus (

r_{max}

is the distance of highest probability). The curve starts at zero at the nucleus, rises to a maximum and then falls slowly. The electron cloud is densest where the probability is highest.

If we could see an atom, we would not see a sharp boundary between the nucleus and the electron cloud. Instead we would see a fuzzy cloud of electrons surrounding the nucleus. This cloud shows the probable locations of the electrons at any given time. The nucleus would appear as a small, dense core within the cloud.

The electrons are organized into different energy levels or shells around the nucleus. These shells are not physical tracks or orbits like those of the planets in the solar system; they stand for the energy levels that electrons can occupy. The closer a shell is to the nucleus, the lower the energy of the electrons in it. Electrons can move between energy levels by absorbing or releasing energy. When an electron absorbs energy it jumps to a higher level; when it releases energy it drops to a lower level. The energy can come from various sources, such as heat, light or electric currents.

21.2Nuclear representationsOften examined

In describing the atomic nucleus we use the following quantities.

Atomic number $Z$ (charge number, proton number): The number of protons in the nucleus.
Mass number $A$ (nucleon number): The number of nucleons (protons plus neutrons) in the nucleus.
Neutron number $N$: The number of neutrons in the nucleus.
Nucleon: A particle found inside the nucleus: a proton or a neutron.
Nuclide: The kind of atom whose nucleus has a specific number of protons and neutrons, that is, given values of $A$ and $Z$ .

N = A - Z

neutron number

To show how many protons and neutrons are present in a nucleus, we use a system of symbols. The nuclide is written $_{Z}^{A} X$ , where $X$ is the chemical symbol of the element, $Z$ is the number of protons and $A$ is the number of nucleons (protons and neutrons together), as shown in Figure 21.5. The number of neutrons is found by subtracting the number of protons from the number of nucleons: $N = A - Z$ .

Figure 21.5: Nuclear representation.

A

(top left) is the mass number,

Z

(bottom left) is the atomic number. The example is the nuclide carbon-12,

_{6}^{12} C

Reading nuclide symbols: the neutron number is $A - Z$
Nuclide	$Z$ (protons)	$A$ (nucleons)	$N = A - Z$ (neutrons)
$_{6}^{12} C$	6	12	6
$_{6}^{14} C$	6	14	8
$_{27}^{60} Co$	27	60	33
$_{53}^{131} I$	53	131	78
$_{92}^{235} U$	92	235	143
$_{92}^{238} U$	92	238	146

21.3Isotopes and radioisotopesOften examined

Isotopes: Atoms of an element with the same atomic number (the same number of protons) but a different mass number (a different number of neutrons).

The different numbers of neutrons in the isotopes of an element give a different mass to each isotope. An element can exist in various isotopic forms because the number of neutrons in the nucleus of an atom of that element can vary while the number of protons stays the same.

For example, consider the three isotopes of hydrogen shown in Figure 21.6. The simple hydrogen atom $^{1} H$ (protium) has only one proton in its nucleus. $^{2} H$ (deuterium) has one proton and one neutron, whereas $^{3} H$ (tritium) has one proton and two neutrons. The figure also shows a helium atom, which is not an isotope of hydrogen because it has a different number of protons (and therefore of electrons).

Figure 21.6: Three isotopes of hydrogen and a helium atom. The neutrons (coloured) are the only difference between the hydrogen isotopes.

The isotopes of hydrogen
Isotope	Symbol	Protons	Neutrons	Nucleons $A$
Protium	$_{1}^{1} H$	1	0	1
Deuterium	$_{1}^{2} H$	1	1	2
Tritium	$_{1}^{3} H$	1	2	3

Isotopes of the same element have the same number of electrons, so they have identical chemical properties. Isotopes cannot be separated by chemical methods, but only by methods based on their mass differences, such as mass spectrometry.

Isotopes can exist naturally and can also be produced artificially. Some of them are unstable and emit radiations to reach stability. These isotopes are called radioisotopes.

Radioisotope: An isotope that is unstable and emits radiation to attain stability.
Relative abundance: The percentage of atoms with a specific atomic mass found in a naturally occurring sample of an element.

The relative abundance of $^{20} Ne$ is $90.48%$ , that of $^{21} Ne$ is $0.27%$ and that of $^{22} Ne$ is $9.25%$ (Figure 21.7). Natural uranium contains two isotopes: $^{238} U$ ( $99%$ ) and $^{235} U$ (a little less than $1%$ ). $^{235} U$ is used as a fuel in nuclear reactors and in atomic bombs, so techniques have been developed to separate $^{235} U$ from natural uranium.

Figure 21.7: Isotopes of neon. All have 10 protons and 10 electrons; only the number of neutrons changes.
Isotope of neon	Protons	Neutrons	Electrons
Neon-20 (natural)	10	10	10
Neon-21 (natural)	10	11	10
Neon-22 (natural)	10	12	10
Neon-24 (the most stable artificial isotope)	10	14	10

The first 80 elements of the periodic table have stable isotopes. Over 3,000 isotopes either exist in nature or have been made artificially in particle accelerators, and only 257 of them are stable. The remaining isotopes are unstable and are called radioisotopes. Radioisotopes are produced artificially and safely in research reactors and accelerators. One of the many uses of radioisotopes is to manage cancer and chronic diseases by radioisotope therapy, which treats cancerous cells in a safe and effective manner.

Point to ponder: isobars and isotones

Is it possible to have atoms with the same nucleon number but different proton numbers? Yes.

Isobars: Atomic species with the same mass number $A$ but a different atomic number $Z$ . They should not be confused with isotopes, which share the same atomic number and so belong to the same element. Examples: $^{14} C$ , $^{14} N$ , $^{14} O$ .
Isotones: Two or more species of atoms or nuclei that have the same number of neutrons. Chlorine-37 and potassium-39 are isotones: the nucleus of chlorine-37 has 17 protons and 20 neutrons, and that of potassium-39 has 19 protons and 20 neutrons.

Worked example: isotopes, isobars and isotones

Five nuclides are

_{17}^{35} Cl

_{17}^{37} Cl

_{19}^{39} K

_{18}^{40} Ar

and

_{20}^{40} Ca

. (a) Find the number of protons and neutrons in each. (b) Name the pairs or groups that are isotopes, isobars and isotones.

Protons are $Z$ (the lower number) and neutrons are $N = A - Z$ .
Work out $N$ for each nuclide: $35 - 17 = 18$ , $37 - 17 = 20$ , $39 - 19 = 20$ , $40 - 18 = 22$ , $40 - 20 = 20$ .

Nuclide	Protons $Z$	Nucleons $A$	Neutrons $N = A - Z$
$_{17}^{35} Cl$	17	35	18
$_{17}^{37} Cl$	17	37	20
$_{19}^{39} K$	19	39	20
$_{18}^{40} Ar$	18	40	22
$_{20}^{40} Ca$	20	40	20

Isotopes (same $Z$ , different $A$ ): chlorine-35 and chlorine-37, both with $Z = 17$ .
Isobars (same $A$ , different $Z$ ): argon-40 and calcium-40, both with $A = 40$ .
Isotones (same $N$ ): chlorine-37, potassium-39 and calcium-40, each with $N = 20$ .
Check: potassium-39 and calcium-40 are neither isotopes nor isobars, because both $Z$ and $A$ differ.

I so t o p es : C l - 35 an d C l - 37. I so ba r s : A r - 40 an d C a - 40. I so t o n es : C l - 37, K - 39 an d C a - 40 (20 n e u t r o n s) .

21.4Radioactivity

Henri Becquerel made a groundbreaking discovery in 1896 when he found radioactivity by accident. He observed that crystals of uranium salt gave out invisible radiations that could affect a photographic plate. This led him to propose that radioactivity came from the breakdown of unstable nuclei. He also noted that these radiations could ionize a gas.

Other researchers did further experiments showing that different substances emit similar radiations. Marie Curie and her husband Pierre made the remarkable discovery of two new elements, polonium and radium, which emitted these radiations. Marie Curie introduced the term radioactivity for this emission of radiation by certain elements.

Radioactivity: The natural emission of radiations from unstable nuclei.
Radioactive elements: The elements which emit such radiations.

Radioactivity was found not to be controlled or influenced by physical factors such as heating and cooling, and it remained unchanged even after strong chemical reactions. It was therefore recognized as a fundamental property of atoms: as radioactive atoms decompose, their nuclei emit the radiations.

When Rutherford and his colleagues passed a beam of radioactive emission through an electric field and a magnetic field (Figure 21.8), they found that it split into three types, at first called alpha particles, beta particles and gamma rays. The amount and direction of the deflection depend on their charge and mass. A neutral radiation, like gamma rays, is not deflected at all.

Figure 21.8: The three types of nuclear radiation in (a) an electric field (from the bottom plate to the top plate) and (b) a magnetic field into the page. Alpha particles bend slightly one way, beta particles bend strongly the other way, and gamma rays go straight on.

21.4.1Nature of nuclear emissionsOften examined

All three nuclear radiations were observed to have a distinct nature.

Alpha ( $α$ ) emissions

Alpha particles are equivalent to helium nuclei (two protons and two neutrons bound together) emitted from the nucleus. Certain atoms restore the balance between the neutrons and protons in the nucleus by emitting alpha particles. Alpha emission occurs in very large atoms, that is, those with high atomic numbers.

In an electric field, alpha particles are attracted towards the negative plate. Their deflection is small because they are heavy. In a magnetic field they move in a curved path whose direction is given by Fleming's left-hand rule. Since alpha particles are positive, the force acts in the same direction as the thumb shows.

Beta ( $β$ ) emissions

Beta particles are mostly fast-moving electrons emitted from the nucleus, and they carry a negative charge. Beta emission occurs when the ratio of neutrons to protons in the nucleus is too high. In that case an excess neutron changes into a proton and an electron.

In an electric field, the negatively charged beta particles are attracted to the positive plate. Because they are very light, they are deflected much more than alpha particles. In a magnetic field they also move in curved paths determined by Fleming's left-hand rule; since their charge is negative, they bend the opposite way to alpha particles (Figure 21.8).

Gamma ( $γ$ ) emissions

Gamma rays are electromagnetic radiations emitted from the nucleus, so they are neutral and have no charge. Gamma emission occurs when the nucleus of a radioactive atom has too much energy.

Gamma radiation is not affected by electric or magnetic fields. It passes straight through without any deflection, which makes gamma rays easy to tell apart from charged particles.

Nature of the three radiations and their behaviour in fields
	Alpha ( $α$ )	Beta ( $β$ )	Gamma ( $γ$ )
What it is	Helium nucleus (2 protons, 2 neutrons)	Fast electron	Short-wavelength electromagnetic radiation
Charge	Positive	Negative	None
In an electric field	Slight bend towards the negative plate	Larger bend towards the positive plate	Not deflected
In a magnetic field	Slight curve (Fleming's left-hand rule)	Larger curve, opposite sense to alpha	Not deflected

21.4.2Order of strength: penetrating power and ionizing effectOften examined

The strength of nuclear radiation can be understood in two key ways: its penetrating power and its ionizing effect.

Penetrating power

Penetrating power is how far radiation can travel through matter. It is characterized by the loss of energy when the radiation interacts with a substance. Beta radiation is more penetrating than alpha radiation. It can pass through the skin, but it is absorbed by a few centimetres of body tissue or a few millimetres of aluminium. Gamma rays are the most penetrating form of radiation and can pass through most materials; thick concrete or lead is needed to shield against them effectively.

Ionizing effects

Ionization is the process in which an electron is given enough energy to break away from an atom. Two charged particles, or ions, are formed. All types of nuclear radiation can ionize atoms and molecules, which means they knock electrons off, and this can damage living cells. The relative ionizing ability compares the ionization caused by $α$ , $β$ and $γ$ radiations.

Alpha particles cause the most ionization per unit path because of their large mass and charge. Beta and gamma rays cause less ionization but can still be harmful, especially because they can travel farther and penetrate deeper into living tissue.

Figure 21.9: The three types of nuclear emission. A sheet of paper stops alpha rays, thin aluminium stops beta rays, and thick lead is needed to stop gamma rays.

Table 21.1: Radioactive emissions
Emission	Nature of radiation	Charge	Relative penetrating power	Relative ionizing ability
alpha ( $α$ )	Helium nuclei (two protons and two neutrons)	$+ 2$	Stopped by a sheet of paper	Highest
beta ( $β$ )	Electrons	$- 1$	Several millimetres of aluminium	Medium
gamma ( $γ$ )	Short-wavelength electromagnetic radiation	$0$	Several centimetres of lead	Lowest

Worked example: identifying radiations from their behaviour

A radioactive source gives out three radiations X, Y and Z in a beam moving from left to right through a magnetic field pointing into the page, as in Figure 21.8(b). X bends slightly upwards, Y bends strongly downwards and Z goes straight on. A sheet of paper stops X only, a few millimetres of aluminium stop Y, and Z is reduced only by thick lead. (a) Name X, Y and Z. (b) Which one is the most strongly ionizing? (c) Sketch the paths when the field points out of the page.

(a) Z is not deflected by the field, so it is neutral: it is gamma radiation. This agrees with its great penetration, since only thick lead reduces it.
X bends slightly in the way Figure 21.8(b) shows for alpha particles (positive and heavy), and paper stops it: X is an alpha particle.
Y bends the opposite way and much more (negative and very light), and a few millimetres of aluminium stop it: Y is a beta particle (an electron).
(b) Alpha particles cause the most ionization per unit path because of their large mass and charge, so X is the most ionizing.
(c) Reversing the field reverses the sideways force on every charged particle. X (positive) now bends slightly downwards, Y (negative) bends strongly upwards, and Z (neutral) is still undeflected.

Paths when the magnetic field points out of the page (dots). The directions of the alpha and beta deflections are swapped; the gamma ray still goes straight.

X = alpha, Y = beta, Z = gamma; alpha is the most ionizing; with the field reversed, alpha bends down and beta up.

21.5Nuclear decay and its conservation lawsOften examined

Nuclear disintegration follows several conservation principles, including conservation of mass-energy, momentum and charge. To these laws we add one more for radioactive decay, the conservation of nucleons. It states that the total number of nucleons (the mass number $A$ ) must stay constant during nuclear reactions or decay.

Radioactive decay is a change in an unstable nucleus. Most commonly it results in the emission of an alpha particle, a beta particle or gamma radiation. The original nuclide is called the parent nuclide and the nuclide that is formed is the daughter nuclide.

Radioactive decay: The process by which the nucleus of an unstable atom loses mass and energy by emitting ionizing particles or radiation.
Parent nuclide: The original, unstable nuclide before the decay.
Daughter nuclide: The nuclide formed after the decay.

How the nucleon number $A$ and the proton number $Z$ change in each decay
Decay	Particle emitted	Change in $A$	Change in $Z$	Daughter
Alpha	$_{2}^{4} He$ ( $α$ )	$- 4$	$- 2$	$_{Z - 2}^{A - 4} Y$
Beta (electron)	$_{- 1}^{0} β$	$0$	$+ 1$	$_{Z + 1}^{A} Y$
Beta plus (positron)	$_{+ 1}^{0} β$	$0$	$- 1$	$_{Z - 1}^{A} Y$
Gamma	$γ$ (no particle)	$0$	$0$	$_{Z}^{A} X$ (same nuclide, lower energy)

To write a decay equation, follow these steps.

Write the parent nuclide with its $A$ and $Z$ on the left.
Write the emitted particle on the right with its own $A$ and $Z$ (alpha $_{2}^{4} He$ , beta $_{- 1}^{0} β$ , gamma $γ$ ).
Choose the daughter so that the totals of $A$ and of $Z$ on the two sides are equal (conservation of nucleons and of charge).
Use the periodic table to name the daughter element from its atomic number $Z$ .

21.5.1Alpha decayOften examined

In alpha decay, the original (parent) nuclide changes into a (daughter) nuclide by emitting an alpha particle, which is a helium nucleus. When the reaction is balanced, the daughter nuclide has a nucleon number reduced by four and a proton number decreased by two. This can be written as

_{Z}^{A} X \to_{Z - 2}^{A - 4} Y + α

21.1

For example, radon-222 changes into polonium-218 by releasing an alpha ( $α$ ) particle:

_{86}^{222} Rn \to_{84}^{218} Po +_{2}^{4} He (α)

Polonium-218 is unstable and changes into lead-214, again by releasing an alpha particle, as shown in Figure 21.10.

Figure 21.10: Conversion of

^{218} Po

^{214} Pb

by alpha decay:

_{84}^{218} Po \to_{82}^{214} Pb +_{2}^{4} He (α)

. The daughter has 4 fewer nucleons and 2 fewer protons.

Example 21.1: alpha decay of uranium-238

Find the atomic mass and atomic number of the nucleus that comes after alpha decay of uranium (U-238).

The alpha decay reaction is $_{Z}^{A} X \to_{Z - 2}^{A - 4} Y + α$ . For uranium-238, $A = 238$ and $Z = 92$ .
$_{92}^{238} U \to_{92 - 2}^{238 - 4} Y +_{2}^{4} He$
Subtract: the mass number becomes $234$ and the atomic number becomes $90$ .
$_{92}^{238} U \to_{90}^{234} Y +_{2}^{4} He$
Every element has a unique atomic number. The periodic table shows that the element with $Z = 90$ is thorium (Th). The completed nuclear reaction is:
$_{92}^{238} U \to_{90}^{234} Th +_{2}^{4} He (α)$

mass number 234, atomic number 90 :_{92}^{238} U \to_{90}^{234} Th +_{2}^{4} He

21.5.2Beta decayOften examined

In beta decay, the parent nuclide changes into a daughter nuclide by emitting a beta particle (an electron). This electron is thrown out of the nucleus when a neutron changes into a proton, and the proton stays in the nucleus. When the reaction is balanced, the nucleon number of the daughter nuclide is unchanged, while its charge (proton) number increases by one. This can be written as

_{Z}^{A} X \to_{Z + 1}^{A} Y +_{- 1}^{0} β

21.2

_{0}^{1} n \to_{1}^{1} p +_{- 1}^{0} β

what happens inside the nucleus

The nucleon number $1 = 1 + 0$ and the charge $0 = + 1 + (- 1)$ are both conserved.

For example, molybdenum-99 decays by beta decay because it has too many neutrons. A neutron in the nucleus changes into a proton and a beta particle is emitted:

_{42}^{99} Mo \to_{43}^{99} Tc +_{- 1}^{0} β

Technetium-99 is unstable again and emits beta radiation, the release of an energized electron from the nucleus, as it decays into its daughter ruthenium-99 (Figure 21.11).

Figure 21.11: Conversion of

^{99} Tc

^{99} Ru

by beta decay:

_{43}^{99} Tc \to_{44}^{99} Ru +_{- 1}^{0} β

. One neutron has become a proton: the nucleon number stays 99, the proton number rises from 43 to 44.

Example 21.2: beta decay of cobalt-60

Cobalt-60 undergoes beta decay. Write the reaction equation and determine the identity of the daughter nucleus.

