Nuclear physics

Class 12 Physics, unit 22, book pages 141–169

The nucleus is held together by the strong nuclear force, and the energy that binds it shows up as a loss of mass: the mass defect, linked to the binding energy by $E = Δ m c^{2}$ . This unit uses that idea to explain radioactive decay (alpha, beta and gamma), the exponential decay law and half-life, and the huge energies released in fission and fusion. It then applies them to nuclear reactors, uranium enrichment, particle-antiparticle annihilation and medical imaging with tracers and PET scanners.

Learning outcomes (22)

Recognise the equivalence between energy and mass as represented by $E = Δ m c^{2}$ , and state and use this equation.
Define and use the terms mass defect and binding energy.
Sketch the variation of binding energy per nucleon with nucleon number.
Recall what is meant by nuclear fusion and nuclear fission.
Explain the relevance of binding energy per nucleon to nuclear reactions, including nuclear fusion and nuclear fission.
Explain how the neutrons produced in fission create a chain reaction and how this is controlled in a nuclear reactor (including the action of coolant, moderators and control rods).
Calculate the energy released in nuclear reactions using $E = Δ m c^{2}$ .
Explain that fluctuations in count rate provide evidence for the random nature of radioactive decay.
Explain that radioactive decay is both spontaneous and random.
Define activity and decay constant, and state and use $A = λ N$ .
Explain half-life with examples.
Use $λ = 0.693/ T_{1/2}$ to solve numerical problems.
State the exponential nature of radioactive decay.
Use the relationship $x = x_{0} e^{- λ t}$ , where $x$ could represent activity, number of undecayed nuclei or received count rate, to solve problems analytically and graphically.
Describe the function of the principal components of a water-moderated power reactor (core, fuel rods, moderator, control rods, heat exchanger, safety rods and shielding).
Explain why uranium fuel needs to be enriched before use.
Compare the amount of energy released in a fission reaction with the (given) energy released in a chemical reaction.
Explain what a medical tracer is: a substance containing radioactive nuclei that can be introduced into the body and is then absorbed by the tissue being studied.
Explain annihilation reactions: they occur when a particle interacts with its antiparticle, and mass-energy and momentum are conserved in the process.
Illustrate how PET scanning works: positrons emitted by the decay of the tracer annihilate with electrons in the tissue, producing a pair of gamma-ray photons travelling in opposite directions.
Calculate the energy of the gamma-ray photons emitted during the annihilation of an electron-positron pair.
Explain that the gamma-ray photons from an annihilation event travel outside the body and can be detected, and that an image of the tracer concentration in the tissue is created by processing the arrival times of the photons.

22The nucleus: nucleons, atomic number and mass number

Nuclear physics is the branch of physics that studies the atomic nucleus: what it is made of, what holds it together and how it changes. You met the nucleus in grade 10. It is the tiny, dense centre of the atom and it contains protons (charge $+ e$ ) and neutrons (no charge). Protons and neutrons together are called nucleons.

Nucleon: A proton or a neutron, the particles found in the nucleus.
Atomic number $Z$: The number of protons in the nucleus. It fixes the element and the charge of the nucleus, $+ Z e$ .
Mass number $A$: The total number of protons and neutrons (nucleons) in the nucleus.
Neutron number $N$: The number of neutrons in the nucleus.
Nuclide: A particular kind of nucleus with a definite $Z$ and $A$ , written $X_{Z}^{A} X_{2 Z}^{2 A} X$ where X is the chemical symbol.
Isotopes: Nuclides of the same element (same $Z$ ) with different numbers of neutrons (different $A$ ), for example uranium-235 and uranium-238.

A = Z + N

mass number = atomic number + neutron number

A nucleus is written with the mass number at the top left and the atomic number at the bottom left of the symbol. For uranium-235:

X_{92}^{235} X_{292}^{2235} U : Z = 92, A = 235, N = A - Z = 235 - 92 = 143

Nuclear physics has many applications: nuclear power stations, nuclear medicine and nuclear weapons. It also explains astrophysics, because nuclear reactions produce the energy of the Sun and the other stars.

22.1Mass defectOften examined

Put 10 balls of mass $100 g$ each into a bag of negligible mass. The full bag has a mass of $1000 g$ : normally the mass of a whole is the sum of the masses of its parts. A nucleus breaks this rule. The mass of a nucleus is always less than the total mass of its separate protons and neutrons.

For example, a carbon-12 atom contains 6 protons, 6 neutrons and 6 electrons. Each proton and each neutron has a mass slightly greater than $1 u$ , so the total should be more than $12 u$ . Yet the carbon-12 atom has a mass of exactly $12 u$ . Some mass is missing. Figure 22.1 shows the same effect for deuterium, the simplest nucleus with more than one nucleon.

Figure 22.1 (redrawn): a proton, a neutron and an electron have more mass when they are separate than when they are bound together as a deuterium atom. The difference is the mass defect.

Mass defect: The difference between the sum of the masses of the constituent particles (protons and neutrons) of a nucleus and the mass of the nucleus itself. Symbol $Δ m$ ; unit $kg$ or $u$ .

For deuterium ( $X_{1}^{2} X_{21}^{22} H$ : one proton and one neutron) the mass defect is:

Δ m = (m_{n} + m_{p}) - m_{nucleus}

A nucleus $X_{Z}^{A} X_{2 Z}^{2 A} X$ contains $Z$ protons and $(A - Z)$ neutrons, so in general:

Δ m = [(A - Z) m_{n} + Z m_{p}] - m_{nucleus} (22.1)

$m_{p} = 1.007276 u$ (proton), $m_{n} = 1.008665 u$ (neutron), $m_{nucleus}$ : mass of the nucleus.

Example 22.1: mass defect of helium-4

Find the mass defect in the formation of a helium-4 nucleus. The mass of He-4 is

4.002603 u

, the mass of a proton is

1.007276 u

and the mass of a neutron is

1.008665 u

Given

$m_{He} = 4.002603 u$
$m_{p} = 1.007276 u$
$m_{n} = 1.008665 u$

To find

mass defect $Δ m$

Helium-4 is $X_{2}^{4} X_{22}^{24} He$ : $Z = 2$ protons and $A - Z = 4 - 2 = 2$ neutrons. Use equation (22.1).
$Δ m = [(A - Z) m_{n} + Z m_{p}] - m_{He}$
Substitute the masses.
$Δ m = [2 (1.008665) + 2 (1.007276)] - 4.002603 = 4.031882 - 4.002603$
Subtract.
$Δ m = 0.029279 u$

Δ m = 0.029279 u

Assignment 22.1: mass defect of tritium

Find the mass defect of the tritium nucleus. The mass of tritium is

5.0083 \times 1 0^{- 27} kg

Given

$m_{T} = 5.0083 \times 1 0^{- 27} kg$ tritium, $X_{1}^{3} X_{21}^{23} H$
$m_{p} = 1.007276 u$
$m_{n} = 1.008665 u$