Beta decay is given by $_{Z}^{A} X \to_{Z + 1}^{A} Y +_{- 1}^{0} β$ . For cobalt-60, $A = 60$ and $Z = 27$ .
$_{27}^{60} Co \to_{27 + 1}^{60} Y +_{- 1}^{0} β$
Add one to the atomic number: $Z = 28$ and $A$ stays $60$ .
$_{27}^{60} Co \to_{28}^{60} Y +_{- 1}^{0} β$
Every element has a unique atomic number. The periodic table shows that the element with $Z = 28$ is nickel (Ni). The completed equation is:
$_{27}^{60} Co \to_{28}^{60} Ni +_{- 1}^{0} β$

The daughter is nickel-60:

_{27}^{60} Co \to_{28}^{60} Ni +_{- 1}^{0} β

Beta plus decay

Beta radiation also includes positrons, which are emitted from the nucleus in some beta decays. A positron (or antielectron) is a particle with the same mass as an electron but a positive electric charge of the same size (like a proton); it is the antiparticle of the electron. When a positron is emitted, the process is called beta plus decay, to tell it apart from the emission of an electron. It is caused by the change of a proton into a neutron, which stays in the nucleus.

As a result, the proton number decreases by one and the nucleon number does not change in beta plus decay:

_{Z}^{A} X \to_{Z - 1}^{A} Y +_{+ 1}^{0} β

_{1}^{1} p \to_{0}^{1} n +_{+ 1}^{0} β

what happens inside the nucleus

For example, the emission of a beta plus particle by chromium-46 forms vanadium-46, and the emission of a beta plus particle changes fluorine-18 into oxygen-18:

_{24}^{46} Cr \to_{23}^{46} V +_{+ 1}^{0} β

_{9}^{18} F \to_{8}^{18} O +_{+ 1}^{0} β

Comparison of the two kinds of beta decay
	Beta (electron) decay	Beta plus (positron) decay
Change inside the nucleus	neutron $\to$ proton	proton $\to$ neutron
Particle emitted	electron, $_{- 1}^{0} β$	positron, $_{+ 1}^{0} β$
Proton number $Z$	increases by 1	decreases by 1
Nucleon number $A$	unchanged	unchanged
Example	$_{42}^{99} Mo \to_{43}^{99} Tc +_{- 1}^{0} β$	$_{24}^{46} Cr \to_{23}^{46} V +_{+ 1}^{0} β$

21.5.3Gamma decay

Gamma radiation is a form of high-energy (high-frequency, short-wavelength) electromagnetic radiation. In most cases the emission of an \alpha or \beta particle from the nucleus leaves it in an excited state. Such nuclei reach further stability by giving out the extra energy in the form of gamma rays. This can be written as

{}^{A}_{Z}\mathrm{X}^{*}\to{}^{A}_{Z}\mathrm{X}+\gamma

21.3

The star shows the excited (high-energy) state of the nucleus. For example:

{}^{60}_{27}\mathrm{Co}^{*}\to{}^{60}_{27}\mathrm{Co}+\gamma

Neither A nor Z changes in gamma decay: the nuclide stays the same, only its energy falls. Gamma rays carry no charge and no mass.

21.5.4Stability through nuclear transmutationOften examined

When a particle is emitted from a parent nucleus during radioactive decay, the ratio of neutrons to protons changes, and the resulting daughter nucleus tends to become more stable. This process of changing one element into another is called nuclear transmutation. Typically a series of successive decays takes place until a stable, non-radioactive arrangement is reached.

Nuclear transmutation: The change of one element into another when a nucleus decays.

Nuclear decay is a spontaneous and random process, governed by the inherent characteristics of atomic nuclei. The imbalance between protons and neutrons, along with the energy differences within the nucleus, drives the process without any outside intervention, much like a ball rolling down a hill because of gravity. Just as we cannot predict exactly when the ball will start to move, it is impossible to say the exact moment when a particular unstable nucleus will decay. This uncertainty is governed by the principles of quantum mechanics, which include probability.

Spontaneous: The decay needs no outside cause: heating, cooling or chemical reactions do not start it or change it.
Random: It cannot be predicted which nucleus will decay next or when; only the chance of decay is fixed.

Worked example: a decay chain

Lead-214 ({}^{214}_{82}\mathrm{Pb}) decays by beta emission. Its daughter also decays by beta emission, and the next daughter then emits an alpha particle. Write the three equations, name the nuclides formed and state the final nuclide. Check that the whole chain is balanced.

Step 1, beta decay. A stays 214 and Z rises by one, from 82 to 83. The periodic table gives bismuth (Bi) for Z=83.
{}^{214}_{82}\mathrm{Pb}\to{}^{214}_{83}\mathrm{Bi}+{}^{0}_{-1}\beta
Step 2, beta decay again. A stays 214 and Z rises from 83 to 84: polonium (Po).
{}^{214}_{83}\mathrm{Bi}\to{}^{214}_{84}\mathrm{Po}+{}^{0}_{-1}\beta
Step 3, alpha decay. A falls by 4 to 210 and Z falls by 2 to 82: lead (Pb) again.
{}^{214}_{84}\mathrm{Po}\to{}^{210}_{82}\mathrm{Pb}+{}^{4}_{2}\mathrm{He}
Check the whole chain. Start: A=214. End: 210 (lead) +\,4 (alpha) +\,0+0 (two betas) =214. Charge: start 82; end 82 (lead) -1-1 (two betas) +2 (alpha) =82. Both totals agree.

The chain: two beta decays (nucleon number unchanged, proton number up by 1 each time) followed by one alpha decay (nucleon number down by 4, proton number down by 2).

{}^{214}_{82}\mathrm{Pb}\xrightarrow{\beta}{}^{214}_{83}\mathrm{Bi}\xrightarrow{\beta}{}^{214}_{84}\mathrm{Po}\xrightarrow{\alpha}{}^{210}_{82}\mathrm{Pb}

21.6Half-lifeOften examined

Radioactivity is an unpredictable and spontaneous process: we cannot foresee which nucleus will decay next. This is why we use the idea of half-life to measure the time taken for the overall disintegrations.

Half-life (T_{1/2}): The time in which half of the unstable radioactive nuclei in a sample disintegrate into daughter nuclei (by the emission of \alpha, \beta or \gamma radiation).

The value of the half-life depends on the nature of the radioactive nucleus and not on any other variable such as temperature, pressure or humidity. For example, iodine-131 has a half-life of 8 days, because it takes this time for half of a given quantity of the isotope to disintegrate. In 8 days half of the sample will be left. In the next 8 days only \tfrac14 of the original sample will be left, and so on. In general, let N_0 be the number of nuclei in the sample at the start (it may also be written in units of mass) and N the number present at any time.

Using an equation to model radioactive decay

Quantities whose rate of decay depends on the amount of substance present are said to undergo exponential decay. The amount of a radioactive isotope in a given sample decreases with time as shown in Figure 21.12. The number of nuclei at time t=0 is N=N_0, the number at t=T_{1/2} is N=N_0/2, the number at t=2T_{1/2} is N=N_0/4, and so on.

Radioactive decay

Figure 21.12: Percentage of the sample remaining against time in half-lives. The amount halves in every half-life: 100%, 50%, 25%, 12.5%, 6.25%. The curve gets closer and closer to zero but, in theory, never reaches it.

After each half-life the amount is multiplied by one half
Half-lives passed	Time	Fraction left	Amount N
0	0	1	N_0
1	T_{1/2}	\tfrac12	\tfrac12N_0
2	2T_{1/2}	\tfrac12\times\tfrac12=\left(\tfrac12\right)^2	\left(\tfrac12\right)^2N_0
3	3T_{1/2}	\left(\tfrac12\right)^3	\left(\tfrac12\right)^3N_0
4	4T_{1/2}	\left(\tfrac12\right)^4	\left(\tfrac12\right)^4N_0
n	nT_{1/2}	\left(\tfrac12\right)^n	\left(\tfrac12\right)^nN_0

In general, after n half-lives:

N=\left(\dfrac12\right)^{n}N_0=\dfrac{N_0}{2^{n}}

21.4

The number of half-lives n is found from the total time passed and the half-life:

n=\dfrac{\text{total time passed}}{\text{half-life}}=\dfrac{\Delta t}{T_{1/2}}

21.5

Substituting this n into equation 21.4 gives the final equation:

N=\dfrac{N_0}{2^{\Delta t/T_{1/2}}}

21.6

Table 21.2: Half-lives of selected radioisotopes
Radioisotope	Half-life	Radioisotope	Half-life
Uranium-238	4.5 billion years	Iodine-131	8.02 days
Uranium-235	704 million years	Carbon-14	5,730 years
Plutonium-239	24,110 years	Tritium (hydrogen-3)	12.3 years
Cesium-137	30 years	Radon-222	3.82 days
Strontium-90	28.8 years	Polonium-210	138 days

Example 21.3: lead-212

You have a 50\,\mathrm{mg} sample of lead-212. It decays by beta and gamma emission with a half-life of 10.6 hours. How much of the pure sample is left after 53 hours?

Given

N_0 = 50\,\mathrm{mg}
T_{1/2} = 10.6\,\mathrm{h}
\Delta t = 53\,\mathrm{h}

To find

quantity left N

Find the number of half-lives.
n=\dfrac{\Delta t}{T_{1/2}}=\dfrac{53\,\mathrm{h}}{10.6\,\mathrm{h}}=5
The original sample remaining after n half-lives is N=\left(\tfrac12\right)^nN_0.
N=\left(\dfrac12\right)^{5}\times50\,\mathrm{mg}=\dfrac{50}{32}\,\mathrm{mg}=1.56\,\mathrm{mg}
So 1.56\,\mathrm{mg} of lead-212 is present after 53 hours, which is 5 half-lives.

N=1.56\,\mathrm{mg}

Until 1934 it was thought impossible to create radioactive materials artificially. In that year Frederic Joliot and Irene Curie succeeded in producing new radioactive materials using the methods of nuclear physics. Otto Hahn and Fritz Strassmann later achieved the fission of uranium nuclei in 1938/39. The importance of fission was recognized by Lise Meitner and Otto Frisch, which led to attempts to create super-heavy elements beyond uranium. This search has produced 26 elements that do not exist naturally.

Worked example: iodine-131 in a hospital

A hospital receives a 200\,\mathrm{mg} sample of iodine-131, whose half-life is 8 days. Find the amount left after (a) 24 days, (b) 20 days. (c) After how many days is only 12.5\,\mathrm{mg} left? Plot the decay curve to check your answers.

Given

N_0 = 200\,\mathrm{mg}
T_{1/2} = 8\,\mathrm{days}

To find

(a) N at 24 days
(b) N at 20 days
(c) \Delta t when N=12.5\,\mathrm{mg}

(a) The number of half-lives is n=24/8=3.
N=\dfrac{N_0}{2^{n}}=\dfrac{200}{2^{3}}=\dfrac{200}{8}=25\,\mathrm{mg}
(b) 20 days is not a whole number of half-lives: n=20/8=2.5. Equation 21.6 still works, because 2^{2.5}=4\times\sqrt2=5.657.
N=\dfrac{200}{2^{2.5}}=\dfrac{200}{5.657}=35.4\,\mathrm{mg}
This lies between 50\,\mathrm{mg} (after 16 days) and 25\,\mathrm{mg} (after 24 days), as it should.
(c) Divide the original amount by the amount left to find the power of 2: 200/12.5=16=2^{4}, so n=4.
\Delta t=nT_{1/2}=4\times8=32\,\mathrm{days}

Decay of iodine-131

The curve N=200/2^{t/8}. Every 8 days the amount halves: 200,\ 100,\ 50,\ 25,\ 12.5\,\mathrm{mg}. The 20 day point lies between the 16 day and 24 day points.

N(24\,\mathrm{d})=25\,\mathrm{mg},\quad N(20\,\mathrm{d})\approx35.4\,\mathrm{mg},\quad \Delta t=32\,\mathrm{days}

21.6.2Activity

The level of radioactivity in a substance can be thought of as the number of nuclear decays per unit time. A substance with high radioactivity decays rapidly, which gives a short half-life. If a sample initially contains N unstable nuclei, the rate of decay is \Delta N/\Delta t (taken as its size, since N is falling).

Activity (A): The number of nuclear decays (disintegrations or counts) per second in a radioactive sample.

A=\dfrac{\Delta N}{\Delta t}\qquad(\text{decays per second})

The SI unit of activity (decay rate) is the becquerel (Bq), named after Henri Becquerel: 1\,\mathrm{Bq} is 1 disintegration (count) per second. It is a very small unit. For example, 1.0\,\mathrm{g} of radium has an activity of 3.7\times10^{10}\,\mathrm{Bq}. The kilobecquerel (kBq) and megabecquerel (MBq) are therefore commonly used: the activity of 1.0\,\mathrm{g} of radium is 3.7\times10^{4}\,\mathrm{MBq}.

Another common unit of activity is the curie (Ci), roughly the activity of one gram of radium. It is defined as 3.70\times10^{10} decays per second:

1\,\mathrm{Ci}=3.70\times10^{10}\,\mathrm{Bq}=3.70\times10^{10}\ \text{decays per second}

Units of activity
Unit	Symbol	Meaning
Becquerel	\mathrm{Bq}	1 decay per second (SI unit)
Kilobecquerel	\mathrm{kBq}	10^{3}\,\mathrm{Bq}
Megabecquerel	\mathrm{MBq}	10^{6}\,\mathrm{Bq}
Curie	\mathrm{Ci}	3.70\times10^{10}\,\mathrm{Bq}

21.6.3Radioactive datingOften examined

Archaeologists and geologists use radioactive dating to estimate the age of ancient objects. One common procedure uses carbon-14, which has a half-life of 5730 years. The element carbon, as found in nature, consists of 98.89\% {}^{12}\mathrm{C} atoms and only about 1.11\% {}^{13}\mathrm{C} atoms. Carbon in living organisms also contains a trace, but a measurable amount, of {}^{14}\mathrm{C}: about 1.3\times10^{-12} of the carbon atoms.

Figure 21.13: Isotopes of carbon. Carbon-14 is the radioactive one.
Isotope of carbon	Protons	Neutrons	Electrons
Carbon-12	6	6	6
Carbon-13	6	7	6
Carbon-14	6	8	6

Neutrons from deep space collide with nitrogen atoms in the atmosphere. A proton is knocked off the nucleus of {}^{14}\mathrm{N} and the neutron is absorbed, forming a {}^{14}\mathrm{C} atom:

{}^{1}_{0}\mathrm{n}+{}^{14}_{7}\mathrm{N}\to{}^{14}_{6}\mathrm{C}+{}^{1}_{1}\mathrm{H}

carbon-14 is formed in the atmosphere

This carbon atom can react with oxygen in the atmosphere to form a \mathrm{CO_2} molecule. Through photosynthesis plants take in some of this radioactive carbon as they make their food, and animals that eat the plants also take in radioactive carbon. As long as an organism is alive, it keeps absorbing and accumulating carbon-14. When it dies, it no longer takes in carbon-14, and the proportion of carbon-14 in it gradually decreases (Figure 21.14).

Figure 21.14: Radiocarbon dating. Cosmic rays make carbon-14 in the air, living things take it in, and after death the carbon-14 decays at a known rate; the fraction left in a specimen shows its age.

The amount of carbon-14 in the total carbon is very small, yet it is still measurable. By measuring the activity, one can estimate the age of a specimen. Carbon-14 dating is effective for biological tissue up to 50 or 60 thousand years old, but it is most accurate for younger samples because of their higher {}^{14}\mathrm{C} abundance. The age of geologic formations can be found using materials with longer half-lives. For instance, uranium-238, with a half-life of 4.53\times10^{9} years, can date even the oldest deposits on Earth.

Example 21.4: age of an animal fossil

In an animal fossil the amount of {}^{14}\mathrm{C} is found to be one eighth of that in the bone of a living animal. The half-life of {}^{14}\mathrm{C} is 5730 years. How old is the fossil?

Given

N/N_0 = \tfrac18
T_{1/2} = 5730\ \text{years}

To find

time \Delta t

Use the half-life equation in the form \dfrac{N}{N_0}=\dfrac{1}{2^{\Delta t/T_{1/2}}} and substitute \tfrac18=\tfrac1{2^3}.
\dfrac{1}{2^{3}}=\dfrac{1}{2^{\Delta t/5730}}
The bases are equal, so the powers must be equal.
3=\dfrac{\Delta t}{5730\ \text{years}}
Solve for the time.
\Delta t=3\times5730\ \text{years}=17\,190\ \text{years}
So the fossil is 17\,190 years old (three half-lives).

\Delta t=17\,190\ \text{years}

Worked example: dating a wooden tool, and the limit of the method

A wooden tool contains \tfrac1{16} of the carbon-14 found in living wood. The half-life of carbon-14 is 5730 years. (a) How old is the tool? (b) Explain why carbon-14 would not be a suitable clock for a rock that is 100 million years old.

(a) Write the fraction left as a power of 2: \tfrac1{16}=\tfrac1{2^{4}}, so n=4 half-lives have passed.
\Delta t=nT_{1/2}=4\times5730=22\,920\ \text{years}
Check with the graph below: the \tfrac1{16} level (6.25\%) is reached at 22\,920 years, one half-life after the \tfrac18 level of Example 21.4.
(b) After 60\,000 years the number of half-lives is 60\,000/5730\approx10.5, so only about 1/2^{10.5}\approx1/1400 (roughly 0.07\%) of the carbon-14 is left. That is too small to measure reliably, so the method works only up to about 50 to 60 thousand years.
A 100 million year old rock is about 17\,000 half-lives of carbon-14, so essentially none is left. Also, a rock was never a living organism that took in carbon-14. For rocks and geologic formations a material with a much longer half-life is used, such as uranium-238 (4.53\times10^{9} years).

Carbon-14 left against age

The fraction of carbon-14 left halves every 5730 years. The two marked points are the \tfrac18 level of Example 21.4 (17\,190 years) and the \tfrac1{16} level of the wooden tool (22\,920 years).

\text{(a) }22\,920\ \text{years;}\quad\text{(b) too little }{}^{14}\mathrm{C}\text{ is left and a rock was never alive}

21.7Ionizing nuclear radiations

Radiation, or radiant energy, is energy in the form of waves or particles moving through space. Visible light, heat, radio waves, alpha particles and beta particles are all examples of radiation. When we feel warmth from the sunlight, we are absorbing the radiant energy emitted by the Sun.

Radiation can be classed as ionizing or nonionizing, according to how it interacts with matter. Normally an atom has an equal number of protons and electrons. Through a process called ionization, atoms can lose or gain electrons. Certain types of radiation can ionize atoms by knocking off electrons, leaving the atom in an ionized state. Hydrogen, with the smallest atomic number, needs the least energy (13.6\,\mathrm{eV}) to eject its K-shell electron. Radiation with energy above 13.6\,\mathrm{eV} is called ionizing radiation. Ionizing radiation can change the chemical state of matter and cause biological damage, so it is potentially harmful to people.