To find

$Δ m$

Tritium $X_{1}^{3} X_{21}^{23} H$ has 1 proton and $3 - 1 = 2$ neutrons. Convert its mass to atomic mass units with the precise value $1 u = 1.6605 \times 1 0^{- 27} kg$ . The result agrees with Table 22.1, which lists tritium as $3.016049 u$ .
$m_{T} = \frac{5.0083 \times 1 0 ^{- 27}}{1.6605 \times 1 0 ^{- 27}} u = 3.0161 u (= 3.016049 u)$
Add the masses of the separate nucleons.
$Z m_{p} + (A - Z) m_{n} = 1.007276 + 2 (1.008665) = 3.024606 u$
Subtract the mass of tritium.
$Δ m = 3.024606 - 3.016049 = 0.008557 u$
Convert to kilograms if needed.
$Δ m = 0.008557 \times 1.66 \times 1 0^{- 27} kg = 1.42 \times 1 0^{- 29} kg$

Δ m = 0.008557 u \approx 1.42 \times 1 0^{- 29} kg

22.2Binding energy and mass-energy equivalenceOften examined

For some time the mass defect was a puzzle. The answer came from Einstein's mass-energy relation: the missing mass has been converted into energy. When a nucleus forms from its nucleons, this energy is released. To break the nucleus apart again, the same amount of energy must be supplied. This energy is the binding energy of the nucleus.

Binding energy: The minimum energy required to break a nucleus into its separate protons and neutrons. It equals the energy released when the nucleus is formed from these nucleons.

E_{b} = Δ m c^{2} (22.2)

$E_{b}$ : binding energy (the book writes it simply as $E$ ), $Δ m$ : mass defect in kilograms, $c = 3 \times 1 0^{8} m s^{- 1}$ : speed of light.

The relation shows that mass and energy are inter-convertible: mass can be converted into energy, and energy back into mass. For nuclear calculations the mass must be in kilograms to give the energy in joules.

Figure 22.2 (a) redrawn: the binding energy must be supplied to pull the nucleons apart; the same energy is released when they come together.

Binding energy of helium-4 in joules

Use the mass defect of helium-4 from Example 22.1,

Δ m = 0.029279 u

, to find its binding energy in joules.

Convert the mass defect to kilograms, using $1 u = 1.66 \times 1 0^{- 27} kg$ .
$Δ m = 0.029279 \times 1.66 \times 1 0^{- 27} kg = 4.860 \times 1 0^{- 29} kg$
Apply $E_{b} = Δ m c^{2}$ (equation 22.2).
$E_{b} = (4.860 \times 1 0^{- 29}) (3 \times 1 0^{8})^{2} = 4.374 \times 1 0^{- 12} J$

This is the energy needed to break one helium-4 nucleus. It looks small, but for one mole of helium ( $6.02 \times 1 0^{23}$ nuclei) it is about $2.6 \times 1 0^{12} J$ , far more than any chemical bond energy.

E_{b} \approx 4.37 \times 1 0^{- 12} J

Energy equivalent of one atomic mass unit

Nuclear masses are given in $u$ and nuclear energies are usually given in $MeV$ ( $1 MeV = 1 0^{6} eV = 1.6 \times 1 0^{- 13} J$ ). A direct conversion factor saves time.

Derivation $1 u$ is equivalent to $931.5 MeV$

Aim. Find the energy equivalent of a mass of $1 u$ in joules and in electronvolts.

Assumptions.

Use $E = m c^{2}$ with the precise values $m = 1 u = 1.66054 \times 1 0^{- 27} kg$ and $c = 2.9979 \times 1 0^{8} m s^{- 1}$ .
$1 eV = 1.6022 \times 1 0^{- 19} J$ , the energy gained by an electron accelerated through $1 V$ .

Energy equivalent of $1 u$ in joules.
$E = m c^{2} = (1.66054 \times 1 0^{- 27}) (2.9979 \times 1 0^{8})^{2} = 1.4924 \times 1 0^{- 10} J$
To change joules into electronvolts, divide by the energy of $1 eV$ .
$E = \frac{1.4924 \times 1 0 ^{- 10}}{1.6022 \times 1 0 ^{- 19}} eV = 9.315 \times 1 0^{8} eV$
Write the result in MeV, using $1 MeV = 1 0^{6} eV$ .
$E = 931.5 \times 1 0^{6} eV = 931.5 MeV$

Result

1 u \equiv 931.5 MeV

A mass defect in $u$ can be turned straight into $MeV$ without converting to kilograms: $E (MeV) = Δ m (u) \times 931.5$ .

Table 22.1: rest masses of some particles and their energy equivalents. The book prints the electron mass as $9.1164 \times 1 0^{- 31} kg$ ; the correct value is $9.109 \times 1 0^{- 31} kg$ .
Particle	Mass (kg)	Mass (u)	Energy (MeV)
Electron	$9.109 \times 1 0^{- 31}$	0.0005485	0.511
Proton	$1.6726 \times 1 0^{- 27}$	1.007276	938.27
Neutron	$1.6749 \times 1 0^{- 27}$	1.008665	939.57
Hydrogen atom	$1.6735 \times 1 0^{- 27}$	1.007825	938.78
Deuterium atom	$3.344 \times 1 0^{- 27}$	2.014102	1876.12
Tritium atom	$5.0083 \times 1 0^{- 27}$	3.016049	2809.43
Helium-4 atom	$6.646 \times 1 0^{- 27}$	4.002603	3728.40
Carbon-12 atom	$1.9926 \times 1 0^{- 26}$	12.000000	11177.9

Example 22.2: binding energy of tritium

Find the energy required to break a tritium nucleus (mass

3.016049 u

) in joules and in eV. The masses of the proton and neutron are

1.007276 u

and

1.008665 u

Given

$M = 3.016049 u$ tritium
$Z = 1$
$A - Z = 2$ neutrons
$m_{p} = 1.007276 u$
$m_{n} = 1.008665 u$

To find

binding energy $E_{b}$

First find the mass defect with equation (22.1).
$Δ m = (2) (1.008665) + (1) (1.007276) - 3.016049 = 0.008557 u$
Convert to kilograms.
$Δ m = 0.008557 \times 1.66 \times 1 0^{- 27} kg = 1.420 \times 1 0^{- 29} kg$
Binding energy in joules from $E_{b} = Δ m c^{2}$ .
$E_{b} = (1.420 \times 1 0^{- 29}) (3 \times 1 0^{8})^{2} = 1.278 \times 1 0^{- 12} J$
Convert to electronvolts by dividing by $1.60 \times 1 0^{- 19} J$ .
$E_{b} = \frac{1.278 \times 1 0 ^{- 12}}{1.60 \times 1 0 ^{- 19}} eV = 7.99 \times 1 0^{6} eV \approx 8.0 MeV$
Check with the shortcut: $0.008557 \times 931.5 = 7.97 MeV$ .

E_{b} = 1.28 \times 1 0^{- 12} J \approx 7.99 \times 1 0^{6} eV

Assignment 22.2: binding energy of deuterium

Find the binding energy of the deuterium nucleus. The mass of deuterium is

3.344 \times 1 0^{- 27} kg

Given

$M_{D} = 3.344 \times 1 0^{- 27} kg = 2.014102 u$ deuterium atom, Table 22.1
$m_{e} = 0.000549 u$
$m_{p}, m_{n} = 1.007276 u, 1.008665 u$

To find

$E_{b}$

The given mass is that of the deuterium atom (Table 22.1). As in Figure 22.1, remove the one electron to get the nucleus.
$m_{nucleus} = 2.014102 - 0.000549 = 2.013553 u$
Mass defect for 1 proton and 1 neutron.
$Δ m = (1.007276 + 1.008665) - 2.013553 = 0.002388 u$
Binding energy in MeV, using $1 u = 931.5 MeV$ .
$E_{b} = 0.002388 \times 931.5 = 2.22 MeV$
In joules, using $1 MeV = 1.6 \times 1 0^{- 13} J$ .
$E_{b} = 2.22 \times 1.6 \times 1 0^{- 13} = 3.56 \times 1 0^{- 13} J$

E_{b} \approx 2.22 MeV = 3.56 \times 1 0^{- 13} J

22.2Binding energy per nucleonOften examined

The total binding energy grows with the number of nucleons, so on its own it cannot tell us which nucleus is more stable. Uranium-235 has a much larger binding energy (1786\,\mathrm{MeV}) than iron-56 (486\,\mathrm{MeV}), yet iron is far more stable. A fair comparison uses the binding energy per nucleon: the average energy needed to remove one nucleon from the nucleus.

Binding energy per nucleon: The total binding energy of a nucleus divided by the number of nucleons in it (its mass number A). The greater the binding energy per nucleon, the more stable the nucleus.

f=\dfrac{E_b}{A}=\dfrac{\left[(A-Z)m_n+Zm_p-m_{\text{nucleus}}\right]c^{2}}{A}\qquad(22.3)

Unit: \mathrm{MeV} per nucleon (or \mathrm{J} per nucleon).

Table 22.2: binding energies of some nuclei (book values). Accepted values are 7.68\,\mathrm{MeV} for \ce{^{12}C} and 8.79\,\mathrm{MeV} for \ce{^{56}Fe}; the curve in the next section uses these.
Nucleus	Binding energy (MeV)	Binding energy per nucleon (MeV)
\ce{^{2}H}	2.23	1.12
\ce{^{4}He}	28.3	7.07
\ce{^{12}C}	89.1	7.4
\ce{^{56}Fe}	486.24	8.68
\ce{^{206}Pb}	1622	7.87
\ce{^{235}U}	1786	7.6

Example 22.3: binding energy per nucleon of carbon-12

Find the packing fraction (binding energy per nucleon) for the C-12 nucleus. The mass of C-12 is 12.000\,\mathrm{u}, the mass of a proton is 1.007276\,\mathrm{u} and the mass of a neutron is 1.008665\,\mathrm{u}.

Given

m_{\text{nucleus}} = 12.000\,\mathrm{u}
Z = 6
A-Z = 6
m_p,\ m_n = 1.007276\,\mathrm{u},\ 1.008665\,\mathrm{u}

To find

f=E_b/A

Mass defect from (22.1).
\Delta m=6(1.008665)+6(1.007276)-12.000=0.095646\,\mathrm{u}
Binding energy in MeV.
E_b=0.095646\times931.5=89.1\,\mathrm{MeV}
Divide by the number of nucleons, A=12.
f=\dfrac{89.1}{12}=7.4\,\mathrm{MeV\ per\ nucleon}

f\approx7.4\,\mathrm{MeV\ per\ nucleon}

Assignment 22.3: binding energy per nucleon of nitrogen

Find the binding energy per nucleon for the nitrogen nucleus having mass 2.3258\times10^{-26}\,\mathrm{kg} (the book says nitrogen-13). The mass of a proton is 1.007276\,\mathrm{u} and of a neutron 1.008665\,\mathrm{u}.

Convert the mass to \mathrm{u} with 1\,\mathrm{u}=1.6605\times10^{-27}\,\mathrm{kg}.
M=\dfrac{2.3258\times10^{-26}}{1.6605\times10^{-27}}=14.0066\,\mathrm{u}
A mass of about 14\,\mathrm{u} means A=14: this is nitrogen-14 (Z=7, N=7), not nitrogen-13 (whose mass is 13.006\,\mathrm{u}).
\ce{^{14}_{7}N}:\quad Z=7,\quad A-Z=7
Mass defect.
\Delta m=7(1.007276)+7(1.008665)-14.0066=14.1116-14.0066=0.1050\,\mathrm{u}
Binding energy and binding energy per nucleon.
E_b=0.1050\times931.5=97.8\,\mathrm{MeV},\qquad \dfrac{E_b}{A}=\dfrac{97.8}{14}=7.0\,\mathrm{MeV}

\dfrac{E_b}{A}\approx7.0\,\mathrm{MeV\ per\ nucleon}\ (1.12\times10^{-12}\,\mathrm{J})

22.2The binding energy curveOften examined

A graph of binding energy per nucleon against mass number A for all nuclides is called the binding energy curve (Figure 22.2 (b)). Its shape explains why some nuclei are stable, and why both fission and fusion release energy.

Binding energy per nucleon (in \mathrm{MeV}) against mass number

The curve rises steeply for light nuclei, peaks near A=56 (iron, about 8.8\,\mathrm{MeV} per nucleon) and falls slowly to about 7.6\,\mathrm{MeV} for uranium. Moving towards the peak (fusion from the left, fission from the right) releases energy.

Features of the curve

For the lightest nuclei the value is small (1.1\,\mathrm{MeV} for \ce{^{2}H}) and it rises steeply as A increases.
Helium-4 stands out at 7.07\,\mathrm{MeV}, much higher than its neighbours. This is why the alpha particle behaves as a very stable unit.
The curve has a broad maximum for medium-sized nuclei, A\approx50 to 100. This is the region of the most stable nuclei, which the book calls the valley of stability. The peak is near iron-56, about 8.8\,\mathrm{MeV} per nucleon, so \ce{^{56}_{26}Fe} is the most stable nucleus.
Beyond A\approx60 the value falls slowly, to about 7.6\,\mathrm{MeV} for uranium-235 and uranium-238. Heavy nuclei are less tightly bound.