If electrons are raised to higher energy levels without being ejected from the atoms, the process is called excitation, and the atom is said to be "excited".

Radiation with energy less than 13.6\,\mathrm{eV} is called nonionizing radiation, because it cannot eject even this most easily removed electron. Nonionizing radiation bounces off matter or passes through it without displacing electrons. Examples include visible light and radio waves. At present it is unclear whether nonionizing radiation is harmful to human health.

Ionization: The process in which an atom loses or gains electrons (usually an electron is knocked out of an atom), forming ions.
Excitation: Raising electrons to higher energy levels without ejecting them from the atom.
Ionizing radiation: Radiation with energy above 13.6\,\mathrm{eV}, able to eject electrons from atoms; it can change the chemical state of matter and cause biological damage.
Nonionizing radiation: Radiation with energy below 13.6\,\mathrm{eV}, which cannot eject electrons; examples are visible light and radio waves.

Ionizing and nonionizing radiation
	Ionizing radiation	Nonionizing radiation
Energy	Above 13.6\,\mathrm{eV}	Below 13.6\,\mathrm{eV}
Effect on atoms	Ejects electrons (ionization)	Cannot eject electrons; passes through or bounces off
Possible harm	Can alter chemical state and damage living cells	Unclear whether it harms health
Examples	Alpha, beta and gamma radiation	Visible light, radio waves

21.7.1Effects of ionizing nuclear radiations on living things

Ever since ionizing radiation was first discovered, its ability to cause harmful effects has been recognized. These effects occur when the human body absorbs energy from the radiation. This absorbed energy can lead to ionization (charged particles) and excitation within the body, as well as chemical changes through the creation of free radicals. In the end this creates the potential for biological changes or damage.

Radiation interacts with matter by transferring energy to the molecules of the material that absorbs it. Since water is the most abundant molecule in a cell, the main result is an ionized water molecule (\mathrm{H_2O^{\bullet+}}). The ionized water molecules then react with other water molecules to form hydroxyl radicals (\mathrm{{}^{\bullet}OH}), as in Figure 21.15. The ejected electrons may have enough energy to cause further ionizations until their energy is used up. They end up as an aqueous electron (\mathrm{e^-_{aq}}) or combine with other species to form reducing species such as hydrogen atoms (\mathrm{H^{\bullet}}) or superoxide (\mathrm{O_2^{\bullet-}}).

Ionizing radiation can harm all types of molecules in a cell, but DNA is a main target. The highly reactive \mathrm{{}^{\bullet}OH} radicals, along with less reactive species, can damage DNA. Radiation can cause several types of damage to DNA, such as breaks in the strands, damage to bases or sugars, and cross-links between molecules (like DNA-DNA or DNA-protein cross-links). Overall, the most harmful effect of radiation is believed to be the DNA double-strand break (DSB).

Figure 21.15 (as a flow chart): ionizing radiation ionizes water in the cell, the reactive hydroxyl radicals attack DNA, and the damage leads either to cell death or to a mutation.

21.7.2Harmful effects of nuclear radiations

Cell death

Cell death by radiation happens when ionizing radiation damages the vital parts of a cell, especially its DNA. The damage can be so severe that the cell cannot repair it and eventually dies. There are two main types of radiation-induced cell death: mitotic cell death and apoptosis. Mitotic cell death happens when cells with damaged DNA try to divide, which leads to their breakdown. Apoptosis, or programmed cell death, is a controlled process in which cells destroy themselves when they detect significant damage. Both types protect the body by removing damaged cells that could otherwise cause harm.

Mutations

Sometimes the damage caused by ionizing radiation is not fatal to the cell. However, it can alter the cell's DNA and lead to mutations. Mutations are changes in the DNA sequence of a cell's genetic material. They can be passed on to daughter cells during cell division. Mutations can have a variety of effects, some harmless, but some can disturb the normal working of the cell. Worryingly, some mutations can give a cell the ability to grow uncontrollably, which is a hallmark of cancer.

Cancer

As mentioned above, mutations caused by ionizing radiation can increase the risk of cancer. Cancer is a disease in which cells lose control of their growth and division, forming tumours that can invade healthy tissue. The risk of developing cancer from radiation exposure depends on factors like the amount of radiation received, the age of the person exposed and other individual factors.

Effects on people

Nuclear radiation can harm humans by changing the structure of molecules in living cells through ionization. This can lead to cell death and possibly harm the whole organism. In addition, if radioactive material is swallowed or breathed in, it causes internal damage through contamination, because the emitted radiation affects cells from inside the body.

Radiation exposure can have serious health consequences, including radiation sickness, cancer and genetic mutations. These effects can be immediate or can develop over time, even from low-level exposure. Radiation sickness is the name for the acute effects of radiation exposure. Its symptoms can include nausea, vomiting, fever, diarrhoea and hair loss, and it can even lead to death, depending on the severity of the dose. Long-term effects of radiation exposure may include chronic health problems such as hair loss, eye cataracts, various types of cancer and genetic defects that can be passed down to future generations.

Harmful effects of ionizing radiation
Type of effect	Examples
Cellular	Mitotic cell death, apoptosis (programmed cell death), DNA double-strand breaks
Immediate (acute)	Radiation sickness: nausea, vomiting, fever, diarrhoea, hair loss; death at high doses
Long term	Chronic health problems, eye cataracts, various cancers
Genetic	Mutations in DNA passed to daughter cells and to future generations

Not everyone is equally sensitive to radiation exposure. Children, pregnant women and people with weakened immune systems are more vulnerable. Radioactive materials can also contaminate the environment and put wildlife and future generations at risk. For this reason, a comprehensive set of rules and guidelines has been made all over the world to ensure the safe and secure use of nuclear technology. These "nuclear regulations" are usually enforced by independent regulatory bodies. They may be national agencies, like the Office for Nuclear Regulation (ONR) in the UK or the Pakistan Nuclear Regulatory Authority (PNRA), or international organizations, like the International Atomic Energy Agency (IAEA).

21.7.3Safe handling of radioactive materials

When working near ionizing radiation, remember the mantra: time, distance and shielding. Reduce the time you spend near the source: the less time you are exposed, the lower the dose. Increase the distance between yourself and the source: the intensity of radiation weakens rapidly, so even a few extra steps make a big difference. Whenever possible, use barriers such as lead or concrete to block the radiation.

Time, distance and shielding
Rule	What to do	Why it helps
Time	Spend as little time as possible near the source	A shorter exposure gives a lower dose
Distance	Stay as far from the source as possible	The intensity of the radiation weakens rapidly with distance
Shielding	Place a barrier such as lead or concrete between you and the source	The barrier absorbs the radiation

In a radiation laboratory, extra care is needed. Always wear personal protective equipment such as lab coats, shoes and safety glasses. Keep the workspace clean and organized by avoiding unnecessary materials, and carefully keep a record of all radioactive sources in the laboratory so that they are handled and disposed of properly.

Radiation symbols are important wherever there are radiation sources, containers or products with radioactive parts. These symbols are found on safety documents, warning signs and permits for handling radioactive materials. They are essential for warning people of possible radiation dangers and for clear communication in radiation safety. Figure 21.16 shows two common radiation symbols.

International (trefoil) symbol of radiation. This sign must be displayed where radioactive materials are handled or where equipment that produces radiation is used. It is used as a warning to protect people from being exposed to radioactivity.
Supplementary symbol. In 2007 the International Atomic Energy Agency (IAEA) launched a new symbol for "Ionizing-Radiation Warning". It is meant to supplement the existing, well-known trefoil symbol. The new symbol is designed to give the message "Danger: stay away" to anyone who sees it, whatever their age, education or cultural background.

Figure 21.16: The international trefoil symbol for radiation. The book also shows the supplementary IAEA symbol, a triangle that warns "danger, stay away".

21.7.4Dose descriptors

We measure radiation to understand its effects on living things such as humans. This helps us set safety limits to protect people from radiation exposure. Several biological measures are commonly used to judge the effects of radiation exposure.

Absorbed dose: The amount of energy absorbed by a tissue or organ exposed to radiation. It is usually measured in gray (Gy) or milligray (mGy).
Equivalent dose: A dose that takes into account the type of radiation and the sensitivity of the tissue or organ exposed. It is measured in sievert (Sv) or millisievert (mSv).
Effective dose: A measure of the overall radiation dose to the whole body, taking into account the dose to different tissues and organs and their sensitivities to radiation. It is also measured in sievert (Sv) or millisievert (mSv).

Table 21.3: Radiation dose limits
Radiation dose limit	Description
1\,\mathrm{mSv} per year	Annual dose limit for members of the general public
20\,\mathrm{mSv} per year	Annual dose limit for radiation workers
1\,\mathrm{mSv} per year	Maximum dose limit for frequent exposure of the public, such as patients undergoing medical procedures
50\,\mathrm{mSv} in a single year	Maximum dose limit for radiation workers in emergency situations
100\,\mathrm{mSv} in a single year	Maximum dose limit for radiation workers in life-saving emergency situations

21.8Applications of radiation

Radiation has a wide range of applications in many fields.

Smoke alarm (21.8.1)

The radioactive material used in smoke detectors is typically americium-241, a man-made element that emits alpha particles. These particles collide with air molecules in the detector and make ions, which allow an electric current to flow. When smoke particles enter the detector, they disturb this flow of ions, which sets off the alarm and warns the people in the building of a possible fire (Figure 21.17). This technology has saved countless lives by giving early warning of fires in homes and buildings.

Figure 21.17: Smoke alarm. (a) Alpha particles from americium-241 ionize the air between two plates, so a steady current flows. (b) Smoke particles disturb the ions, the current falls and the alarm is triggered.

Food irradiation (21.8.2)

Food irradiation is a method that uses radiation to remove bacteria, mould and insects, which helps to prolong the life of food. It is often used to sterilize spices, meat and other kinds of food. Even when the food is packaged, gamma rays can still pass through the packaging and kill the harmful micro-organisms (Figure 21.18). Although this technique greatly extends the shelf life of food, it may sometimes affect the taste.

Figure 21.18: Food irradiation. Gamma rays pass through the packaging and kill micro-organisms in the food.

Equipment sterilization (21.8.3)

Radiation can destroy bacteria and other tiny organisms, which makes it a valuable tool for sterilizing medical equipment, food and many other items. The use of radiation in health care is especially helpful because it keeps equipment sterile and so lowers the risk of infection. This is especially important for plastic equipment such as syringes, which could be damaged by high temperatures.

Controlling thickness (21.8.4)

The choice of isotope for measuring and controlling the thickness of a material depends on the radiation it emits and on its half-life. The main points in this choice are the ability of the radiation to penetrate the material and the stability of the isotope over time. Beta emitters with moderate penetration are used for thin materials, while gamma emitters with high penetration are used for thicker materials. The half-life makes sure that the source stays active for as long as the application needs.

Figure 21.20: Controlling thickness. A beta source and detector check a thin sheet; a gamma source and detector check a thick sheet. The detector reading is used to adjust the rollers.

Diagnosis and treatment of cancer (21.8.5)

Radiopharmaceuticals play a crucial role in diagnosing and treating various diseases. These drugs contain radioactive material and can be given by injection, inhalation or by mouth. They are used both for medical imaging of internal organs and body processes and for treatment. Ionizing radiation serves two distinct purposes in medicine: diagnosis and therapy.

Medical diagnostics

Each organ in our bodies works differently from a chemical point of view. Scientists and medical professionals have found specific chemicals that are absorbed by certain organs. For instance, the thyroid absorbs iodine, while the brain uses glucose, among others. Using this knowledge, radiopharmacists can attach different radioisotopes to biologically active substances. These radioisotopes make it possible to detect disease and its stage accurately in vital organs through imaging techniques such as the gamma camera or PET scans in nuclear diagnostics. When choosing a radioisotope for diagnosis, it is essential that it emits gamma rays with enough energy to leave the body and has a half-life short enough for it to decay soon after the imaging is finished.

Radiation therapy

Radiation treatment uses high-energy radiation to target specific tissues, such as cancerous tumours. One common source of radiation is cobalt-60, which emits beta particles and high-energy gamma rays that are effective in treating various types of cancer. Similarly, high-energy gamma rays produced by linear accelerators (Linacs) can be used for the same purpose. Radiation therapy with gamma rays targets cancerous tissue with high-energy radiation to destroy or damage cancer cells and so prevent them from growing and dividing. The two common sources of gamma rays in radiation therapy are cobalt-60 and linear accelerators.

Cobalt-60 is an isotope that emits gamma rays and beta particles. It is held in a machine known as a cobalt unit (Figure 21.21). The gamma rays are aimed at cancerous tissue and penetrate deep into the body because of their high penetrating power. These rays ionize molecules in cells and create free radicals that damage the DNA of cancer cells. This damage stops the cells from replicating and they eventually die. The cobalt unit rotates around the patient to deliver radiation from various angles, focusing on the tumour while keeping damage to healthy tissue as small as possible.

Figure 21.21: The cobalt unit. The gamma source moves round the patient, so beams arrive from several directions and meet at the tumour; healthy tissue receives only a part of the dose.

Another approach in radiation therapy uses radioisotopes that are absorbed by specific organs. This allows the radiation to be concentrated on the affected tissue. For example, iodine-131 is a radioisotope that is specifically absorbed by the thyroid gland, which makes it an effective treatment for thyroid cancer. By focusing the radiation on the cancerous tissue, these treatments cause the least damage to the surrounding healthy tissue and make cancer treatment more effective.

Choosing a radiation or isotope for the job: the type of radiation and the half-life decide
Application	Radiation or isotope	Why it suits the job
Smoke alarm	Americium-241 (alpha)	Alpha particles ionize air; smoke disturbs the ion current
Food irradiation, sterilizing equipment	Gamma rays	Penetrate packaging and kill bacteria and micro-organisms
Thickness of thin sheets	Beta emitter	Moderate penetration, so the reading changes with a thin sheet
Thickness of thick sheets	Gamma emitter	High penetration, so it passes through thicker material
Medical diagnosis (imaging)	Gamma emitter with a short half-life, e.g. technetium-99m (6 h)	Gamma rays leave the body; the isotope decays soon after imaging
Radiation therapy	Cobalt-60 (gamma), linear accelerators; iodine-131 for the thyroid	Gamma rays penetrate deeply and damage cancer cells; iodine collects in the thyroid

21.9Background radiation

Background radiation: The ubiquitous ionizing radiation that everyone is exposed to, coming from natural and artificial sources. It is a measure of the level of ionizing radiation present in the environment at a particular place which is not due to the deliberate introduction of radiation sources.

Natural background radiation

Since the beginning of time, all living things on Earth have been bathed in a sea of radiation from natural sources. This radiation, called background radiation, comes from three main sources.

Cosmic rays. These are high-energy particles that bombard the Earth from outer space. The amount of cosmic radiation we receive depends on where we live: higher altitudes and areas with weaker magnetic fields receive slightly higher doses.
Terrestrial radiation. Radioactive materials such as uranium and thorium exist naturally in soil, rocks and even water. These materials decay over time and release radiation. The amount of terrestrial radiation also changes with the local geology.
Internal radiation. Surprisingly, we even carry small amounts of radioactive materials inside our bodies. Elements such as potassium-40 and carbon-14 occur naturally and emit low levels of radiation.

Terrestrial radiation has three main parts:

Radon. Radon is a radioactive gas produced by the decay of uranium and thorium in the Earth's crust. It can enter homes through cracks in the foundation, especially in basements. Radon is a major source of terrestrial radiation and a primary cause of lung cancer in non-smokers, because its decay products are breathed in.
Rocks and buildings. Various rocks and minerals contain radioactive elements such as uranium, thorium and potassium-40. The radiation from these substances adds to the terrestrial dose people receive, especially in regions with high concentrations of these elements.
Food and drink. Food and drink naturally contain trace amounts of radioactive elements, namely potassium-40 and carbon-14. These elements are found in all living organisms, including plants and animals, and so are present in our food. Eating them leads to internal exposure to radiation, which is a part of terrestrial radiation.

Figure 21.22: Percentage contribution of the various sources to the total radiation dose per person (the book draws this as a pie chart). Radon gas is the largest single source; natural sources together are 48+14+12+10=84\% and artificial sources 16\%.

Artificial background radiation

Artificial sources of background radiation come from medical procedures, industrial activities and historical nuclear events.

Medical radiation. This comes from medical procedures and treatments: diagnostic imaging (such as X-rays and CT scans) and therapy (such as radiation therapy). It plays a major part in the radiation exposure of people who undergo these procedures.
Industrial radiation. This arises from industrial activities and products, including the radiation emitted by nuclear power plants, radioactive materials used in industry, and consumer products such as smoke detectors and some types of luminous watches. Its contribution to the overall background radiation is generally low, but it can be noticeable near particular industrial sites.
Fallout radiation. This comes from nuclear weapons testing and nuclear accidents, which release leftover radioactive particles into the environment during nuclear explosions or accidents such as Chernobyl and Fukushima. The extent of fallout radiation varies greatly with location and history. In general it is a minor part of the present background radiation.

Sources of background radiation (Figure 21.22)
Source	Type	Share of the total dose
Radon gas from the ground	Natural	48\%
Ground and buildings	Natural	14\%
Food and drink	Natural	12\%
Cosmic rays	Natural	10\%
Medical	Artificial	14\%
Industrial and others	Artificial	1.5\%
Nuclear power and weapon tests	Artificial	0.5\%

21.10Nuclear fissionOften examined

Nuclear fission was discovered on 17 December 1938 by Otto Hahn and his assistant Fritz Strassmann, and explained theoretically in January 1939 by Lise Meitner and her nephew Otto Robert Frisch. They found that a uranium nucleus, after absorbing a low-energy neutron (a thermal or slow neutron), splits into two fragments of intermediate size.

Nuclear fission: The process of splitting a heavy and unstable nucleus into comparatively smaller nuclei by absorbing a neutron, with the emission of more neutrons and a large amount of energy.

It can be represented by the nuclear reaction

{}^{1}_{0}\mathrm{n}+{}^{A}_{Z}\Delta\to{}^{A+1}_{Z}\Delta^{*}\to\mathrm{X}+\mathrm{Y}+\text{neutrons}

21.7

Here \Delta may be uranium-235 (U-235), plutonium-239 (Pu-239) or uranium-233 (U-233), and \Delta^{*} is an intermediate excited state that lasts only a very short time before splitting into the fission fragments \mathrm{X} and \mathrm{Y}. Different fission reactions give different possible fragments. On average about 2.5 neutrons are released per fission, along with a large amount of energy.

A typical fission reaction of this type is the following (Figure 21.23):

{}^{1}_{0}\mathrm{n}+{}^{235}_{92}\mathrm{U}\to{}^{236}_{92}\mathrm{U}^{*}\to{}^{141}_{56}\mathrm{Ba}+{}^{92}_{36}\mathrm{Kr}+3\,{}^{1}_{0}\mathrm{n}+Q

The nucleon numbers balance: 1+235=236=141+92+3, and so do the charges: 92=56+36 (neutrons carry no charge). Q is the energy released.

Figure 21.23: Nuclear fission. U-235 absorbs a slow neutron to become the unstable U-236, which splits into barium-141, krypton-92 and three neutrons, releasing energy Q.

In this reaction the total mass of the fission fragments is less than the original mass of the parent nucleus. This reduced mass (the mass defect) is released as thermal energy, according to Einstein's famous mass-energy relation:

E=mc^{2}

where E is the energy released, m is the mass converted into energy and c=3\times10^{8}\,\mathrm{m\,s^{-1}} is the speed of light. Matter can be changed into energy and energy into matter; in this way the law of conservation of energy (mass-energy) still holds.

Measurements show that about 200\,\mathrm{MeV} of energy is released in a single fission event. This is a large amount compared with the energy released in chemical processes. For example, 1\,\mathrm{kg} of uranium gives as much energy as about 3000 tons of coal.

Chain reaction

The neutrons given out in a fission reaction may cause further fission reactions in other nuclei in the sample. If each of the neutrons emitted in the fission of uranium-235 is absorbed by another uranium-235 nucleus and so causes another fission, even more neutrons are released, followed by more fissions, and so on. This process is called a fission chain reaction (Figure 21.24), and different fission fragments may be produced.

Figure 21.24: A fission chain reaction. Each fission releases neutrons that cause more fissions, so the number of fissions grows with every generation (here 1, 3, 9).

As long as the average number of neutrons available to produce new fissions is greater than 1 per reaction, the number of fissions grows with time. If this happens in an uncontrolled way, a huge amount of energy can be produced in a very short time. This chain-reaction principle is used in the atomic bomb, where a huge amount of energy is released in an uncontrolled fission chain reaction. In contrast, nuclear reactors release the energy of nuclear fission in a controlled manner, which is used for useful purposes.

Example 21.5: mass converted into energy

During a reaction, 5\times10^{18}\,\mathrm{J} of energy is released. Calculate the mass of the matter converted into energy.

Given

E = 5\times10^{18}\,\mathrm{J}
c = 3\times10^{8}\,\mathrm{m\,s^{-1}}

To find

mass m

By Einstein's mass-energy relation E=mc^{2}, so m=E/c^{2}.
m=\dfrac{E}{c^{2}}
Put in the values.
m=\dfrac{5\times10^{18}}{(3\times10^{8})^{2}}=\dfrac{5\times10^{18}}{9\times10^{16}}=55.56\,\mathrm{kg}
So a mass of 55.56\,\mathrm{kg} is converted into energy during the reaction.

m=55.56\,\mathrm{kg}

Worked example: balancing a fission equation and finding the energy

(a) In the fission {}^{1}_{0}\mathrm{n}+{}^{235}_{92}\mathrm{U}\to{}^{144}_{56}\mathrm{Ba}+{}^{A}_{36}\mathrm{Kr}+3\,{}^{1}_{0}\mathrm{n} find the mass number A of the krypton nucleus. (b) In a reactor, 0.50\,\mathrm{g} of mass is converted into energy. How much energy is released?

(a) The total number of nucleons is conserved. On the left: 1+235=236. On the right: 144+A+3\times1=147+A.
147+A=236\quad\Rightarrow\quad A=89
Check the charge: left 0+92=92; right 56+36+0=92. Both sides agree, so the nucleus is krypton-89, {}^{89}_{36}\mathrm{Kr}.
(b) Change the mass to kilograms first.
m=0.50\,\mathrm{g}=0.50\times10^{-3}\,\mathrm{kg}=5.0\times10^{-4}\,\mathrm{kg}
Use E=mc^{2}.
E=(5.0\times10^{-4})(3\times10^{8})^{2}=5.0\times10^{-4}\times9\times10^{16}=4.5\times10^{13}\,\mathrm{J}
Even half a gram of mass gives an enormous amount of energy because c^{2} is so large.

A=89\ (\text{krypton-89}),\qquad E=4.5\times10^{13}\,\mathrm{J}

21.11Nuclear fusionOften examined

Nuclear fusion: The process in which two light nuclei combine to form a comparatively heavy nucleus.

When two nuclei join to form a larger nucleus, the mass of the larger nucleus is less than the total mass of the nuclei that formed it. This loss of mass (the mass defect) appears in the form of energy.

A self-sustaining fusion reaction is also possible, but the energy needed to start a fusion reaction is available only in the surroundings of stars, including the Sun. One such cycle is the proton-proton cycle: the direct collision of protons produces a heavier nucleus, and its collision in turn produces a helium nucleus, as shown in Figure 21.25.

Figure 21.25: The proton-proton chain. Step 1 and step 2 each happen twice, giving two helium-3 nuclei; in step 3 they fuse to one helium-4 nucleus and two protons are released.

Step 1. A proton reacts with another proton to form a deuterium nucleus, with the emission of a beta (positron) particle and some energy:

{}^{1}_{1}\mathrm{H}+{}^{1}_{1}\mathrm{H}\to{}^{2}_{1}\mathrm{H}+{}^{0}_{+1}\beta

Step 2. A proton reacts with the deuterium nucleus and forms a helium-3 nucleus, with the emission of some energy:

{}^{1}_{1}\mathrm{H}+{}^{2}_{1}\mathrm{H}\to{}^{3}_{2}\mathrm{He}+\gamma

Step 3. Two helium-3 nuclei fuse together to form an alpha particle (helium-4), with the release of two protons and some energy:

{}^{3}_{2}\mathrm{He}+{}^{3}_{2}\mathrm{He}\to{}^{4}_{2}\mathrm{He}+{}^{1}_{1}\mathrm{H}+{}^{1}_{1}\mathrm{H}+\gamma

21.8

The "energy" on the right-hand side of the equations simply means that the particles on the right have more kinetic and radiant energy than the particles on the left. In the reaction, mass has been converted into energy in agreement with Einstein's equation E=mc^{2}, and the total energy released per fusion is about 26.3\,\mathrm{MeV}. The energy that comes from the Sun and the stars is believed to be the result of the fusion of hydrogen nuclei into helium nuclei, with release of energy.

Fission and fusion compared
	Nuclear fission	Nuclear fusion
What happens	A heavy nucleus splits into smaller nuclei	Two light nuclei combine into a heavier nucleus
How it starts	A heavy nucleus absorbs a slow neutron	Needs the conditions found in stars such as the Sun
Example	{}^{1}_{0}\mathrm{n}+{}^{235}_{92}\mathrm{U}\to{}^{141}_{56}\mathrm{Ba}+{}^{92}_{36}\mathrm{Kr}+3\,{}^{1}_{0}\mathrm{n}+Q	{}^{1}_{1}\mathrm{H}+{}^{2}_{1}\mathrm{H}\to{}^{3}_{2}\mathrm{He}+\gamma
Energy	About 200\,\mathrm{MeV} per fission	About 26.3\,\mathrm{MeV} per fusion in the proton-proton chain
Where the energy comes from	The mass of the products is less than the mass of the reactants: mass defect \to E=mc^{2}	The same: mass defect \to E=mc^{2}
Used in	Nuclear reactors, atomic bomb	Energy source of the Sun and stars

21.11The Sun

The Sun is a medium-sized star. Like other stars, it is a self-luminous sphere of hot gas held together by gravity and powered by nuclear reactions in its core.

The bright, visible "surface" of the Sun is called the photosphere. The temperature of the photosphere is about 6000\,\mathrm{K}. Analysis of the Sun's spectrum shows that its photosphere is made up, by mass, of about 75\% hydrogen, 25\% helium and less than 1\% heavier elements, the most abundant of which are oxygen, carbon, neon and iron.

A distinct feature of the Sun's surface is the periodic appearance of sunspots: huge regions of cooler material on the surface of the Sun where the magnetic field is very intense. Other distinct features of the Sun's surface are solar flares and prominences. Flares are sudden, bright, explosive events that start on the Sun's surface; they are releases of energy connected with the Sun's magnetic field. Prominences are enormous filaments of excited gas arching over the surface and usually reaching hundreds of thousands of kilometres outwards. For us, one of the most important features of the Sun is that it radiates energy. It radiates most of its energy in the infrared, visible and ultraviolet regions of the electromagnetic spectrum.

Not until 1938 did scientists understand that the Sun radiates energy because of nuclear fusion. The Sun's core is the site of several nuclear chain reactions that power its energy output. The proton-proton chain is the dominant fusion reaction in the Sun and converts hydrogen into helium. Fusion in the Sun is started by the high temperatures and pressures in the core, which make hydrogen nuclei collide with enough energy to overcome their electrostatic repulsion and combine into a single helium nucleus. The reaction is highly efficient, with only a small fraction of the hydrogen being converted into helium at any given time.

The Sun in numbers
Quantity	Value
Photosphere temperature	About 6000\,\mathrm{K}
Composition of photosphere (by mass)	75\% hydrogen, 25\% helium, less than 1\% heavier elements
Hydrogen in the Sun	About 10^{30}\,\mathrm{kg}
Time the Sun has been fusing hydrogen	About 5 billion years (and about 5 billion more to come)
Hydrogen converted each second	About 6.0\times10^{11}\,\mathrm{kg}
Helium formed each second	5.96\times10^{11}\,\mathrm{kg}
Energy released each second	4.0\times10^{26}\,\mathrm{J}
Solar constant (energy reaching the Earth)	About 1400\,\mathrm{J\,s^{-1}\,m^{-2}}

The Sun contains about 10^{30}\,\mathrm{kg} of hydrogen and has been fusing it into helium for about 5 billion years. Every second in the Sun's interior about 6.0\times10^{11}\,\mathrm{kg} of hydrogen is converted into 5.96\times10^{11}\,\mathrm{kg} of helium and 4.0\times10^{26}\,\mathrm{J} of energy. Even at this rate, the Sun is expected to radiate energy from hydrogen fusion for another 5 billion years.

The amount of energy that reaches the Earth from the Sun is about 1400 joules per second per square metre (\mathrm{J\,s^{-1}\,m^{-2}}). This amount of energy is called the solar constant. Nearly all life on the planet Earth depends on energy from the Sun, because this energy controls our whole environment: the atmosphere, food supplies, water sources and so on.

21.12Dark matter

The universe is mostly made of dark matter and dark energy. Dark matter is an elusive, invisible substance that is five times more abundant than regular matter and makes up about 27\% of the universe. Its exact composition is unknown, but we know that it interacts through gravity and not through electromagnetic forces.

Unlike regular matter, dark matter does not emit, absorb or reflect light, which makes it extremely difficult to detect. This mysterious substance is called "dark matter" because it cannot be seen. It takes up space and has mass, just like regular matter, but its presence is inferred rather than directly observed.

One piece of evidence for the existence of dark matter comes from studying the rotation of galaxies. When we measure the orbital speed of matter at various distances from the centre of a galaxy, we find that the speed does not decrease with distance as expected. Instead, the rotation speeds stay roughly constant. Figure 21.27 shows the distance in kiloparsecs (kpc), equal to 3261.56 light years, and the rotation speed in kilometres per second, and there is a clear difference between the observed and the expected graph.

Galaxy rotation: observed and expected

Figure 21.27 (schematic): Rotation speed v against distance R from the centre of a galaxy. The speed expected from the visible matter falls away at large R, but the observed speed stays high.

Galaxies rotate so fast that the gravitational force from their visible matter alone should be too weak to hold them together; they should have been torn apart long ago. This difference suggests that a large amount of unseen matter lies at the edges of galaxies and beyond their visible parts.

The same effect is seen in clusters of galaxies, which suggests that it reaches beyond single galaxies. Scientists propose that an undetected form of matter (dark matter) gives these galaxies the extra mass that provides the gravitational pull needed to keep their structure. According to current scientific understanding, dark matter forms a vast web-like structure through the universe and acts as a gravitational framework that attracts most of the normal matter of the cosmos.

21.13Falsifiability

What decides whether an idea is truly scientific? Falsifiability is a key idea in the philosophy of science. A scientific theory is considered valid only if it makes predictions or statements that can be tested and could be proved wrong.

Falsifiability: The idea that, for a theory to be scientific, there must be some possible observation or experiment that could show the theory to be false. A hypothesis can be accepted as a scientific one only if it can, in principle, be disproved.

If no such test can be thought of, the theory is not scientific because it cannot be tested by experiment. This principle helps to tell scientific theories from pseudoscience, encourages rigorous testing, supports objective evaluation and improves predictive power. The idea is crucial in every field of physics, including nuclear physics and the study of dark matter.

Examples from this unit

Nuclear fusion in the Sun. Observations of the Sun's energy output and analysis of its spectrum can test the theory. If the Sun's energy output did not match the predictions of nuclear fusion models, the theory would be questioned.
Dark matter. By measuring the rotation curves of galaxies, scientists can test the existence of dark matter. If observations showed that galaxies could be explained by visible matter alone, without any need for dark matter, the theory would be falsified.

Falsifiability is crucial in science because it allows scientific theories to be tested and possibly disproved. This process is essential for scientific progress: scientists propose theories, carry out experiments to test them, and then refine or reject them according to the results. If a theory cannot be tested or disproved, it is not considered scientific, because it lacks empirical support.

Falsifiability makes sure that theories can be tested and are open to refutation. However, it is not the only test of a scientific theory. Scientists also consider other factors, such as explanatory power, how well a theory fits with existing knowledge, its predictive accuracy and its simplicity, often called Occam's razor. Together these criteria help to make sure that scientific theories are strong and reliable.

Formula sheet

Neutron number

N=A-Z

N: number of neutrons; A: mass (nucleon) number; Z: atomic (proton) number

a number (no unit)

Alpha decay (21.1)

{}^{A}_{Z}\mathrm{X}\to{}^{A-4}_{Z-2}\mathrm{Y}+{}^{4}_{2}\mathrm{He}\,(\alpha)

parent X, daughter Y; A falls by 4, Z falls by 2

Beta decay (21.2)

{}^{A}_{Z}\mathrm{X}\to{}^{A}_{Z+1}\mathrm{Y}+{}^{0}_{-1}\beta

a neutron changes into a proton; A unchanged, Z rises by 1

Beta plus decay

{}^{A}_{Z}\mathrm{X}\to{}^{A}_{Z-1}\mathrm{Y}+{}^{0}_{+1}\beta

a proton changes into a neutron; A unchanged, Z falls by 1

Gamma decay (21.3)

{}^{A}_{Z}\mathrm{X}^{*}\to{}^{A}_{Z}\mathrm{X}+\gamma

\mathrm{X}^{*}: nucleus in an excited state; A and Z unchanged

Amount left after n half-lives (21.4)

N=\left(\dfrac12\right)^{n}N_0=\dfrac{N_0}{2^{n}}

N_0: initial number of nuclei (or mass); N: amount left; n: number of half-lives

same unit as N_0

Number of half-lives (21.5)

n=\dfrac{\Delta t}{T_{1/2}}

\Delta t: total time passed; T_{1/2}: half-life (same unit as \Delta t)

a number (no unit)

Half-life equation (21.6)

N=\dfrac{N_0}{2^{\Delta t/T_{1/2}}}

N_0: initial amount; \Delta t: time passed; T_{1/2}: half-life

same unit as N_0

Activity

A=\dfrac{\Delta N}{\Delta t}

number of decays \Delta N in time \Delta t (size of the rate of decay)

Unit: \mathrm{Bq} (\mathrm{s^{-1}})

The curie

1\,\mathrm{Ci}=3.70\times10^{10}\,\mathrm{Bq}

\mathrm{Ci}: curie; \mathrm{Bq}: becquerel (1 decay per second)

Unit: \mathrm{Bq}

Mass-energy relation

E=mc^{2}

E: energy released; m: mass converted; c=3\times10^{8}\,\mathrm{m\,s^{-1}}: speed of light

Units: E in \mathrm{J}, m in \mathrm{kg}

Fission of uranium-235 (typical)

{}^{1}_{0}\mathrm{n}+{}^{235}_{92}\mathrm{U}\to{}^{236}_{92}\mathrm{U}^{*}\to{}^{141}_{56}\mathrm{Ba}+{}^{92}_{36}\mathrm{Kr}+3\,{}^{1}_{0}\mathrm{n}+Q

Q: energy released (about 200\,\mathrm{MeV} per fission)

Proton-proton chain, step 3 (21.8)

{}^{3}_{2}\mathrm{He}+{}^{3}_{2}\mathrm{He}\to{}^{4}_{2}\mathrm{He}+{}^{1}_{1}\mathrm{H}+{}^{1}_{1}\mathrm{H}+\gamma

steps 1 and 2: {}^{1}_{1}\mathrm{H}+{}^{1}_{1}\mathrm{H}\to{}^{2}_{1}\mathrm{H}+{}^{0}_{+1}\beta and {}^{1}_{1}\mathrm{H}+{}^{2}_{1}\mathrm{H}\to{}^{3}_{2}\mathrm{He}+\gamma; about 26.3\,\mathrm{MeV} per fusion

Key terms

Nucleus: The region at the centre of the atom consisting of protons and neutrons.
Nucleons: The particles found inside the nucleus, that is, protons and neutrons.
Nuclide: A type of atom whose nucleus has specific numbers of protons and neutrons.
Atomic number (Z): The number of protons in the nucleus (also called the charge number or proton number).
Mass number (A): The number of nucleons (protons plus neutrons) in the nucleus.
Neutron number (N): The number of neutrons in the nucleus, N=A-Z.
Isotopes: Atoms of the same element with the same number of protons but different numbers of neutrons.
Radioisotope: An unstable isotope that emits radiation to attain stability.
Relative abundance: The percentage of atoms with a specific atomic mass found in a naturally occurring sample of an element.
Isobars: Nuclides with the same mass number but different atomic numbers.
Isotones: Nuclides with the same number of neutrons.
Radioactivity: The natural emission of radiations from unstable nuclei.
Radioactive substance: A substance or object that emits nuclear radiation.
Nuclear radiation: The rays that originate in the nuclei of atoms.
Radioactive decay: The process by which the nucleus of an unstable atom loses mass and energy by emitting ionizing particles.
Alpha decay: A type of radioactive decay in which a nucleus emits an alpha particle (a helium nucleus).
Beta decay: A type of radioactive decay in which a nucleus emits a beta particle (an electron).
Beta plus decay: Beta decay in which a proton changes into a neutron and a positron is emitted.
Gamma decay: A type of radioactive decay in which an excited nucleus emits gamma radiation.
Nuclear transmutation: The change of one element into another when a nucleus decays.
Penetrating power: How far radiation can travel through matter before it loses its energy.
Half-life: The time needed for half of the original nuclei of a sample of a radioactive substance to decay.
Activity: The number of nuclear decays per second in a radioactive sample; the SI unit is the becquerel (Bq).
Radioactive dating: An application of radioactive decay in which the age of a material is found from the amount of a particular radioactivity left in it.
Carbon-14 dating: A radioactive dating technique based on the radioactivity of carbon-14.
Ionization: The process in which an atom loses or gains electrons, forming ions.
Excitation: The raising of electrons to higher energy levels without ejecting them from the atom.
Ionizing radiation: Radiation with energy above 13.6\,\mathrm{eV} that can eject electrons from atoms.
Absorbed dose: The energy absorbed by a tissue or organ exposed to radiation (gray, Gy).
Equivalent dose: A dose that takes into account the type of radiation and the sensitivity of the tissue (sievert, Sv).
Effective dose: The overall radiation dose to the whole body, allowing for different tissues and their sensitivities (sievert, Sv).
Shielding: A technique to limit radiation exposure by putting a barrier such as lead or concrete in the way.
Background radiation: The level of ionizing radiation present in the environment at a place which is not due to the deliberate introduction of radiation sources.
Nuclear fission: A nuclear reaction in which a heavy nucleus splits into smaller nuclei with the emission of a large amount of energy.
Fission chain reaction: A series of fissions in which the neutrons released by one fission cause further fissions.
Nuclear fusion: A nuclear reaction in which two nuclei are combined to form a larger nucleus with the emission of a large amount of energy.
Mass defect: The loss of mass in a nuclear reaction; the missing mass appears as energy (E=mc^{2}).
Dark matter: Invisible matter believed to make up most of the matter in the universe; its existence is inferred from its gravity.
Falsifiability: The idea that a theory is scientific only if it makes assertions that can, in principle, be disproved.