Why the curve has this shape

The strong nuclear force is very strong but acts only over about 10^{-15}\,\mathrm{m}, so each nucleon attracts only its nearest neighbours. In a small nucleus a large fraction of the nucleons sit at the surface with few neighbours, so the binding per nucleon is low. As the nucleus grows, more nucleons are fully surrounded and E_b/A rises.

The electrostatic repulsion between protons has a long range: every proton repels every other proton. In large nuclei this repulsion grows roughly as Z^{2}, while the nuclear attraction grows only as A. The repulsion is not balanced by the strong nuclear force, so E_b/A falls for heavy nuclei.

Relevance to fission and fusion

Energy is released in any nuclear change that produces nuclei with a higher binding energy per nucleon, that is, any change that moves nuclei towards the peak of the curve. The products are more tightly bound, so their total mass is smaller, and the lost mass appears as energy (E=\Delta mc^{2}).

Fusion (left side of the curve): light nuclei such as hydrogen isotopes join to form helium and climb the steep part of the curve. For \ce{^{2}H + ^{3}H -> ^{4}He + n} the energy released is 28.3-(2.2+8.5)=17.6\,\mathrm{MeV}.
Fission (right side of the curve): a heavy nucleus such as U-235 (about 7.6\,\mathrm{MeV} per nucleon) splits into two medium nuclei (about 8.5\,\mathrm{MeV} per nucleon). Each of the 236 nucleons gains about 0.9\,\mathrm{MeV}, so about 236\times0.9\approx200\,\mathrm{MeV} is released.
Nuclei at the peak (iron and nickel) cannot release energy by either process.

22.3RadioactivityOften examined

In 1896 Henri Becquerel discovered that uranium salts give out invisible radiation all by themselves. Marie Curie studied this radioactivity in uranium and thorium. Using an electrometer, she found that the air around radioactive samples became charged and conducting, because the radiation ionises the air.

Radioactivity: The natural (spontaneous) emission of radiation from unstable nuclei.
Spontaneous nuclear decay: The natural process in which an unstable nucleus breaks down into a more stable configuration by emitting radiation, without any external influence or trigger.
Parent and daughter nuclei: The decaying nucleus is the parent. The nucleus formed after the decay is the daughter; it is often a different element.

An unstable nucleus becomes more stable by emitting radiation. There are three main types of radioactive decay: alpha decay, beta decay and gamma decay. Other radiations, such as X-rays and neutrons, can also be produced in nuclear processes.

Main features of radioactivity

It is a property of the nucleus, not of the electrons. So it is not affected by chemical combination, temperature, pressure, or electric and magnetic fields.
It is spontaneous and random (Section 22.3.4).
Alpha and beta decay change the element (transmutation); gamma decay only removes energy.
Each radioactive isotope decays at its own fixed rate, described by its half-life.
The radiation ionises the matter it passes through. This is how it is detected, and also why it can harm living cells.

Summary of the types of radioactive decay.
Decay	Particle emitted	Change in Z	Change in A	Example
Alpha (\alpha)	helium nucleus \ce{^{4}_{2}He}	-2	-4	\ce{^{238}_{92}U -> ^{234}_{90}Th}
Beta-minus (\beta^-)	electron \ce{^{0}_{-1}e} (+ antineutrino)	+1	0	\ce{^{14}_{6}C -> ^{14}_{7}N}
Beta-plus (\beta^+)	positron \ce{^{0}_{+1}e} (+ neutrino)	-1	0	\ce{^{10}_{6}C -> ^{10}_{5}B}
Gamma (\gamma)	photon	0	0	\ce{^{60}_{27}Co^{*} -> ^{60}_{27}Co}

22.3.1Alpha decayOften examined

In alpha decay a nucleus emits an alpha particle: a helium-4 nucleus made of 2 protons and 2 neutrons (\ce{^{4}_{2}He}, charge +2e). The nucleus loses 2 protons and 2 neutrons, so its atomic number falls by 2 and its mass number falls by 4.

\ce{^{A}_{Z}X -> ^{A-4}_{Z-2}Y + ^{4}_{2}He} + Q

X is the parent nucleus, Y is the daughter nucleus and Q is the energy released, mostly as the kinetic energy of the alpha particle. For example, uranium-238 (92 protons, 146 neutrons) decays into thorium-234 (90 protons, 144 neutrons):

\ce{^{238}_{92}U -> ^{234}_{90}Th + ^{4}_{2}He} + Q

Figure 22.3 (redrawn): alpha decay of uranium-238.

Alpha decay happens mainly in heavy nuclei (Z>82), which have too many protons for the strong nuclear force to hold comfortably. Emitting an alpha particle reduces both the size and the charge of the nucleus. Another example is radium-226, which produces radioactive radon gas:

\ce{^{226}_{88}Ra -> ^{222}_{86}Rn + ^{4}_{2}He} + Q

22.3.2Beta decayOften examined

In beta decay a nucleus emits a beta particle, which is either an electron (\beta^-) or a positron (\beta^+). A positron is the antiparticle of the electron: it has the same mass but charge +e. The beta particle is created inside the nucleus at the moment of decay. Beta decay changes the atomic number by 1 but never changes the mass number, because the total number of nucleons stays the same.

Beta-minus decay

In a nucleus with too many neutrons, a neutron changes into a proton. An electron and an antineutrino (\bar\nu) are emitted:

\ce{^{1}_{0}n -> ^{1}_{1}p + ^{0}_{-1}e} + \bar{\nu}

The nucleus gains one proton and loses one neutron, so Z increases by 1 and A is unchanged:

\ce{^{A}_{Z}X -> ^{A}_{Z+1}Y + ^{0}_{-1}e} + \bar{\nu}\qquad\text{e.g.}\qquad \ce{^{14}_{6}C -> ^{14}_{7}N + ^{0}_{-1}e} + \bar{\nu}

Figure 22.4 (redrawn): beta-minus decay of carbon-14. Inside the nucleus a neutron becomes a proton.

Beta-plus decay

In a nucleus with too many protons, a proton changes into a neutron. A positron and a neutrino (\nu) are emitted:

\ce{^{1}_{1}p -> ^{1}_{0}n + ^{0}_{+1}e} + \nu

Now Z decreases by 1 and A is unchanged:

\ce{^{A}_{Z}X -> ^{A}_{Z-1}Y + ^{0}_{+1}e} + \nu\qquad\text{e.g.}\qquad \ce{^{10}_{6}C -> ^{10}_{5}B + ^{0}_{+1}e} + \nu

Beta-plus emitters such as carbon-11 and fluorine-18 are the tracers used in PET scanning (Section 22.8.2).