Multiple choice questionsp. 346–347

Charge on \alpha-particle is how many times of the charge on proton?

Atwo timesAnswer
Bthree times
Cfour times
Done time

Why

An \alpha-particle is a helium nucleus with two protons and no electrons, so its charge is +2e. A proton has charge +e, so the ratio is \dfrac{2e}{e}=2. The \alpha-particle carries two times the charge of a proton.

Option D (one time) would need one proton's charge, which is only the case for a single proton. Options B and C (three and four times) would need three or four protons, which an \alpha-particle does not have.

Uranium with atomic number 92 has an isotope {}^{239}U, the number of neutrons in it are:

A92
B239
C147Answer
D331

Why

The number of neutrons is the nucleon number minus the proton number: N=A-Z=239-92=147. The answer is 147, option C.

Option A (92) is the proton number Z. Option B (239) is the nucleon number A, which includes both protons and neutrons. Option D (331) adds A and Z, which has no meaning here.

Which of the following is deflected at large angles by electric or magnetic field?

A\alpha-particles
B\beta-particlesAnswer
C\gamma-rays
Dneutrons

Why

\beta-particles are fast-moving electrons. They are charged and very light, so they are deflected more than \alpha-particles in electric and magnetic fields (book, Figure 21.8). The answer is **\beta-particles**, option B.

\alpha-particles (A) are charged, but they are heavy, so their deflection is small. \gamma-rays (C) and neutrons (D) carry no charge, so they are not deflected at all.

Beta (\beta) particles are fast moving:

AelectronsAnswer
Bphotons
Chydrogen nuclei
Dhelium nuclei

Why

A \beta-particle is an electron that is emitted from the nucleus at high speed. The answer is electrons, option A.

Photons (B) are the quanta of light and of \gamma-radiation, not beta particles. Hydrogen nuclei (C) are protons and helium nuclei (D) are \alpha-particles.

Cancer treatment involves targeting tumors with radiation. Which type of radiation is most commonly used for its ability to penetrate deep tissues?

A\alpha-particles
B\beta-particles
C\gamma-raysAnswer
Dprotons

Why

\gamma-rays are the most penetrating of the three nuclear radiations. The book describes them as penetrating deep within the body, which is why cobalt-60 and linear accelerators produce them for radiation therapy. The answer is **\gamma-rays**, option C.

\alpha-particles (A) are stopped by a sheet of paper and \beta-particles (B) by a few centimetres of body tissue. Protons (D) are not one of the nuclear emissions of this unit.

Which of the following decay modes is due to a transition between states of the same nucleus?

Aalpha decay
Bbeta decay
Cgamma decayAnswer
Dall of these

Why

When an \alpha or \beta emission leaves the nucleus in an excited state, the extra energy is released as a \gamma-ray. The nucleus stays the same nuclide, with the same Z and A, so this is a transition between energy states of one nucleus. The answer is gamma decay, option C.

Alpha decay (A) and beta decay (B) each change the nucleus into a different nuclide, so neither is a transition within the same nucleus. Option D is wrong because it includes A and B.

During \beta-decay, the nucleon number:

Adecreases by 4
Bdoes not changeAnswer
Cdecreases by 1
Dincreases by 1

Why

In \beta-decay a neutron changes into a proton and an electron is emitted. The number of nucleons (protons plus neutrons) stays the same, so the nucleon number does not change, option B. Only the proton number increases by one.

Decreasing by 4 (A) is what happens in \alpha-decay. Decreasing by 1 (C) is not a change that occurs in any decay of this unit. Increasing by 1 (D) is the change in the proton number, not the nucleon number.

A radioactive element with a half-life of two years, the sample remained after 6 years is

A75\%
B50\%
C25\%
D12.5\%Answer

Why

The number of half-lives is n=\dfrac{\Delta t}{T_{1/2}}=\dfrac{6}{2}=3. The fraction remaining is \left(\dfrac{1}{2}\right)^{3}=\dfrac{1}{8}=12.5\,\%. The answer is 12.5%, option D.

50% (B) is what remains after one half-life (2 years) and 25% (C) after two half-lives (4 years), so both are too large. 75% (A) is not a power of one half, so it cannot be the result of whole half-lives.

The half life of stable isotope is theoretically:

Azero
B1 s
C11 day
DinfiniteAnswer

Why

Half-life is the time in which half of the nuclei of a sample decay. A stable isotope does not decay, so the time for half of its nuclei to decay is infinite, option D.

Zero (A) and 1 s (B) would mean the nuclei decay almost at once, which is not stable. 11 days (C) is the half-life of a radioisotope, not a stable one.

The unit for absorbed radiation dose is:

ABecquerel (Bq)
BCurie (Ci)
CGray (Gy)Answer
DSievert (Sv)

Why

Absorbed dose is the energy absorbed per unit mass of a tissue or organ. It is measured in gray (Gy) or milligray (mGy), so the answer is Gray (Gy), option C.

Becquerel (A) and curie (B) measure activity, which is the number of decays per second. Sievert (D) measures equivalent or effective dose, which also takes account of the type of radiation and the sensitivity of the tissue.

Which of the following is the largest contributor to natural background radiation exposure for most people?

ACosmic radiation
BRadon gasAnswer
CFood
Dnuclear fallout

Why

Figure 21.22 shows that radon gas from the ground gives 48% of the total dose per individual, which is the largest single share. The answer is radon gas, option B.

Cosmic rays (A) give 10%, food and drink (C) give 12% and nuclear fallout (D) gives only 0.5%. Fallout is also an artificial source, not a natural one.

Radiocarbon dating is a technique used to estimate the age of organic materials. What is the approximate maximum age limit for reliable dating using this method?

A10,000 years
B50,000 yearsAnswer
C1 million years
D5 billion years

Why

The book states that carbon-14 dating is effective for biological tissues up to about 50 or 60 thousand years old. The answer is 50,000 years, option B. Beyond that, too little carbon-14 is left to measure reliably.

10,000 years (A) is well inside the range, not its limit. 1 million years (C) and 5 billion years (D) are far beyond the method; uranium-238, with a half-life of 4.53\times10^{9} years, is used for the oldest rocks.

Food irradiation eliminates bacteria to extend shelf life. Which isotope would be ideal for this purpose?

AStrontium-90 (beta emitter, 28.8 years half-life)
BCobalt-60 (gamma emitter, 5.27 years half-life)Answer
CIodine-131 (beta emitter, 8.02 days half-life)
DPlutonium-239 (alpha emitter, 24,100 years half-life)

Why

Food irradiation needs radiation that passes through the packaging and kills micro-organisms (book, Figure 21.18). That is a gamma emitter, and cobalt-60 is one, with a half-life of 5.27 years, long enough to keep a source working. The answer is cobalt-60, option B.

Strontium-90 (A) and iodine-131 (C) are beta emitters, which are stopped by a few millimetres of metal and do not reach through packaged food. Plutonium-239 (D) is an alpha emitter, which cannot pass through the food at all.

A worker is preparing to use a radioactive source for an experiment. Which of the following is the PRIMARY reason to maintain a safe distance from the source during use?

ATo conserve the radioactive material for future experiments.
BTo minimize the risk of contamination from the source.
CTo reduce the intensity of radiation reaching the worker.Answer
DTo prevent accidental activation of the source.

Why

Radiation spreads out from a source, and its intensity weakens rapidly with distance (book, safety rule of distance). Keeping away from the source therefore reduces the intensity of radiation reaching the worker, option C.

Conserving the material (A) and preventing activation of the source (D) are not reasons that distance addresses. Keeping distance does not stop contamination (B), which is controlled by containment and handling.

The energy of the sun is released due to:

Anuclear fission
Bnuclear fusionAnswer
Creduction
Doxidation

Why

The Sun is powered by nuclear fusion, in which hydrogen nuclei combine to form helium nuclei and release energy (book, Section 21.11). The answer is nuclear fusion, option B.

Nuclear fission (A) splits heavy nuclei and is used in reactors, not by the Sun. Reduction (C) and oxidation (D) are chemical processes, which release far less energy per kilogram than nuclear reactions.

The existence of dark matter is primarily inferred from its:

ALight emitted from distant galaxies
BGravitational influence on visible matterAnswer
CChemical reactions with normal matter
DAbundance in meteorites

Why

Dark matter does not emit, absorb or reflect light, so it cannot be seen. Its existence is inferred from its gravitational effect on visible matter, such as the rotation speeds of galaxies that stay high at large distances. The answer is gravitational influence on visible matter, option B.

Light from distant galaxies (A) is the light dark matter does not give out. Chemical reactions (C) and abundance in meteorites (D) are not the evidence used to infer dark matter.

Constructed response questionsp. 347–348

Complete the following nuclear decay equations.

Solution

(a) Lead-210 converts into bismuth-210.

{}^{210}_{82}\mathrm{Pb}\rightarrow{}^{210}_{83}\mathrm{Bi}+\square

Proton number: the parent has Z=82 and the daughter has Z=83, so the particle has Z=82-83=-1.
82=83+Z \Rightarrow Z=-1
Nucleon number: the parent and daughter both have A=210, so the particle has A=0.
210=210+A \Rightarrow A=0
A particle with A=0 and Z=-1 is an electron emitted from the nucleus, which is a beta particle.

{}^{0}_{-1}\beta

(b) Radon-222 emits an alpha particle and transforms into polonium (Po).

{}^{222}_{86}\mathrm{Rn}\rightarrow\square+{}^{4}_{2}\mathrm{He}

Nucleon number: 222=A+4, so the polonium nucleus has A=218.
222=A+4 \Rightarrow A=218
Proton number: 86=Z+2, so Z=84, which is polonium.
86=Z+2 \Rightarrow Z=84
The product is therefore polonium-218.

{}^{218}_{84}\mathrm{Po}

{}^{220}_{86}\mathrm{Rn}\rightarrow{}^{216}_{84}\mathrm{Po}+\square

The printed equation starts from radon-220 and ends in polonium-216. The missing particle has A=220-216=4 and Z=86-84=2.
A=4,\quad Z=2
A particle with A=4 and Z=2 is an alpha particle, a helium nucleus.
Radium-224 (Z=88) decays to radon-220 by the same process: {}^{224}_{88}\mathrm{Ra}\rightarrow{}^{220}_{86}\mathrm{Rn}+{}^{4}_{2}\mathrm{He}, which is the reaction the question describes.

{}^{4}_{2}\mathrm{He}\;(\alpha)

(d) Neptunium-239 releases a beta particle and converts into plutonium (Pu).

{}^{239}_{93}\mathrm{Np}\rightarrow\square+{}^{0}_{-1}\beta

Nucleon number is unchanged in beta decay, so the plutonium nucleus has A=239.
Proton number: 93=Z-1, so Z=94, which is plutonium.
93=Z-1 \Rightarrow Z=94
The product is plutonium-239.

{}^{239}_{94}\mathrm{Pu}

(e) Iodine-131 emits a gamma ray.

{}^{131}_{53}\mathrm{I}\rightarrow\square+\gamma

Gamma emission does not change the proton number or the nucleon number, so the nucleus stays the same nuclide.
The gamma ray carries away the extra energy that the excited nucleus has after an alpha or beta emission, so the nucleus goes from an excited state to its ground state.
{}^{131}_{53}\mathrm{I}^{*}\rightarrow{}^{131}_{53}\mathrm{I}+\gamma

{}^{131}_{53}\mathrm{I}

Note. Three printing points. In (a) the printed sign is an equals sign and it is read here as a decay arrow. In (c) the printed equation starts from radon-220, while the text says radium-224; the blank is an alpha particle either way. In (e) the parent is shown without the excited-state asterisk, which a gamma emission needs.

Which isotope X is needed to complete:

Solution

(a) The fission reaction.

{}^{1}_{0}\mathrm{n}+{}^{235}_{92}\mathrm{U}\rightarrow\mathrm{X}+{}^{134}_{54}\mathrm{Xe}+2\,{}^{1}_{0}\mathrm{n}

A. {}^{102}_{38}\mathrm{Sr}
B. {}^{101}_{37}\mathrm{Rb}
C. {}^{102}_{36}\mathrm{Kr}
D. {}^{100}_{38}\mathrm{Sr}

Nucleon number on each side: 1+235=A_X+134+2\times1, so A_X=236-136=100.
1+235=A_X+134+2 \Rightarrow A_X=100
Proton number on each side: 0+92=Z_X+54+2\times0, so Z_X=38.
0+92=Z_X+54 \Rightarrow Z_X=38
The element with Z=38 is strontium, so X is strontium-100. This is option D.

Answer: D

X={}^{100}_{38}\mathrm{Sr}

(b) The fusion reaction.

{}^{2}_{1}\mathrm{H}+\mathrm{X}\rightarrow{}^{4}_{2}\mathrm{He}+{}^{1}_{0}\mathrm{n}

A. {}^{1}_{1}\mathrm{H}
B. {}^{2}_{1}\mathrm{H}
C. {}^{3}_{1}\mathrm{H}
D. {}^{3}_{2}\mathrm{He}

Nucleon number: 2+A_X=4+1, so A_X=3.
2+A_X=4+1 \Rightarrow A_X=3
Proton number: 1+Z_X=2+0, so Z_X=1.
1+Z_X=2 \Rightarrow Z_X=1
A nucleus with Z=1 and A=3 is tritium, the third isotope of hydrogen. This is option C.

Answer: C

X={}^{3}_{1}\mathrm{H}

Short response questionsp. 348–349

Imagine you're explaining the difference between a nucleon and a nuclide to a younger sibling. How would you describe each term in a way they could understand?

Solution

A nucleon is one of the particles inside the nucleus of an atom. There are two kinds: the proton, which has a positive charge, and the neutron, which has no charge. The total number of nucleons is the nucleon number (mass number) A.

A nuclide is a particular kind of nucleus, identified by how many protons and how many neutrons it has. Think of nucleons as single bricks and a nuclide as a wall built from a set number of bricks of the two kinds. For example, carbon-12 is a nuclide with 6 protons and 6 neutrons, written {}^{12}_{6}\mathrm{C}.

Think about the everyday materials we encounter. How does the presence or absence of radioactivity fundamentally change the properties of a material?

Solution

Radioactivity comes from the nucleus, not from the electrons that take part in chemical reactions. The book notes that radioactivity remains unchanged even after strong chemical reactions, so it does not change the chemical behaviour of a material. A material with radioactive atoms reacts chemically in the same way as one without them.

What radioactivity does change is the radiation the material gives out. A radioactive material emits alpha, beta or gamma radiation, which ionizes the atoms it passes through and can damage living cells. So a radioactive material needs precautions such as time, distance and shielding, while a material without radioactivity gives out no nuclear radiation.

Our bodies are made up of countless atoms. Is there a perfect balance between the number of neutrons and protons in every atom within our bodies? Explain your reasoning.

Solution

No. The number of neutrons and protons is not balanced in every atom. Stable nuclei have a suitable neutron to proton ratio for their size. Light stable nuclei have about equal numbers of neutrons and protons, while heavy stable nuclei have more neutrons than protons. Hydrogen-1 has one proton and no neutron at all, and it is stable.

Many atoms have an unstable ratio. The book notes that more than 3000 isotopes are known, of which only 257 are stable, and the rest are radioisotopes. Carbon-14 has 6 protons and 8 neutrons, and it is unstable. It decays by emitting a beta particle until its nucleus reaches a more stable balance. So the atoms in our bodies include some that are not in balance, such as carbon-14.

Why can a beta particle travel farther through air than an alpha particle, even if they have the same energy level?

Solution

An alpha particle is a helium nucleus with two protons and two neutrons, so it is heavy and has a charge of +2e. It collides strongly with the air molecules along its path and ionizes many of them. Each collision takes away a large part of its energy, so it uses up its energy over a short distance.

A beta particle is an electron with a much smaller mass and a charge of -e. It ionizes fewer atoms per unit path and loses its energy more slowly, so with the same energy it travels much farther. This is consistent with Table 21.1 of the book, where alpha is stopped by a sheet of paper and beta by several millimetres of aluminium.

Alpha particles are known to cause more ionization in solids than beta particles. What explains this difference in their ability to interact with matter?

Solution

Ionization happens when a radiation particle knocks an electron off an atom. An alpha particle is heavy and carries a charge of +2e, so it exerts a strong electric force on the electrons near its path. It knocks off many electrons close to its track, so it causes the most ionization per unit path length, as shown by the highest ionizing ability in Table 21.1.

A beta particle is lighter and carries only -e, so each interaction transfers less energy and it ionizes fewer atoms per unit path. It passes further through the solid before it loses its energy. Gamma rays are neutral and cause the least ionization of the three, but they penetrate the most.

Radium-226 undergoes alpha decay. Explain the process of alpha decay using a balanced nuclear equation and describe the resulting daughter nuclide.

Solution

In alpha decay the parent nucleus emits an alpha particle, which is a helium nucleus. The nucleon number falls by 4 and the proton number falls by 2. For radium-226 (Z=88):

{}^{226}_{88}\mathrm{Ra}\rightarrow{}^{222}_{86}\mathrm{Rn}+{}^{4}_{2}\mathrm{He}

The equation balances: nucleon number 226=222+4 and proton number 88=86+2. The daughter nuclide is radon-222, with 86 protons and 136 neutrons. It is itself radioactive, with a half-life of 3.82 days (Table 21.2).

Why is carbon dating technique primarily limited to dating materials that were once living organisms?

Solution

Carbon dating compares the amount of carbon-14 left in a sample with the amount in living tissue. A living organism keeps taking in carbon-14 from the atmosphere through photosynthesis and through its food, so the ratio of carbon-14 to carbon-12 in living tissue is known. When the organism dies it stops taking in carbon-14, and the carbon-14 it contains decays with a half-life of 5730 years.

The method therefore works only for materials that were once part of a living organism, such as wood, bone or shell. Rocks and minerals never took in carbon-14 in this way, so there is no starting ratio to compare with. The method is also limited to about 50 to 60 thousand years, because after that too little carbon-14 is left to measure.

What makes Cobalt-60 a suitable choice in radiotherapy for treating cancer compared to other isotopes?

Solution

Cobalt-60 emits high-energy gamma rays and some beta particles. Gamma rays have a high penetrating power, so they pass through the skin and body tissue to reach a tumour deep inside the body. There they ionize the molecules of the cancer cells, damaging their DNA so that the cells cannot divide. Alpha particles would be stopped by the skin and beta particles are absorbed within a few centimetres of tissue, so neither reaches a deep tumour.

Cobalt-60 also has a half-life of about 5.27 years, which is long enough for a cobalt unit to be used for years before the source must be replaced. The unit can rotate around the patient, so the beam can reach the tumour from many angles while the healthy tissue receives less. Iodine-131 is useful only for tissues that absorb iodine, such as the thyroid gland.

Some food is irradiated with gamma rays to kill bacteria. While gamma rays are powerful, why is there no concern about people consuming food containing residual gamma radiation?

Solution

The gamma rays come from a source outside the food. They pass through the food and transfer their energy to the micro-organisms, which kills them, but the food does not keep any radiation. Once the source is switched off or removed, the radiation stops, so no gamma radiation remains in the food.

The food also does not become radioactive. The gamma rays are electromagnetic waves, not particles of material that can settle in the food and decay there. The food may change in taste, as the book notes, but the process does not leave radioactive material in it, so consuming it does not expose a person to residual radiation.

Standing too close to a bonfire can burn your skin. How is this similar to the potential dangers of radiation exposure? Explain the fundamental mechanisms behind both phenomena.

Solution

Both can damage the skin and the cells inside it because energy is transferred into the tissue. A bonfire transfers heat, and the heat raises the temperature of the skin cells so much that their proteins are damaged and the cells die. The closer you stand, the more energy reaches you, so the burn is worse.

Radiation exposure works in a different way. Ionizing radiation transfers its energy to the molecules of the body and ionizes them, forming free radicals. In water, the most abundant molecule in cells, this produces hydroxyl radicals that damage DNA, which can lead to cell death or mutations. Like the heat of a bonfire, the dose falls rapidly with distance, which is why distance is a safety rule for both.

We are constantly exposed to low levels of ionizing radiation from various sources. What practical steps could you take to reduce your exposure to this radiation?

Solution

The book gives three rules for reducing exposure: time, distance and shielding. Spend as little time as possible near a radiation source, because the dose grows with the time of exposure. Keep as much distance as possible from the source, because radiation intensity weakens rapidly with distance. Place barriers such as lead or concrete between yourself and the source to block the radiation.

In daily life, a person can also avoid unnecessary medical procedures that use radiation, since medical sources are the largest artificial contribution (Figure 21.22). Radon gas is the largest natural contribution, and it enters buildings through cracks in the foundation, especially in basements, so keeping basements well ventilated lowers the exposure.

How does the presence of dark matter significantly contribute to our understanding of the universe's structure and evolution?

Solution

Dark matter makes up about 27% of the universe, which is about five times more than ordinary matter. It interacts through gravity but not with electromagnetic forces, so it does not emit or absorb light. Its gravity explains why galaxies rotate at high speeds without flying apart, and why galaxy clusters stay together.

The book suggests that dark matter forms a vast, web-like structure across the universe. This structure acts as a gravitational framework that attracts most of the normal matter. Including dark matter therefore helps explain how galaxies and galaxy clusters formed and how the large-scale structure of the universe took its present shape.

Long response questionsp. 349–350

How did the discovery of the atomic nucleus transform our understanding of atomic structure and what implications did it have for modern physics?

Solution

Thomson's model of the atom

In 1898 J.J. Thomson proposed that an atom is a uniform sphere of positive charge with negatively charged electrons scattered through it, like raisins in a pudding. The model explained that atoms are neutral and contain electrons. It predicted that alpha particles would pass through a thin metal foil with very little deflection.

Rutherford's scattering experiment

Rutherford and Geiger directed a beam of alpha particles at a very thin gold foil and detected them on a screen that lights up at each impact. Most alpha particles passed straight through, a few were deflected through large angles, and about 1 in 20,000 was deflected by more than 90^\circ. Rutherford compared the rebounds to a 15-inch shell bouncing back from a sheet of tissue paper. Thomson's model could not produce such strong repulsion, because its positive charge was spread too thinly.

Figure 21.6: Rutherford's scattering experiment. Most alpha particles pass straight through; a very few rebound, so the positive charge must be concentrated in a small nucleus.

Rutherford's conclusions

The atom has a very small nucleus surrounded by mostly empty space, because most alpha particles pass through without deflection.
The nucleus contains most of the mass of the atom, because only a few alpha particles are turned back.
The nucleus is positively charged, so it repels alpha particles only when they come very close.
Light, negatively charged electrons move around the nucleus, held by electrostatic attraction, as planets move around the Sun.

Implications for modern physics

The nuclear model shows that the mass and positive charge of an atom are concentrated in a tiny nucleus. Later, the nucleus was found to be made of protons and neutrons, and the nucleus became the starting point for the study of radioactivity, nuclear fission and nuclear fusion. Its size, about 10,000 times smaller than the atom, is why most of the atom is empty space.

The orbit picture was later replaced by quantum mechanics. Electrons occupy energy levels and their positions are described by probabilities, shown as an electron cloud around a small, dense nucleus (book, Figure 21.4). The notation {}^{A}_{Z}\mathrm{X} for nuclides, which is used throughout modern nuclear physics, also comes from this nuclear picture.

In what ways do stable and unstable isotopes of an element differ in terms of their nuclear properties and applications in science and industry?

Solution

Isotopes of one element

Isotopes of an element have the same number of protons and so the same chemical properties, because the chemistry depends on the electrons. They differ in the number of neutrons, which changes the nucleon number and the nuclear properties. Of the more than 3000 known isotopes, only 257 are stable; the rest are radioisotopes that decay until they reach a stable form.

Nuclear properties

Stable and unstable isotopes compared.
Property	Stable isotope	Unstable isotope (radioisotope)
Balance of neutrons and protons	In balance for its size	Out of balance, so the nucleus changes
Decay	Does not decay; half-life is effectively infinite	Emits alpha, beta or gamma radiation
Half-life	Not applicable	Fixed for each radioisotope, for example 8.02 days for iodine-131
Example	Neon-20, neon-21, neon-22; hydrogen-1 and hydrogen-2	Hydrogen-3 (tritium), half-life 12.3 years; carbon-14, half-life 5730 years

Isotopes can be separated only by methods based on their mass, such as mass spectrometry, because their chemistry is identical. Natural uranium contains about 99% ^{238}U and a little less than 1% ^{235}U, which is separated because ^{235}U is the fuel used in nuclear reactors and atomic bombs.

Applications

Applications of isotopes in science and industry.
Application	Isotope	Why this isotope is used
Radioactive dating of once-living material	Carbon-14	Half-life of 5730 years suits samples up to about 50 to 60 thousand years old
Geological dating	Uranium-238	Half-life of 4.5\times10^{9} years dates the oldest Earth deposits
Fuel for nuclear reactors and bombs	Uranium-235	Undergoes fission when it absorbs a slow neutron
Treatment of thyroid cancer	Iodine-131	Absorbed by the thyroid gland, so radiation is concentrated on the affected tissue
Medical imaging	Technetium-99m	Emits low-energy gamma rays that leave the body and have a short half-life of six hours
Thickness control in industry	Beta and gamma emitters	Beta emitters for thin sheets, gamma emitters with high penetration for thicker materials

Summary

Stable isotopes are used where their mass or their fixed properties are needed, for example in separation and mass measurement. Unstable isotopes are used where their radiation is useful, such as dating, medical diagnosis and therapy, and industrial gauges. Their half-life decides how long the source stays useful and how quickly it becomes safe.

How do the characteristics of alpha, beta, and gamma radiation influence their use in medical treatments, industrial applications, and scientific research?

Solution

Characteristics of the three radiations

Radioactive emissions (book, Table 21.1).
Radiation	Nature	Charge	Penetrating power	Ionizing ability
\alpha	Helium nucleus, two protons and two neutrons	+2	Stopped by a sheet of paper	Highest
\beta	Fast-moving electron	-1	Several millimetres of aluminium	Medium
\gamma	Short-wavelength electromagnetic radiation	0	Several centimetres of lead	Lowest

In electric and magnetic fields the three radiations behave differently. Alpha particles bend a little towards the negative plate, beta particles bend much more towards the positive plate, and gamma rays are not deflected at all. The bending depends on the charge and the mass of the particle, so the lighter beta particle bends more than the heavy alpha particle.

The three radiations in an electric field. The beta path bends most because it is light and negatively charged; the gamma path is straight.

Medical uses

Diagnosis: technetium-99m emits gamma rays with enough energy to leave the body, and a short enough half-life to decay soon after the scan. Gamma cameras and PET scans detect these rays.
Therapy: cobalt-60 emits beta particles and high-energy gamma rays. The gamma rays penetrate deep into the body to destroy cancer cells, and the cobalt unit rotates around the patient.
Positron emission tomography: fluorine-18 emits positrons, and the gamma rays from their collisions with electrons give a 3D image of metabolic activity.

Industrial uses

Smoke detectors: americium-241 emits alpha particles, which ionize the air in the detector. Alpha particles cannot leave the detector, so the detector is safe to use in homes.
Thickness control: beta emitters, which have moderate penetration, control thin materials; gamma emitters with high penetration control thicker materials.
Food irradiation and sterilization: gamma rays pass through packaging and kill bacteria, mould and insects, and they can sterilize plastic syringes that heat would damage.

Research

The different penetrating powers and deflections are used to identify radiation. In the laboratory, a beam is passed through electric and magnetic fields to separate alpha, beta and gamma radiation, which is how Rutherford and his colleagues identified the three types.

What are the potential consequences of alpha decay on the stability of an atomic nucleus, and how does this process compare to other types of radioactive decay?

Solution

Alpha decay

Alpha emission occurs in very large atoms, which have high atomic numbers. The book explains that when the ratio of neutrons to protons in a large nucleus is too low, the atom restores the balance by emitting an alpha particle, which is a helium nucleus.

{}^{A}_{Z}\mathrm{X}\rightarrow{}^{A-4}_{Z-2}\mathrm{Y}+{}^{4}_{2}\alpha

Example: uranium-238 emits an alpha particle and becomes thorium-234, with Z=90 (book, Example 21.1).

{}^{238}_{92}\mathrm{U}\rightarrow{}^{234}_{90}\mathrm{Th}+{}^{4}_{2}\mathrm{He}

Consequences for the nucleus

The nucleon number falls by four and the proton number falls by two, so the nucleus becomes smaller.
Two protons leave the nucleus, which reduces the electrostatic repulsion between the protons in a very large nucleus.
The daughter nucleus is a different element, and it may still be radioactive. A series of decays continues until a stable configuration is reached (book, Section 21.5.4).

Comparison with other decays

Changes produced by each type of decay.
Decay	Change in A	Change in Z	Change in neutrons N	Typical example
Alpha	-4	-2	-2	{}^{238}_{92}\mathrm{U}\rightarrow{}^{234}_{90}\mathrm{Th}+\alpha
Beta (electron)	0	+1	-1	{}^{99}_{42}\mathrm{Mo}\rightarrow{}^{99}_{43}\mathrm{Tc}+\beta
Beta plus (positron)	0	-1	+1	{}^{18}_{9}\mathrm{F}\rightarrow{}^{18}_{8}\mathrm{O}+{}^{0}_{+1}\beta
Gamma	0	0	0	{}^{60}_{27}\mathrm{Co}^{*}\rightarrow{}^{60}_{27}\mathrm{Co}+\gamma

Alpha decay changes both the mass and the charge of the nucleus, so it changes the element. Beta decay changes only the proton number, so the element changes but the nucleon number does not. Gamma decay changes neither number: it only lowers the energy of an excited nucleus, so the nuclide stays the same.

When a nucleus ejects an electron during beta decay, what fundamental changes occur in the atomic structure, and how does this process affect the element's identity?

Solution

What happens in the nucleus

Beta decay happens when the nucleus has an excess of neutrons. An excess neutron changes into a proton, and the electron it produces is ejected from the nucleus as a beta particle. The electron comes from the nucleus, not from the electron shells around it (book, Section 21.5.2).

{}^{A}_{Z}\mathrm{X}\rightarrow{}^{A}_{Z+1}\mathrm{Y}+{}^{0}_{-1}\beta

Changes to the atom

The proton number increases by one, from Z to Z+1.
The nucleon number A is unchanged, because a neutron has become a proton.
The number of neutrons falls by one, so the ratio of neutrons to protons moves towards a more stable value.

Effect on the element

The element is defined by its proton number. Because Z increases by one, the daughter nucleus is a different element. Molybdenum-99 decays into technetium-99, which decays in turn into ruthenium-99 (book, Figure 21.11):

{}^{99}_{42}\mathrm{Mo}\rightarrow{}^{99}_{43}\mathrm{Tc}+{}^{0}_{-1}\beta,\qquad {}^{99}_{43}\mathrm{Tc}\rightarrow{}^{99}_{44}\mathrm{Ru}+{}^{0}_{-1}\beta

Each beta decay therefore moves the atom one place to the right in the periodic table, while the mass number stays at 99. This is the transmutation of one element into another, which the book describes as a typical part of a series of successive decays until a stable form is reached.

How does the concept of half-life facilitate our understanding of radioactive decay, and what practical implications does it have for fields such as archaeology and geology?

Solution

Half-life

Radioactive decay is random for each nucleus, so we cannot say when a particular nucleus will decay. We can, however, describe a sample. The half-life is the time in which half of the nuclei of a sample decay. The decay rate depends on the amount of substance present, so the amount left falls exponentially with time.

Derivation Amount remaining after n half-lives

Aim. Show how the number of nuclei left, N, depends on the number of half-lives n that have passed.

Assumptions.

Each half-life halves the number of nuclei that remain.
The sample contains many nuclei.

Start with N_0 nuclei. After one half-life half of them remain.
N_1=\tfrac12 N_0
After a second half-life half of those remain.
N_2=\tfrac12\times\tfrac12 N_0=\left(\tfrac12\right)^{2}N_0
Repeating the step for n half-lives gives the general result.
N=\left(\tfrac12\right)^{n}N_0
The number of half-lives is the total time divided by the half-life.
n=\dfrac{\Delta t}{T_{1/2}}
Substitute n into the general result.
N=N_0\left(\tfrac12\right)^{\Delta t/T_{1/2}}=\dfrac{N_0}{2^{\Delta t/T_{1/2}}}

ResultN=\dfrac{N_0}{2^{\Delta t/T_{1/2}}}

The amount left depends only on how many half-lives have passed. The half-life of an isotope is fixed by its nucleus and does not depend on temperature, pressure or humidity.

Remaining fraction of a radioactive sample

The fraction of nuclei remaining falls by half at every half-life. Points mark 1, 2 and 3 half-lives.

Archaeology

Carbon-14 has a half-life of 5730 years. Living organisms keep a steady ratio of carbon-14 to carbon-12, so the amount of carbon-14 left in a fossil or a bone gives its age. Carbon-14 dating is effective for biological tissues up to about 50 to 60 thousand years old, and it is most accurate for younger samples, because the amount of carbon-14 left becomes small.

Geology

Materials with longer half-lives date older deposits. Uranium-238, with a half-life of 4.53\times10^{9} years, can date even the oldest Earth deposits (book, Section 21.8). Because the half-life is so long, the amount of uranium-238 changes very little over human timescales, which makes it suitable for such dating.

In what ways do ionizing nuclear radiations impact biological systems, and what are some of the long-term health effects observed in humans exposed to such radiation?

Solution

How ionizing radiation damages cells

Ionizing radiation transfers energy to the molecules it passes through. Because water is the most abundant molecule in a cell, the first result is usually an ionized water molecule. The ionized water reacts with other water molecules to form hydroxyl radicals, and the ejected electrons can cause further ionizations. Highly reactive hydroxyl radicals and other reactive species then damage DNA (book, Figure 21.15).

Figure 21.15: from radiation absorbed to cell damage. The DNA damage (accent) is the step at which the harm can become permanent.

Radiation can break DNA strands, damage bases or sugars and form cross-links between molecules. The most harmful effect is believed to be the double-strand break (DSB).

Cell death and mutation

If the damage is severe, the cell cannot repair itself and dies. There are two main types: mitotic cell death, when a cell with damaged DNA tries to divide and breaks down, and apoptosis, a controlled self-destruction that removes damaged cells. Both protect the body by removing damaged cells.

If the damage is not fatal, it can change the DNA sequence, which is called a mutation. Mutations can be passed to daughter cells. Most are harmless, but some disrupt the normal functioning of the cell, and some allow a cell to grow uncontrollably, which is the basis of cancer.

Long-term health effects in humans

Radiation sickness: the acute effects of high exposure, with nausea, vomiting, fever, diarrhoea and hair loss. It can be fatal depending on the dose.
Cancer: the risk depends on the amount of radiation received, the age of the person and other individual factors.
Eye cataracts and other chronic health problems that develop over time, even from low-level exposure.
Genetic defects, which can be passed on to future generations.

Children, pregnant women and people with weakened immune systems are more vulnerable. Radioactive materials can also contaminate the environment, which poses risks to wildlife and future generations.

What are the most effective safety protocols for handling radioactive materials, and what strategies can be implemented to ensure safe disposal of radioactive waste?

Solution

Time, distance and shielding

The book gives three rules for working near radiation:

Time: minimize the time spent near the source, because the dose grows with the time of exposure.
Distance: maximize the distance from the source, because radiation intensity weakens rapidly with distance.
Shielding: use barriers such as lead or concrete to block the radiation.

Working practice in a radiation laboratory

Wear personal protective equipment: lab coat, shoes and safety glasses.
Keep the workspace clean and organized, and avoid bringing unnecessary materials into it.
Track all radioactive sources in the laboratory so that each one is handled and disposed of properly.
Use a survey meter to detect radiation levels, so that staff are alerted to dangers.

Warning symbols and monitoring

Warning symbols must be placed where radioactive sources, containers or radiation-producing equipment are used. The international trefoil symbol warns of radiation, and the IAEA's supplementary symbol says 'Danger, stay away' to people of any background. Dose limits in Table 21.3 set the most each person may receive.

Radiation dose limits (book, Table 21.3).
Radiation dose limit	Description
1 mSv per year	Annual dose limit for members of the general public
20 mSv per year	Annual dose limit for radiation workers
1 mSv per year	Maximum dose for frequent exposure to the public, such as patients undergoing medical procedures
50 mSv in a single year	Maximum dose for radiation workers in emergency situations
100 mSv in a single year	Maximum dose for radiation workers in life-saving emergency situations

Safe disposal of radioactive waste

Radioactive sources must be tracked from the moment they are received until they are disposed of, so that none is lost or left unshielded. Waste is kept in a shielded and labelled store until it is disposed of as the regulatory body requires. In Pakistan the Pakistan Nuclear Regulatory Authority (PNRA) makes the rules, grants licences for radioactive material, and inspects facilities to ensure compliance; internationally the IAEA gives guidance.