Electron capture

A nucleus with too many protons can also capture one of the atom's own inner (K-shell) electrons. The electron combines with a proton to form a neutron, and a neutrino is emitted. As in beta-plus decay, Z decreases by 1 and A does not change. The book does not describe this process, but Short Question 2 asks about it.

\ce{^{1}_{1}p + ^{0}_{-1}e -> ^{1}_{0}n} + \nu\qquad\text{e.g.}\qquad \ce{^{7}_{4}Be + ^{0}_{-1}e -> ^{7}_{3}Li} + \nu

The neutrino and the antineutrino are uncharged and have almost no mass. They share the decay energy with the beta particle, which is why beta particles from one isotope come out with a range of energies instead of one fixed energy.

22.3.3Gamma decayOften examined

After an alpha or beta decay the daughter nucleus is often left in an excited state, with its nucleons in higher energy levels (much like an electron in an excited atom). It falls to a lower energy state by emitting a gamma ray: a high-energy photon of electromagnetic radiation that travels at the speed of light.

\ce{^{A}_{Z}X^{*} -> ^{A}_{Z}X + \gamma}

X^{*} is the nucleus in its excited state. The photon energy equals the difference between the two nuclear energy levels, hf=E_2-E_1, usually between a few hundred keV and a few MeV. For example:

\ce{^{60}_{27}Co^{*} -> ^{60}_{27}Co + \gamma}

Figure 22.6 (redrawn): gamma decay. The nucleus loses energy but keeps the same number of protons and neutrons.

No particle with mass is emitted, so Z and A do not change. The laws of conservation of energy, momentum and angular momentum must still be obeyed in the transition. After emitting a photon the nucleus may still be unstable and may decay further until it reaches a stable state.

22.3Properties of alpha, beta and gamma radiationOften examined

Alpha, beta and gamma radiations differ in charge, mass and speed. These differences decide how strongly each one ionises matter and how far it can penetrate.

Table 22.3: properties of nuclear radiations.
Property	Alpha	Beta	Gamma
Nature	helium-4 nucleus	fast electron (or positron)	electromagnetic photon
Charge	+2e	-e (or +e)	0
Mass	4\,\mathrm{u}	0.00055\,\mathrm{u}, very small compared with alpha	no rest mass
Speed	slow, 0.05c to 0.1c	fast, 0.5c to 0.9c	speed of light c
Ionising power	very high	medium	very low
Penetration	stopped by a sheet of paper or a few cm of air	travels several metres in air; stopped by a few mm of aluminium	travels far in air; reduced by several cm of lead or thick concrete
Deflection in fields	small	large, opposite to alpha	none
Typical source	radon-222	strontium-90	cobalt-60

Figure 22.7 (redrawn): penetrating power of the three radiations.

Ionising power and penetrating power

As radiation passes through matter it knocks electrons out of atoms, producing ion pairs. An alpha particle has a large charge, a large mass and a low speed, so it interacts strongly with every atom it passes. It produces about 10^{4} ion pairs per centimetre of air. It therefore has the highest ionising power, but it loses its energy quickly and has the lowest penetrating power. A beta particle ionises less (about 10^{2} ion pairs per cm) and goes further. A gamma photon is uncharged, ionises very weakly (about 1 ion pair per cm) and is the most penetrating.

The three radiations passing between charged plates. Gamma is not deflected, alpha bends slightly towards the negative plate and beta bends strongly towards the positive plate.

In an electric or magnetic field, alpha and beta particles are deflected in opposite directions because their charges are opposite, and gamma rays are not deflected at all. Although an alpha particle has twice the charge of a beta particle, it is about 7300 times heavier. Its charge-to-mass ratio is therefore about 3600 times smaller, so it is deflected much less.

22.3.4Spontaneous and random nature of decay

Radioactive decay is random: it is impossible to predict exactly when a particular nucleus will decay. It may decay the moment you start observing it, or not for many years. Decay is also spontaneous: it happens by itself, because the unstable nucleus moves towards a more stable state. It cannot be started, stopped or speeded up by external factors such as temperature, pressure or chemical change.

Count rate: The number of decays detected per unit time by a detector, for example counts per second or counts per minute.

When a Geiger-Muller counter records the count rate from a source, the reading changes from one interval to the next even though nothing in the experiment changes. These fluctuations happen because each decay is spontaneous, independent of the others, random and unpredictable. They are the evidence for the random nature of radioactive decay.

Counts recorded in equal 10 s intervals

Each point is the number of counts in one 10\,\mathrm{s} interval. The same source, the same distance and the same counter, yet every reading is different. The readings scatter randomly about a steady mean: this is the evidence that decay is random.

Every nucleus of an isotope has the same fixed probability of decaying in each second (this probability per second is the decay constant \lambda of Section 22.4). One nucleus cannot be predicted, but a sample contains an enormous number of nuclei, so the average behaviour is very predictable and follows the half-life. This is why the count rate averaged over a long time is steady, even though counts in short intervals fluctuate.

22.3.4Energy released in a nuclear decayOften examined

The energy released in a decay (or in any nuclear reaction) is called its Q value. It comes from the difference between the total mass before and after the decay: the products have less mass than the parent, and the missing mass appears as energy.

Q=\left[m_{\text{parent}}-\left(m_{\text{daughter}}+m_{\text{emitted particle}}\right)\right]c^{2}

Q\,(\mathrm{MeV})=\Delta m\,(\mathrm{u})\times931.5

Method

Write the balanced decay equation.
Add the masses on each side. Use atomic masses for the parent, the daughter and (in alpha decay) helium, so that the electrons balance.
Find the mass lost: \Delta m= mass before - mass after.
Convert: Q=\Delta m\times931.5\,\mathrm{MeV}, or \Delta m in kg times c^{2} for joules. A decay that happens by itself must have Q>0.

Example 22.4 (corrected): energy of the alpha decay of U-238

Calculate the amount of energy Q released in the reaction \ce{^{238}_{92}U -> ^{234}_{90}Th + ^{4}_{2}He} + Q. The book gives the mass of U-238 as 238.02891\,\mathrm{u}, of Th-234 as 234.04360\,\mathrm{u} and of the alpha particle as 4.001506\,\mathrm{u}.