Note. The book does not describe waste-disposal procedures in detail. This answer covers only the points the book makes: tracking, shielding, regulation by national and international bodies, and the dose limits.

How do the applications of radiation in everyday technologies, like smoke alarms and food irradiation, reflect the balance between safety and utility in modern society?

Solution

Applications

Everyday applications of radiation.
Application	Radiation used	How it works	Why the risk is acceptable
Smoke alarm	Alpha particles from americium-241	Alpha particles ionize the air, which allows a small current to flow. Smoke disrupts this flow and triggers the alarm.	Alpha particles are stopped by the detector casing and by a sheet of paper, so the dose is very small.
Food irradiation	Gamma rays	Gamma rays pass through packaging and kill bacteria, mould and insects, which extends the shelf life.	Food does not keep the radiation, but the taste may be affected, so the process is used when the benefit is clear.
Sterilization of equipment	Gamma rays	Radiation kills bacteria and other organisms on medical equipment.	Plastic syringes are kept sterile without the high temperatures that would damage them.
Medical diagnosis and therapy	Gamma and beta emitters, for example technetium-99m and cobalt-60	Diagnosis shows which organs are active; therapy destroys cancer cells.	The benefit to the patient is large, and doses are limited by the dose limits in Table 21.3.

Balance between safety and utility

Each application uses the radiation that suits its job. Alpha particles are ideal for smoke alarms because they cannot leave the detector, while gamma rays are needed for food and equipment because they reach through packaging. The risk is controlled by choosing the least harmful radiation, shielding the source and keeping the dose within the limits set by regulators.

Society accepts these uses because the benefits are large: fewer deaths from fires, longer-lasting food, safer medical equipment and effective treatment of cancer. The regulatory bodies, such as the PNRA and the IAEA, make sure that the risk to workers and the public stays within agreed limits, so that the utility of radiation is kept while its harm is kept low.

What are the primary sources of background radiation in our environment, and how do they contribute to the overall exposure experienced by living organisms?

Solution

Background radiation

Background radiation is the ionizing radiation that everyone is exposed to all the time, from natural and artificial sources. The book divides the natural sources into cosmic rays, terrestrial radiation and internal radiation.

Natural sources

Cosmic rays: high-energy particles from outer space. Higher altitudes and areas with weaker magnetic fields receive slightly higher doses.
Terrestrial radiation: uranium, thorium and potassium-40 in soil, rocks and water. Radon, a radioactive gas formed from uranium and thorium in the Earth's crust, enters homes through cracks in the foundation and is a major source.
Internal radiation: potassium-40 and carbon-14 in food, drink and the body, which emit low levels of radiation.

Artificial sources

Medical radiation from diagnostic imaging, such as X-rays and CT scans, and from therapy.
Industrial radiation from nuclear power plants, industrial radioisotopes, smoke detectors and luminous watches.
Fallout from nuclear weapons testing and accidents, such as Chernobyl and Fukushima, which now makes a minor contribution.

Overall contribution

Percent contribution of various sources of exposure (book, Figure 21.22).
Source	Share of total dose per individual
Radon gas from the ground	48%
Ground and buildings	14%
Food and drink	12%
Cosmic rays	10%
Artificial sources (medical 14%, industrial and others 1.5%, nuclear power and weapon tests 0.5%)	16%

Natural sources make up 84% of the total dose and artificial sources 16%. Radon is the largest single source. Every living organism is exposed to these sources all the time, so the total exposure is the sum of all the contributions. The dose limit for members of the public from artificial sources is 1 mSv per year (book, Table 21.3).

Can you explain the significance of neutron-induced fission in sustaining a nuclear chain reaction, and what factors can influence the rate of this reaction?

Solution

Neutron-induced fission

A uranium-235 nucleus that absorbs a slow neutron becomes unstable and splits into two fragments of intermediate size. On average about 2.5 neutrons are released per fission, together with about 200 MeV of energy (book, Section 21.10).

{}^{1}_{0}\mathrm{n}+{}^{235}_{92}\mathrm{U}\rightarrow{}^{236}_{92}\mathrm{U}^{*}\rightarrow{}^{141}_{56}\mathrm{Ba}+{}^{92}_{36}\mathrm{Kr}+3\,{}^{1}_{0}\mathrm{n}+Q

Why fission can sustain a chain reaction

The neutrons released by one fission can cause further fissions in other uranium-235 nuclei. If the average number of neutrons that go on to cause a new fission is greater than 1, the number of fissions grows with each generation. This is the fission chain reaction. If it is exactly 1, the number of fissions stays the same, and if it is less than 1, the reaction dies out.

Number of fissions per generation

Illustrative curves for an average of k = 2, k = 1 and k = 0.5 neutrons causing new fissions in each generation. Only k greater than 1 gives a growing reaction.

Factors that influence the rate

The average number of neutrons that cause new fissions. Each fission releases about 2.5 neutrons, but only those absorbed by other uranium-235 nuclei continue the chain.
The speed of the neutrons. Slow neutrons are the ones that cause fission in uranium-235 most readily (book, Section 21.10).
The proportion of uranium-235 in the sample. Natural uranium has a little less than 1% of uranium-235, so techniques are used to increase it before it is used as fuel.
The amount of fissile material in the sample, because the released neutrons can cause further fissions in the other nuclei present.
Control. In a reactor the energy is released in a controlled manner, so the rate is kept steady by regulating how many neutrons go on to cause fission. In an atomic bomb the chain reaction is uncontrolled.

What are the essential conditions required for nuclear fusion to occur, and how do these conditions relate to the processes occurring in stars like the sun?

Solution

Conditions for fusion

Nuclear fusion joins two light nuclei to form a heavier nucleus. Both nuclei are positively charged, so they repel each other electrostatically. They must be pushed together very closely, which requires very high temperatures and pressures so that the nuclei have enough kinetic energy to overcome the repulsion. Once fusion begins, it can be self-sustaining, but the energy needed to start it exists only in the environment of stars, including the Sun.

The proton-proton chain

In the Sun, the dominant fusion reaction is the proton-proton chain, which converts hydrogen into helium in three steps (book, Figure 21.25):

Step 1: two protons combine to form deuterium, with the emission of a positron and some energy.
{}^{1}_{1}\mathrm{H}+{}^{1}_{1}\mathrm{H}\rightarrow{}^{2}_{1}\mathrm{H}+{}^{0}_{+1}\beta
Step 2: a proton combines with the deuterium nucleus to form helium-3, with the emission of a gamma ray.
{}^{1}_{1}\mathrm{H}+{}^{2}_{1}\mathrm{H}\rightarrow{}^{3}_{2}\mathrm{He}+\gamma
Step 3: two helium-3 nuclei fuse to form helium-4 and release two protons and energy.
{}^{3}_{2}\mathrm{He}+{}^{3}_{2}\mathrm{He}\rightarrow{}^{4}_{2}\mathrm{He}+2\,{}^{1}_{1}\mathrm{H}+\gamma

The total energy released per fusion is about 26.3 MeV, and the mass of the products is less than the mass of the reactants. The lost mass appears as energy according to E=mc^{2}.

Energy output of the Sun

Derivation Mass converted each second in the Sun

Aim. Use the book's values to check the energy released by the Sun each second.

The hydrogen used each second is 6.0\times10^{11} kg and the helium produced is 5.96\times10^{11} kg. The mass lost each second is the difference.
\Delta m=6.0\times10^{11}-5.96\times10^{11}=4.0\times10^{9}\ \mathrm{kg}
Convert the lost mass into energy with E=mc^{2}, using c=3\times10^{8}\,\mathrm{m\,s^{-1}}.
E=\Delta m\,c^{2}=4.0\times10^{9}\times(3\times10^{8})^{2}
Evaluate.
E=4.0\times10^{9}\times9\times10^{16}=3.6\times10^{26}\ \mathrm{J}

ResultE\approx3.6\times10^{26}\ \mathrm{J\ per\ second}

This is close to the book's value of 4.0\times10^{26} J per second. The small difference comes from the rounded masses quoted in the book, so the check confirms that the energy comes from the mass lost in fusion.

Relation to stars

The Sun is a self-luminous sphere of hot gas, held together by gravity and energized by nuclear reactions in its core. Its core is at very high temperature and pressure, so hydrogen fuses into helium. The Sun has been fusing hydrogen for about 5 billion years and is expected to continue for another 5 billion years. The same conditions exist in other stars, which is why they shine.

How does the sun's emission of various wavelengths within the electromagnetic spectrum contribute to life on Earth, and what implications does this have for our understanding of solar energy?

Solution

Radiation from the Sun

The Sun radiates most of its energy in the infrared, visible and ultraviolet regions of the electromagnetic spectrum (book, Section 21.12). Its visible surface, the photosphere, is at about 6000 K. The energy comes from fusion in the core, which converts about 6.0\times10^{11} kg of hydrogen into helium each second.

Energy arriving at Earth

The energy that reaches Earth from the Sun is about 1400 joules per second on each square metre. This is called the solar constant. Over one hour, one square metre receives about 1400\times3600\approx5.0\times10^{6} J.

Contribution to life

Nearly all life on Earth depends on this energy, because it controls the total environment: the atmosphere, the water sources and the food supplies. The energy that drives the weather and the water cycle, and the energy that plants capture to make food, all come from the Sun.

Implications for solar energy

The energy supply from the Sun is steady and very large, because it comes from fusion that is expected to continue for billions of years.
Because the energy arrives across the whole electromagnetic spectrum, a solar collector can use infrared, visible and ultraviolet radiation.
The solar constant gives a measured starting point for designing systems that use solar energy. Those systems can be checked against the observed value.

Understanding that the Sun's energy comes from nuclear fusion allows us to test the theory with observations of its energy output and its spectrum. If the Sun's output did not match the predictions of fusion models, the theory would be questioned.

What evidence supports the existence of dark matter in the universe, and how does this hypothesis challenge or reinforce our current understanding of cosmology?

Solution

Evidence from galaxy rotation

The orbital speed of stars and gas is measured at different distances from the galactic centre. The expected speed falls with distance, as in the solar system, because most of the mass is near the centre. The observed speed instead stays roughly constant or rises at large distances (book, Figure 21.27).

Rotation speed of a galaxy

Illustrative curves after Figure 21.27 of the book. The observed speed (accent) stays high at large distances, while the speed expected from visible matter (ink) falls.

Other evidence

The same effect is seen in galaxy clusters, which suggests that the extra unseen mass extends beyond individual galaxies. The galaxies would be torn apart if only visible matter held them together. The best explanation is a large amount of matter that does not emit, absorb or reflect light.

Dark matter in cosmology

Dark matter is about 27% of the universe and about five times more abundant than ordinary matter. Current understanding suggests that it forms a vast, web-like structure, which acts as a gravitational framework that attracts most of the normal matter. This reinforces the gravitational picture of how galaxies and clusters form.

Challenges to current understanding

Dark matter is inferred rather than observed directly. If it were not present, the observed rotation of galaxies would have to be explained by visible matter alone, which the observations show is not possible. The hypothesis therefore challenges models of cosmology that count only the matter we can see. It is a testable claim, since observations that galaxies can be explained without dark matter would falsify it (book, Section 21.13).

Why is falsifiability considered a cornerstone of the scientific method, and how does it influence the development and acceptance of scientific theories?

Solution

What falsifiability means

Falsifiability means that for a theory to be scientific there must be a possible observation or experiment that could show the theory to be false. A scientific theory makes predictions or assertions that can be tested and potentially proven wrong. If no such test can be imagined, the theory cannot be checked by evidence and is not scientific (book, Section 21.13).

Examples from this unit

Nuclear fusion in the Sun: observations of the Sun's energy output and its spectrum can test the theory. If the output did not match the predictions of fusion models, the theory would be questioned.
Dark matter: measuring the rotation curves of galaxies can test the hypothesis. If galaxies could be explained by visible matter alone, dark matter would be falsified.

Why it is a cornerstone

Falsifiability makes theories testable and open to refutation. It separates scientific theories from pseudoscience, promotes rigorous testing and encourages objective evaluation. Scientists propose a theory, test it, and then refine or reject it according to the results. This cycle is essential for scientific progress.

Influence on acceptance

A theory is accepted when it has survived tests that could have shown it to be wrong. Falsifiability is not the only criterion, however. Scientists also consider explanatory power, how well a theory fits with existing knowledge, its predictive accuracy and its simplicity, often called Occam's razor. Combining these criteria makes scientific theories robust and reliable.

Numerical response questionsp. 350

Polonium {}^{209}_{84}\mathrm{Po} is a radioactive nucleus. Determine the number of protons, the number of neutrons and the number of nucleons in it.

Solution

Given

A = 209 nucleon number (top number)
Z = 84 proton number (bottom number)

To find

number of protons Z
number of neutrons N
number of nucleons A

The proton number is the bottom number of the symbol.
Z=84
The nucleon number is the top number of the symbol.
A=209
The neutron number is the nucleon number minus the proton number.
N=A-Z=209-84=125

Z=84,\quad N=125,\quad A=209

Book answer. 84, 125, 209

Americium {}^{241}_{95}\mathrm{Am} undergoes alpha decay and produces an alpha particle. Write down the chemical equation for it and name the nucleus formed.

Solution

Given

A = 241 nucleon number of americium
Z = 95 proton number of americium

To find

the daughter nucleus and its name
the equation of the decay

In alpha decay the nucleon number falls by 4 and the proton number falls by 2.
Nucleon number of the daughter: 241-4=237.
A_d=241-4=237
Proton number of the daughter: 95-2=93, which is neptunium (Np).
Z_d=95-2=93
The alpha particle is a helium nucleus, {}^{4}_{2}\mathrm{He}.

{}^{241}_{95}\mathrm{Am}\rightarrow{}^{237}_{93}\mathrm{Np}+{}^{4}_{2}\mathrm{He}\quad\text{(neptunium-237)}

Book answer. {}^{237}_{93}\mathrm{Np}

A radioactive substance has a half-life of 16 months. In how much time, one-eighth of the substance will be left?

Solution

Given

T_{1/2} = 16\ \mathrm{months}
N/N_0 = \dfrac18

To find

\Delta t

Write the amount left after n half-lives and set it equal to one-eighth.
\left(\tfrac12\right)^{n}=\dfrac18=\left(\tfrac12\right)^{3}
So n=3 half-lives have passed.
n=3
Time is the number of half-lives times the half-life.
\Delta t=n\,T_{1/2}=3\times16\ \mathrm{months}
Evaluate.
\Delta t=48\ \mathrm{months}

\Delta t=48\ \mathrm{months}\;(4\ \mathrm{years})

Book answer. 48 months

In a nuclear reaction, 2\,\mu\mathrm{g} of mass is converted into energy. Find the energy.

Solution

Given

m = 2\,\mu\mathrm{g}=2\times10^{-9}\ \mathrm{kg}
c = 3\times10^{8}\ \mathrm{m\,s^{-1}} speed of light

To find

Einstein's mass-energy relation gives the energy released.
E=mc^{2}
Substitute the values.
E=2\times10^{-9}\times(3\times10^{8})^{2}
Evaluate.
E=2\times10^{-9}\times9\times10^{16}=1.8\times10^{8}\ \mathrm{J}

E=1.8\times10^{8}\ \mathrm{J}=180\ \mathrm{MJ}

Book answer. 180 MJ

In 420 days, one-eighth of polonium (Po) remains un-decayed. Calculate the half-life of polonium.

Solution

Given

\Delta t = 420\ \mathrm{days}
N/N_0 = \dfrac18

To find

T_{1/2}

One-eighth remaining means three half-lives have passed.
\left(\tfrac12\right)^{n}=\dfrac18 \Rightarrow n=3
The half-life is the total time divided by the number of half-lives.
T_{1/2}=\dfrac{\Delta t}{n}
Substitute.
T_{1/2}=\dfrac{420}{3}=140\ \mathrm{days}

T_{1/2}=140\ \mathrm{days}

Book answer. 140 days

The half-life of radium is about 1600 years. In how much time will 250 g remain un-decayed when 1 kg of radium was there initially?

Solution

Given

N_0 = 1\ \mathrm{kg}=1000\ \mathrm{g}
N = 250\ \mathrm{g}
T_{1/2} = 1600\ \mathrm{years}

To find

\Delta t

The fraction remaining is N/N_0=250/1000=1/4.
\dfrac{N}{N_0}=\dfrac14=\left(\tfrac12\right)^{2}
So n=2 half-lives have passed.
n=2
Substitute into the time relation.
\Delta t=n\,T_{1/2}=2\times1600=3200\ \mathrm{years}

\Delta t=3200\ \mathrm{years}

Book answer. 3200 years

Three-fourth of the initial mass of a certain radioactive isotope decays after one hour. Find half-life of isotope in minutes?

Solution

Given

\Delta t = 1\ \mathrm{h}=60\ \mathrm{min}
1-N/N_0 = \dfrac34

To find

T_{1/2} in minutes

Three-fourths decayed, so one-fourth remains.
\dfrac{N}{N_0}=\dfrac14=\left(\tfrac12\right)^{2}
So n=2 half-lives have passed in 60 minutes.
n=2
The half-life is the time divided by the number of half-lives.
T_{1/2}=\dfrac{60}{2}=30\ \mathrm{min}

T_{1/2}=30\ \mathrm{min}

Book answer. 30 min

The half-life of {}^{210}_{83}\mathrm{Bi} is 5 days. If we start with 64,000 nuclei of this isotope, how many nuclei will be left after one month?

Solution

Given

T_{1/2} = 5\ \mathrm{days}
N_0 = 64\,000
\Delta t = 1\ \mathrm{month}=30\ \mathrm{days} one month taken as 30 days

To find

Number of half-lives in 30 days.
n=\dfrac{\Delta t}{T_{1/2}}=\dfrac{30}{5}=6
Number of nuclei left after six half-lives.
N=\dfrac{N_0}{2^{n}}=\dfrac{64\,000}{2^{6}}
Evaluate.
N=\dfrac{64\,000}{64}=1000\ \text{nuclei}

N=1000\ \text{nuclei}

Book answer. 1,000 nuclei

Note. One month is taken as 30 days, which gives the book's answer.

An archaeologist digs up a human skull at a dig site. It was found that concentration of carbon-14 with a half-life of 5730 years is 25% of its initial concentration. What is the age of the skull?

Solution

Given

T_{1/2} = 5730\ \mathrm{years}
N/N_0 = 25\%=\dfrac14

To find

\Delta t

One-quarter remaining means two half-lives have passed.
\left(\tfrac12\right)^{n}=\dfrac14 \Rightarrow n=2
Age of the skull.
\Delta t=n\,T_{1/2}=2\times5730=11\,460\ \mathrm{years}

\Delta t=11\,460\ \mathrm{years}

Book answer. 11,460 years

The C-14 content with a half-life of 5730 years of an ancient piece of wood was found to have three tenths of that in living trees (indicating 70% of the C-14 had decayed). How old is that piece of wood?