With the book's numbers, Q=238.02891-(234.04360+4.001506)=-0.016196\,\mathrm{u}, a negative value. A negative Q would mean that U-238 cannot decay by itself, which is false. The error is in the data: 238.02891\,\mathrm{u} is the average atomic mass of natural uranium from the periodic table, not the mass of the U-238 atom. Also, the atomic mass of thorium must be paired with the atomic mass of helium (4.00260\,\mathrm{u}), not the bare alpha particle, so that the electrons balance (92=90+2). The correct working is:

Given

M_U = 238.05079\,\mathrm{u} U-238 atom
M_{Th} = 234.04360\,\mathrm{u} Th-234 atom
M_{He} = 4.00260\,\mathrm{u} He-4 atom

To find

Total mass on the left.
M_U=238.05079\,\mathrm{u}
Total mass on the right.
M_{Th}+M_{He}=234.04360+4.00260=238.04620\,\mathrm{u}
Mass lost in the decay.
\Delta m=238.05079-238.04620=0.00459\,\mathrm{u}
Convert to energy.
Q=0.00459\times931.5=4.28\,\mathrm{MeV}\;(\approx6.8\times10^{-13}\,\mathrm{J})

Q\approx4.3\,\mathrm{MeV}\ \text{released}

Almost all of Q becomes kinetic energy of the alpha particle, because the heavy daughter recoils slowly (momentum is conserved). A small part may be given out as gamma rays if the daughter is left in an excited state.

Assignment 22.4 (corrected): energy of the beta decay of C-14

Find the energy Q in the reaction \ce{^{14}_{6}C -> ^{14}_{7}N + e-} + Q. The mass of N-14 is 14.00307\,\mathrm{u} and the mass of an electron is 0.0005486\,\mathrm{u}. The book prints 'mass of C-12 is 12.000 u'; this must be the mass of C-14, 14.003242\,\mathrm{u}.

Given

M_C = 14.003242\,\mathrm{u} C-14 atom
M_N = 14.00307\,\mathrm{u} N-14 atom
m_e = 0.0005486\,\mathrm{u}

To find

The masses are atomic masses. Find the nuclear masses by removing the electrons: 6 for carbon, 7 for nitrogen.
m_C=M_C-6m_e,\qquad m_N=M_N-7m_e
The nucleus of C-14 changes into the nucleus of N-14 plus one emitted electron.
\Delta m=m_C-(m_N+m_e)=(M_C-6m_e)-(M_N-7m_e)-m_e=M_C-M_N
The electron masses cancel. Substitute the atomic masses.
\Delta m=14.003242-14.00307=0.000172\,\mathrm{u}
Convert to energy.
Q=0.000172\times931.5=0.16\,\mathrm{MeV}

Q\approx0.16\,\mathrm{MeV}

22.4Half-lifeOften examined

We cannot say when one nucleus will decay, but we can say how long it takes for half of a large number of nuclei to decay. This time is the same for every sample of a given isotope, however large or old the sample is.

Half-life: The time taken for half of the radioactive nuclei present in a sample to decay. Symbol T_{1/2}.

Suppose a sample has 100 undecayed nuclei at t=0. After one half-life 50 have decayed and 50 remain. In the next half-life, half of these 50 decay, leaving 25. After the third and fourth half-lives, 12.5 and 6.25 remain on average. The same fraction, one half, decays in every half-life. For example, cobalt-60 has a half-life of 5.27 years, so 10\,\mathrm{g} of cobalt-60 becomes 5\,\mathrm{g} after 5.27 years, 2.5\,\mathrm{g} after 10.54 years, and so on (Figure 22.8).

Decay of a radioactive sample (Figures 22.8 and 22.9)

In every half-life the number of undecayed nuclei halves: N_0\to N_0/2\to N_0/4\to N_0/8. The curve gets closer and closer to zero but never reaches it.

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

N_0: undecayed nuclei at the start; N: undecayed nuclei after time t; n: number of half-lives in time t (need not be a whole number).

The number of nuclei halves in every half-life but never becomes exactly zero. In principle it would take an infinite time for all the nuclei to decay. Half-lives range from fractions of a second to billions of years.

Half-lives of some isotopes (the book's table). The book prints the daughter of cesium-137 as barium-130; beta decay does not change A, so it is barium-137.
Isotope	Decays into	Half-life
Carbon-14	Nitrogen-14	5730 years
Aluminium-26	Magnesium-26	740000 years
Iodine-129	Xenon-129	17 million years
Uranium-235	Lead-207 (through a chain)	704 million years
Uranium-238	Thorium-234	4.5 billion years
Potassium-40	Argon-40	1.3 billion years
Rubidium-87	Strontium-87	49 billion years
Plutonium-239	Uranium-235	24110 years
Cesium-137	Barium-137	30 years
Strontium-90	Yttrium-90	28.8 years
Radon-222	Polonium-218	3.82 days

Time for the number of nuclei to fall to one eighth

Carbon-14 has a half-life of 5730 years. After how many years will the number of C-14 nuclei in a sample drop to 1/8 of the initial number?

Write 1/8 as a power of 1/2 to find the number of half-lives.
\dfrac{N}{N_0}=\dfrac{1}{8}=\left(\dfrac{1}{2}\right)^{3}\;\Rightarrow\;n=3
Multiply by the half-life.
t=nT_{1/2}=3\times5730=17190\,\text{years}\approx1.72\times10^{4}\,\text{years}

t\approx1.72\times10^{4}\ \text{years}

A time that is not a whole number of half-lives

Iodine-131 has a half-life of 8 days. What fraction of a sample is left after 20 days?

Number of half-lives.
n=\dfrac{t}{T_{1/2}}=\dfrac{20}{8}=2.5
Fraction left.
\dfrac{N}{N_0}=\left(\dfrac{1}{2}\right)^{2.5}=0.177

\dfrac{N}{N_0}\approx0.18\ (18\%)

22.4Rate law, decay constant and activityOften examined

The more undecayed nuclei a sample contains, the more decays happen each second. Double the sample and you double the number of decays per second. This is the rate law of radioactive decay.

Rate law for nuclear decay: The rate at which a radioactive substance decays is directly proportional to the number of radioactive nuclei present at that time.

\dfrac{\Delta N}{\Delta t}\propto-N\qquad\Longrightarrow\qquad\dfrac{\Delta N}{\Delta t}=-\lambda N

\Delta N is the change in the number of undecayed nuclei in a short time \Delta t. The negative sign shows that N decreases with time. The constant of proportionality \lambda is the decay constant:

\lambda=-\dfrac{1}{N}\,\dfrac{\Delta N}{\Delta t}\qquad\left(\text{in size: }\lambda=\dfrac{1}{N}\times\dfrac{\Delta N}{\Delta t}\right)

Decay constant: The fraction of the radioactive nuclei in a sample that decay per unit time. It is the probability per second that any one nucleus decays. SI unit: \mathrm{s^{-1}} (per second). It is a fixed property of each isotope.
Activity: The number of disintegrations (decays) per unit time in a sample, A=\left|\Delta N/\Delta t\right|. SI unit: becquerel (Bq).
Becquerel: The activity of a sample in which one nucleus disintegrates per second: 1\,\mathrm{Bq}=1 decay per second =1\,\mathrm{s^{-1}}.
Curie: The activity of a sample in which 3.70\times10^{10} nuclei disintegrate per second (roughly the activity of 1\,\mathrm{g} of radium-226).

A=\lambda N

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

The becquerel is a very small unit compared with the curie. Radiation is present everywhere (background radiation), but natural levels are low and safe; harmful effects need much larger doses. For example, the radium on a luminous watch dial has an activity of about 4\times10^{4}\,\mathrm{Bq}, while radiotherapy sources are about 4\times10^{13}\,\mathrm{Bq}.

Example 22.5: decay constant and activity

Find the decay constant and activity of an element if 100 out of its 700 nuclei decay in 5 min.

Given

N = 700
\Delta N = 100
\Delta t = 5\,\mathrm{min}=300\,\mathrm{s}

To find

decay constant \lambda
activity A

Decay constant from its definition.
\lambda=\dfrac{1}{N}\times\dfrac{\Delta N}{\Delta t}=\dfrac{1}{700}\times\dfrac{100}{300}=4.76\times10^{-4}\,\mathrm{s^{-1}}
Activity from A=\lambda N.
A=(4.76\times10^{-4})(700)=0.333\,\mathrm{Bq}
In curie, divide by 3.70\times10^{10}.
A=\dfrac{0.333}{3.70\times10^{10}}=9.0\times10^{-12}\,\mathrm{Ci}

\lambda=4.76\times10^{-4}\,\mathrm{s^{-1}},\qquad A=0.333\,\mathrm{Bq}\approx9\times10^{-12}\,\mathrm{Ci}

Assignment 22.5

Find the decay constant and activity of an element if 240 out of its 780 nuclei decay in 12 seconds.

Given

N = 780
\Delta N = 240
\Delta t = 12\,\mathrm{s}

To find

\lambda
A

Decay constant.
\lambda=\dfrac{1}{N}\times\dfrac{\Delta N}{\Delta t}=\dfrac{1}{780}\times\dfrac{240}{12}=2.56\times10^{-2}\,\mathrm{s^{-1}}
Activity.
A=\lambda N=(2.56\times10^{-2})(780)=20\,\mathrm{Bq}=5.4\times10^{-10}\,\mathrm{Ci}

\lambda=2.56\times10^{-2}\,\mathrm{s^{-1}},\qquad A=20\,\mathrm{Bq}

22.4.1Exponential nature of nuclear decayOften examined

Radioactive decay is exponential: in equal intervals of time, the same fraction of the remaining nuclei decays. The rate law \Delta N/\Delta t=-\lambda N leads directly to the exponential decay law.

Derivation Exponential decay law, N=N_0e^{-\lambda t}

Aim. Find the number of undecayed nuclei N left at time t, if there were N_0 undecayed nuclei at t=0.

Assumptions.

The sample contains a very large number of nuclei, so N can be treated as a continuous quantity.
The decay constant \lambda does not change: it is not affected by temperature, pressure or chemical state.

Exponential decay of undecayed nuclei

In every half-life the number of undecayed nuclei halves: N_0\to N_0/2\to N_0/4\to N_0/8. The curve gets closer and closer to zero but never reaches it.

Start from the rate law: the rate of decay is proportional to the number of undecayed nuclei.
\dfrac{\Delta N}{\Delta t}=-\lambda N
For a very small time interval, the ratio of small changes becomes a derivative.
\dfrac{dN}{dt}=-\lambda N
Separate the variables: everything with N on one side, everything with t on the other.
\dfrac{1}{N}\,dN=-\lambda\,dt
Integrate both sides. The integral of 1/N is \ln N.
\int\dfrac{1}{N}\,dN=-\lambda\int dt\;\;\Longrightarrow\;\;\ln N=-\lambda t+C\qquad(\text{i})
Find the constant C from the initial condition: at t=0, N=N_0.
\ln N_0=-\lambda(0)+C\;\;\Longrightarrow\;\;C=\ln N_0
Put C back into (i) and combine the logarithms.
\ln N-\ln N_0=-\lambda t\;\;\Longrightarrow\;\;\ln\dfrac{N}{N_0}=-\lambda t
Take the exponential of both sides to remove the logarithm.
\dfrac{N}{N_0}=e^{-\lambda t}

ResultN=N_0\,e^{-\lambda t}

The number of undecayed nuclei falls exponentially with time: quickly at first, when N is large, then more and more slowly, and it never becomes exactly zero. A larger decay constant means a faster decay.

The same law for activity and count rate

Activity is proportional to N (A=\lambda N), so it follows the same law. Multiply both sides of N=N_0e^{-\lambda t} by \lambda:

A=\lambda N=\lambda N_0e^{-\lambda t}=A_0\,e^{-\lambda t}

The count rate received by a detector is a fixed fraction of the activity, so it decays in the same way. In general, if x stands for the number of undecayed nuclei, the activity, the mass of the radioactive isotope or the received count rate:

x=x_0\,e^{-\lambda t}

Solving decay problems analytically

Activity after a given time

A radioactive sample has an initial activity of 800\,\mathrm{Bq} and a half-life of 2.0\,\mathrm{h}. Find (a) its activity after 5.0\,\mathrm{h}, (b) the time after which its activity is 100\,\mathrm{Bq}.

Decay constant (in \mathrm{h^{-1}}, because all times are in hours).
\lambda=\dfrac{0.693}{T_{1/2}}=\dfrac{0.693}{2.0}=0.3465\,\mathrm{h^{-1}}
(a) Use A=A_0e^{-\lambda t}.
A=800\,e^{-(0.3465)(5.0)}=800\,e^{-1.733}=800\times0.177=141\,\mathrm{Bq}
(b) 100\,\mathrm{Bq} is 1/8 of 800\,\mathrm{Bq}, which is 3 half-lives.
t=3T_{1/2}=3\times2.0=6.0\,\mathrm{h}
Check (b) with logarithms: \ln(A_0/A)=\lambda t.
t=\dfrac{\ln 8}{0.3465}=\dfrac{2.079}{0.3465}=6.0\,\mathrm{h}

(a)\ A\approx141\,\mathrm{Bq}\qquad(b)\ t=6.0\,\mathrm{h}

Solving decay problems graphically

Taking natural logarithms of N=N_0e^{-\lambda t} gives the equation of a straight line:

\ln N=-\lambda t+\ln N_0\qquad(\text{compare with }y=mx+c)

A graph of \ln N (or \ln A, or \ln of the count rate) against t is therefore a straight line with gradient -\lambda and intercept \ln N_0. Real readings scatter because decay is random, so draw the best straight line through the points and find \lambda from its gradient; then T_{1/2}=0.693/\lambda.

Straight-line graph of \ln N against t (time in hours)

Plotting \ln N (or \ln A, or \ln of the count rate) against time turns the decay curve into a straight line. Its gradient is -\lambda and its intercept is \ln N_0. Here the gradient is -2/4=-0.5\,\mathrm{h^{-1}}, so \lambda=0.5\,\mathrm{h^{-1}} and T_{1/2}=0.693/0.5=1.4\,\mathrm{h}.