Solution

Given

T_{1/2} = 5730\ \mathrm{years}
N/N_0 = \dfrac{3}{10}=0.3 70% has decayed, so 30% remains

To find

\Delta t

Set the fraction remaining equal to (1/2)^{n} and solve for n using logarithms.
\left(\tfrac12\right)^{n}=0.3 \Rightarrow n=\dfrac{\ln 0.3}{\ln 0.5}
Evaluate the number of half-lives.
n=\dfrac{-1.20397}{-0.69315}=1.73697
Age of the wood.
\Delta t=n\,T_{1/2}=1.73697\times5730\approx9953\ \mathrm{years}

\Delta t\approx9953\ \mathrm{years}

Book answer. 9953 years

FBISE 2026 questions

Questions from the FBISE SSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

FBISE SSC-II 2026 Annual I, Set B, MCQ 91 mark

Half life of iodine is 8 days. How much of 40 g sample of iodine will be left after 16 days?

A2.5 g
B20 g
C10 gAnswer
D5 g

Why

The number of half-lives is n=\dfrac{\Delta t}{T_{1/2}}=\dfrac{16}{8}=2, so the mass left is N=N_0\left(\tfrac12\right)^{2}=\dfrac{40}{4}=10\,\mathrm{g}. The answer is option C.

Option B (20\,\mathrm{g}) is what is left after only one half-life (8 days), and option D (5\,\mathrm{g}) would need three half-lives (24 days). Do not take 40-16=24\,\mathrm{g} or subtract the days: each half-life halves the amount that is left.

FBISE SSC-II 2026 Annual I, Set B, Q.2 (i)2+1 = 3 marks

What is meant by an isotope? Do the different isotopes of an element have similar chemical properties?

Solution

Isotopes are atoms of the same element that have the same atomic number (the same number of protons) but different mass numbers, because they have different numbers of neutrons.

Example: hydrogen has three isotopes, protium {}^{1}_{1}\mathrm{H} (no neutron), deuterium {}^{2}_{1}\mathrm{H} (one neutron) and tritium {}^{3}_{1}\mathrm{H} (two neutrons). All have one proton.

Chemical properties: yes, the isotopes of an element have similar (identical) chemical properties. Chemical behaviour depends on the electrons, and isotopes of an element have the same number of electrons, since they have the same number of protons. Only their masses differ, so they cannot be separated by chemical methods but only by methods based on mass, such as mass spectrometry.

FBISE SSC-II 2026 Annual I, Set B, Q.2 (ix)2+1 = 3 marks

What is meant by the term half-life of a radioactive substance. Explain briefly with and example.

Solution

The half-life T_{1/2} of a radioactive substance is the time in which half of the unstable nuclei of a sample decay into daughter nuclei.

Example: iodine-131 has a half-life of 8 days. If we start with 40\,\mathrm{g}, then 20\,\mathrm{g} is left after 8 days, 10\,\mathrm{g} after 16 days and 5\,\mathrm{g} after 24 days. The amount left after n half-lives is

N=\left(\dfrac12\right)^{n}N_0,\qquad n=\dfrac{\Delta t}{T_{1/2}}

Time (days)	Half-lives n	Fraction left	Mass of iodine-131 (g)
0	0	1	40
8	1	\tfrac12	20
16	2	\tfrac14	10
24	3	\tfrac18	5

FBISE SSC-II 2026 Annual I, Set B, Q.3 OR(1+1)+3 = 5 marks

a. Briefly describe nuclear fusion. Why it requires very high temperature? b. During a reaction 55.56 kg of mass is converted into energy. Calculate the amount of energy released.

Solution

(a) Nuclear fusion (2 marks)

Nuclear fusion is the process in which two light nuclei combine to form a comparatively heavy nucleus. The mass of the new nucleus is less than the total mass of the nuclei that formed it. This loss of mass (the mass defect) appears as energy, according to E=mc^{2}. An example is the proton-proton chain in the Sun, which converts hydrogen into helium and releases about 26.3\,\mathrm{MeV} per fusion. Its three steps are

Two protons form a deuterium nucleus and a positron.
{}^{1}_{1}\mathrm{H}+{}^{1}_{1}\mathrm{H}\to{}^{2}_{1}\mathrm{H}+{}^{0}_{+1}\beta
A proton joins the deuterium nucleus to form helium-3.
{}^{1}_{1}\mathrm{H}+{}^{2}_{1}\mathrm{H}\to{}^{3}_{2}\mathrm{He}+\gamma
Two helium-3 nuclei fuse to form helium-4 and release two protons.
{}^{3}_{2}\mathrm{He}+{}^{3}_{2}\mathrm{He}\to{}^{4}_{2}\mathrm{He}+{}^{1}_{1}\mathrm{H}+{}^{1}_{1}\mathrm{H}+\gamma

Why very high temperature is needed

Both nuclei are positively charged, so they repel each other strongly. To fuse, they must be brought extremely close together. At very high temperature the nuclei move with very large kinetic energy, so they collide hard enough to overcome their electrostatic repulsion and combine. This is why the energy needed to start fusion is available only in the surroundings of stars, including the Sun.

(b) Energy from 55.56 kg of mass (3 marks)

Given

m = 55.56\ \mathrm{kg}
c = 3\times10^{8}\ \mathrm{m\,s^{-1}} speed of light, from the formula list

To find

The mass converted into energy is given by Einstein's mass-energy relation.
E=mc^{2}
Substitute the values. The mass is already in kilograms.
E=55.56\times(3\times10^{8})^{2}
Evaluate.
E=55.56\times9\times10^{16}=5.0004\times10^{18}\ \mathrm{J}

E\approx5.0\times10^{18}\ \mathrm{J}

FBISE SSC-II 2026 Annual I, Set D, MCQ 91 mark

Which radiation can be stopped by a thin sheet of paper?

AX-rays
BGamma rays
CBeta particles
DAlpha particlesAnswer

Why

An \alpha-particle is a helium nucleus. It is heavy and has a charge of +2e, so it ionizes strongly and loses its energy in a very short distance. A thin sheet of paper (or a few centimetres of air) stops it. The answer is option D.

Beta particles (C) pass through paper and need a few millimetres of aluminium. X-rays (A) and gamma rays (B) are the most penetrating and are only reduced by thick lead or concrete, so a thin sheet of paper does not stop them.

FBISE SSC-II 2026 Annual I, Set D, Q.2 (i)2+1 = 3 marks

Briefly describe \alpha-particles. Give one reaction as example as well.

Solution

**\alpha-particles** are helium nuclei, each made of two protons and two neutrons bound together, {}^{4}_{2}\mathrm{He}. They carry a charge of +2e and are emitted from the nuclei of very large atoms (high atomic number).

They are positively charged, so they are attracted towards the negative plate in an electric field; being heavy, their deflection is small.
They have the strongest ionizing effect but the weakest penetrating power: they are stopped by a thin sheet of paper.

Example of an alpha-decay reaction: radon-222 changes into polonium-218 by releasing an \alpha-particle. The nucleon number falls by 4 and the proton number by 2.

{}^{222}_{86}\mathrm{Rn}\to{}^{218}_{84}\mathrm{Po}+{}^{4}_{2}\mathrm{He}\,(\alpha)

Check: 222=218+4 and 86=84+2, so the equation balances.

FBISE SSC-II 2026 Annual I, Set D, Q.2 (ix)2+1 = 3 marks

What is half life? Why is half-life considered a characteristic property of a radioactive isotope?

Solution

The half-life is the time in which half of the unstable nuclei in a sample of a radioactive substance decay. For example, iodine-131 has a half-life of 8 days: half of any sample is left after 8 days and a quarter after 16 days.

Why it is characteristic of an isotope:

Its value depends only on the nature of the radioactive nucleus, and not on temperature, pressure, humidity or the chemical compound the atoms are in.
Each radioisotope has its own fixed value, from a short time for very unstable nuclei to 4.53\times10^{9} years for uranium-238 (carbon-14 has 5730 years and iodine-131 has 8 days), so the half-life identifies the isotope and tells how quickly it decays.

FBISE SSC-II 2026 Annual I, Set D, Q.3 OR2+3 = 5 marks

a. What is the main energy source of the sun? Give its brief description. b. Explain controlled nuclear fission in nuclear reactors.

Solution

(a) The main energy source of the Sun (2 marks)

The main source of the Sun's energy is nuclear fusion: hydrogen nuclei fuse into helium nuclei in the Sun's core. The Sun is a medium-sized, self-luminous sphere of hot gas, held together by gravity, with a photosphere at about 6000\,\mathrm{K} made of about 75% hydrogen and 25% helium by mass.

The dominant reaction is the proton-proton chain: two protons form deuterium, a proton joins it to form helium-3, and two helium-3 nuclei fuse to give helium-4 and two protons (about 26.3\,\mathrm{MeV} per fusion).
It starts because of the very high temperature and pressure in the core, which make the hydrogen nuclei collide with enough energy to overcome their electrostatic repulsion.
Every second about 6.0\times10^{11}\,\mathrm{kg} of hydrogen is converted into 5.96\times10^{11}\,\mathrm{kg} of helium and 4.0\times10^{26}\,\mathrm{J} of energy. The lost mass appears as energy according to E=mc^{2}.
The Sun has been fusing hydrogen for about 5 billion years and is expected to go on for another 5 billion years.

(b) Controlled nuclear fission in reactors (3 marks)

Each fission of uranium-235 releases about 2.5 neutrons. If every one of them caused a new fission, the number of fissions would grow in each generation and the energy would be released in an instant, as in an atomic bomb (uncontrolled chain reaction). In a nuclear reactor the chain reaction is kept controlled: on average only one neutron from each fission is allowed to cause a new fission, so the number of fissions per second, and the power, stays steady. The energy is then released slowly and can be put to use.

A nuclear reactor in outline. The control rods (accent) absorb neutrons and are moved in or out of the core to keep the chain reaction steady; the moderator slows the neutrons; the coolant carries the heat to the heat exchanger.

Main parts of the reactor

Fuel rods: uranium-235 (natural uranium has a little less than 1% of it, so the proportion is increased before it is used as fuel). The fissions take place here.
Moderator (for example graphite or heavy water): the neutrons from a fission are fast, but slow (thermal) neutrons are the ones that cause fission in uranium-235. The moderator slows them down by collisions.
Control rods (boron or cadmium): they absorb neutrons. Pushing them further in absorbs more neutrons and slows the reaction; pulling them out lets more neutrons cause fission and raises the power. Pushed fully in, they shut the reactor down.
Coolant (water or a gas): it flows through the core, takes away the heat released in fission and carries it to a heat exchanger, where steam is made. The steam turns a turbine and generator to produce electricity.

FBISE SSC-II 2026 Annual I, Set M, Q.2 (i)3 marks

For what facts do the finding's of alpha particle scatter experiment provide evidence? (Any three)

Solution

In Rutherford's experiment a beam of \alpha-particles was fired at a very thin gold foil and the scattered particles were detected on a zinc sulfide screen. The findings gave evidence for the following facts (any three):

Observation	Fact it shows
Most \alpha-particles passed straight through the foil	An atom is mostly empty space, and the nucleus is very small compared with the atom.
A few were deflected through large angles (about 1 in 20 000 by more than 90^\circ), and some almost came straight back	Nearly all the mass of the atom is concentrated in a tiny, dense nucleus at the centre.
Deflection happened only when an \alpha-particle passed very close to the nucleus	The nucleus is positively charged: it repels the positive \alpha-particles.
An \alpha-particle passing an electron is hardly affected	The electrons are very light and they move round the nucleus (the atom is neutral, with negative electrons outside the positive nucleus).

These results disproved Thomson's plum pudding model and led to Rutherford's nuclear model of the atom.

FBISE SSC-II 2026 Annual I, Set M, Q.2 (ix)3 marks

A 50 mg sample of lead-212 decays by beta and gamma emission with a half-life of 10 hours. How much of the pure sample is left after 50 hours?

Solution

Given

N_0 = 50\ \mathrm{mg}
T_{1/2} = 10\ \mathrm{h}
\Delta t = 50\ \mathrm{h}

To find

Number of half-lives in 50 hours.
n=\dfrac{\Delta t}{T_{1/2}}=\dfrac{50}{10}=5
Amount of the lead-212 left after n half-lives.
N=\dfrac{N_0}{2^{n}}=\dfrac{50}{2^{5}}=\dfrac{50}{32}
Evaluate.
N=1.5625\ \mathrm{mg}\approx1.56\ \mathrm{mg}

N\approx1.56\ \mathrm{mg}\ \text{of lead-212 left}

FBISE SSC-II 2026 Annual I, Set M, Q.3 OR(2+1)+2 = 5 marks

What is meant by nuclear fission? Explain controlled nuclear fission in nuclear reactors. Also draw diagram.

Solution

Nuclear fission (2 marks)

Nuclear fission is the process of splitting a heavy and unstable nucleus into comparatively smaller nuclei by absorbing a neutron, with the emission of more neutrons and a large amount of energy. A typical reaction is

{}^{1}_{0}\mathrm{n}+{}^{235}_{92}\mathrm{U}\to{}^{236}_{92}\mathrm{U}^{*}\to{}^{141}_{56}\mathrm{Ba}+{}^{92}_{36}\mathrm{Kr}+3\,{}^{1}_{0}\mathrm{n}+Q

The nucleon numbers balance (1+235=141+92+3) and so do the charges (92=56+36). The mass of the products is less than the mass of the reactants, and the difference appears as energy by E=mc^{2}: about 200\,\mathrm{MeV} per fission, with about 2.5 neutrons released on average.

Chain reaction (1 mark)

The neutrons released can cause further fissions in other uranium-235 nuclei. As long as the average number of neutrons causing new fissions is greater than 1, the number of fissions grows. Uncontrolled, this is the atomic bomb.

Controlled fission in reactors, with diagram (2 marks)

Main parts of the reactor

Fuel rods: uranium-235 (natural uranium has a little less than 1% of it, so the proportion is increased before it is used as fuel). The fissions take place here.
Moderator (for example graphite or heavy water): the neutrons from a fission are fast, but slow (thermal) neutrons are the ones that cause fission in uranium-235. The moderator slows them down by collisions.
Control rods (boron or cadmium): they absorb neutrons. Pushing them further in absorbs more neutrons and slows the reaction; pulling them out lets more neutrons cause fission and raises the power. Pushed fully in, they shut the reactor down.
Coolant (water or a gas): it flows through the core, takes away the heat released in fission and carries it to a heat exchanger, where steam is made. The steam turns a turbine and generator to produce electricity.

Exam focus

Explain controlled nuclear fission in a nuclear reactor and draw a labelled diagram (fuel rods, moderator, control rods, coolant). (long)
2026 Set D Q.3 OR (b) and Set M Q.3 OR (with the definition of fission and a diagram); also long question 11 of the book on the chain reaction.
Go to answer
Half-life numericals: amount left after a given time (remaining fraction, N=N_0/2^{n}). (numerical)
2026 Set B MCQ 9 (40 g of iodine, 16 days), Set M Q.2 (ix) (50 mg of lead-212, 50 hours); numerical questions 3 and 5 to 8 and MCQ 8 of the book use the same method.
Go to answer
Define half-life with an example and say why it is a characteristic property of a radioactive isotope. (short)
2026 Set B Q.2 (ix) and Set D Q.2 (ix); also long question 6 of the book.
Go to answer
Describe nuclear fusion, say why it needs a very high temperature, and name the Sun's energy source (proton-proton chain). (long)
2026 Set B Q.3 OR (a) and Set D Q.3 OR (a); also long questions 12 and 13 of the book.
Go to answer
Mass converted into energy with E=mc^{2} (mass in kilograms). (numerical)
2026 Set B Q.3 OR (b) (55.56 kg); same method as numerical question 4 of the book and the fission example.
Go to answer
What facts does the alpha-particle scattering experiment provide evidence for? (short)
2026 Set M Q.2 (i); also long question 1 of the book.
Go to answer
Define isotopes and say whether they have the same chemical properties; describe alpha particles with an example of alpha decay. (short)
2026 Set B Q.2 (i) (isotopes), Set D Q.2 (i) (alpha particles) and Set D MCQ 9 (paper stops alpha particles).
Go to answer
Complete nuclear decay equations (alpha, beta, gamma) by balancing nucleon and proton numbers. (short)
Not asked in 2026, but this is the core skill of the unit and its first constructed-response question (crq-q1) and numerical questions 1 and 2.
Go to answer

21.1Discovery of the atomic nucleus

Rutherford's alpha-particle scattering experiment

Rutherford's nuclear model of the atom

The atom today

21.2Nuclear representationsOften examined

21.3Isotopes and radioisotopesOften examined

Point to ponder: isobars and isotones

21.4Radioactivity

21.4.1Nature of nuclear emissionsOften examined

Alpha (α) emissions

Beta (β) emissions

Gamma (γ) emissions

21.4.2Order of strength: penetrating power and ionizing effectOften examined

Penetrating power

Ionizing effects

21.5Nuclear decay and its conservation lawsOften examined

21.5.1Alpha decayOften examined

21.5.2Beta decayOften examined

Beta plus decay

21.5.3Gamma decay

21.5.4Stability through nuclear transmutationOften examined

21.6Half-lifeOften examined

Using an equation to model radioactive decay

21.6.2Activity

21.6.3Radioactive datingOften examined

21.7Ionizing nuclear radiations

21.7.1Effects of ionizing nuclear radiations on living things

21.7.2Harmful effects of nuclear radiations

Cell death

Mutations

Cancer

Effects on people

21.7.3Safe handling of radioactive materials

21.7.4Dose descriptors

21.8Applications of radiation

Smoke alarm (21.8.1)

Food irradiation (21.8.2)

Equipment sterilization (21.8.3)

Controlling thickness (21.8.4)

Diagnosis and treatment of cancer (21.8.5)

Medical diagnostics

Radiation therapy

21.9Background radiation

Natural background radiation

Artificial background radiation

21.10Nuclear fissionOften examined

Chain reaction

21.11Nuclear fusionOften examined

21.11The Sun

21.12Dark matter

21.13Falsifiability

Examples from this unit

Formula sheet

Key terms

Multiple choice questionsp. 346–347

Constructed response questionsp. 347–348

Short response questionsp. 348–349

Long response questionsp. 349–350

Thomson's model of the atom

Rutherford's scattering experiment

Rutherford's conclusions

Implications for modern physics

Isotopes of one element

Nuclear properties

Applications

Summary

Characteristics of the three radiations

Medical uses

Industrial uses

Research

Alpha decay

Consequences for the nucleus

Comparison with other decays

What happens in the nucleus

Changes to the atom

Effect on the element

Half-life

Derivation Amount remaining after n half-lives

Archaeology

Geology

Alpha ( $α$ ) emissions

Beta ( $β$ ) emissions

Gamma ( $γ$ ) emissions