The half-life can also be read straight from the ordinary decay curve: choose any value of N and find how long it takes to fall to half of that value. Every such interval equals T_{1/2}.

22.4.2Relation between decay constant and half-lifeOften examined

Consider a sample with N_0 nuclei and a half-life of 4 hours. After 4 hours N_0/2 nuclei remain, after 8 hours (two half-lives) N_0/4, after 12 hours N_0/8, and so on. A graph shows this clearly (Figure 22.9), and it is easy to read the half-life from it.

Undecayed nuclei after each half-life, starting from 10000 (Figure 22.9).
Time	Undecayed nuclei N
0	10000
T_{1/2}	5000
2T_{1/2}	2500
3T_{1/2}	1250
4T_{1/2}	625

Derivation Relation between half-life and decay constant

Aim. Show that T_{1/2}=\dfrac{0.693}{\lambda}.

Assumptions.

The sample decays according to N=N_0e^{-\lambda t}.

Start from the exponential decay law.
N=N_0\,e^{-\lambda t}
By the definition of half-life, when t=T_{1/2} the number left is N=N_0/2. Substitute these values.
\dfrac{N_0}{2}=N_0\,e^{-\lambda T_{1/2}}
Cancel N_0 from both sides.
\dfrac{1}{2}=e^{-\lambda T_{1/2}}
Take the reciprocal of both sides.
2=e^{\lambda T_{1/2}}
Take the natural logarithm of both sides; \ln(e^{x})=x.
\ln2=\lambda T_{1/2}
Rearrange and use \ln2=0.693.
T_{1/2}=\dfrac{\ln2}{\lambda}=\dfrac{0.693}{\lambda}

ResultT_{1/2}=\dfrac{0.693}{\lambda}

Half-life and decay constant are inversely proportional: an isotope with a large decay constant decays quickly and has a short half-life. T_{1/2} does not depend on the number of nuclei present, so it is a fixed characteristic of each isotope.

The half-life of a radioactive material is found by measuring the activity of a sample over a period of time. The longer the half-life, the slower the decay, and the fewer radiations a given number of nuclei emits per second.

Example 22.6: cesium-137

Cesium-137 (a beta emitter used for treating cancer and in industrial devices that measure the thickness of materials) has a half-life of 9.5\times10^{8}\,\mathrm{s}. We have one mole of cesium at the start. What will be the decay constant and the activity?

Given

T_{1/2} = 9.5\times10^{8}\,\mathrm{s}
N_0 = 6.02\times10^{23} one mole

To find

\lambda
A

Decay constant from T_{1/2}=0.693/\lambda.
\lambda=\dfrac{0.693}{T_{1/2}}=\dfrac{0.693}{9.5\times10^{8}}=7.3\times10^{-10}\,\mathrm{s^{-1}}
Activity from A=\lambda N.
A=(7.3\times10^{-10})(6.02\times10^{23})=4.4\times10^{14}\,\mathrm{Bq}
In curie.
A=\dfrac{4.4\times10^{14}}{3.70\times10^{10}}=1.19\times10^{4}\,\mathrm{Ci}

\lambda=7.3\times10^{-10}\,\mathrm{s^{-1}},\qquad A\approx4.4\times10^{14}\,\mathrm{Bq}\approx1.19\times10^{4}\,\mathrm{Ci}

Assignment 22.6: strontium-91

The half-life of strontium-91 is 9.70 hours. Find its decay constant.

Given

T_{1/2} = 9.70\,\mathrm{h}=9.70\times3600\,\mathrm{s}=3.49\times10^{4}\,\mathrm{s}

To find

\lambda

Use T_{1/2}=0.693/\lambda with the half-life in seconds.
\lambda=\dfrac{0.693}{3.492\times10^{4}}=1.98\times10^{-5}\,\mathrm{s^{-1}}
In per hour: \lambda=0.693/9.70=0.0714\,\mathrm{h^{-1}}.

\lambda\approx1.98\times10^{-5}\,\mathrm{s^{-1}}

22.5Nuclear reactions

In radioactive decay a nucleus changes by itself. In a nuclear reaction a nucleus is changed by something that hits it: an incident nucleus, particle or photon collides with a target nucleus, and new nuclides are produced.

Nuclear reaction: A process in which atomic nuclei interact, resulting in a change in the nucleus (such as fusion, fission or the formation of a new nuclide), often releasing or absorbing energy.

The bombarding particle may be an alpha particle, a gamma ray photon, a neutron, a proton or a heavy ion. A charged particle needs enough energy to overcome the electrostatic repulsion of the positive nucleus and come within range of the strong nuclear force. A neutron is uncharged, so even a slow neutron can enter a nucleus.

If a light particle a hits a target nucleus X and produces a nucleus Y and a light particle b:

a+X\longrightarrow Y+b

The first artificial transmutation was observed by Rutherford in 1919 (Figure 22.10). Alpha particles passing through nitrogen produced oxygen-17 and protons:

\ce{^{4}_{2}He + ^{14}_{7}N -> ^{17}_{8}O + ^{1}_{1}H} + Q

Q value: The energy released or absorbed in a nuclear reaction. It equals the difference between the rest-mass energies of the reactants and the products: Q=\left[(m_a+m_X)-(m_Y+m_b)\right]c^{2}.
Exothermic reaction: A reaction with positive Q: energy is released, as kinetic energy of the products or as gamma rays.
Endothermic reaction: A reaction with negative Q: energy must be supplied, usually as kinetic energy of the incoming particle.
Thermal neutron: A slow neutron whose kinetic energy is about the same as that of the surrounding atoms, about 0.025\,\mathrm{eV} at room temperature.

When a thermal neutron is absorbed by nitrogen-14, radioactive carbon-14 and a proton are produced and energy is released. This reaction, caused by cosmic-ray neutrons, makes the carbon-14 used in radiocarbon dating:

\ce{^{1}_{0}n + ^{14}_{7}N -> ^{14}_{6}C + ^{1}_{1}H} + Q

A deuteron (charge +e) needs a high energy to cause a reaction such as:

\ce{^{2}_{1}H + ^{55}_{25}Mn -> ^{56}_{26}Fe + ^{1}_{0}n} + Q

Conservation laws in nuclear reactions

Charge: the total of the lower numbers (Z) is the same on both sides.
Nucleon number: the total of the upper numbers (A) is the same on both sides.
Mass-energy: the total of rest energy and kinetic energy is conserved; the Q value accounts for the change in rest mass.
Momentum (and angular momentum) is conserved.

22.5.1Nuclear fusionOften examined

The energy of the Sun and the other stars, and so almost all the energy that reaches the Earth, comes from a nuclear reaction called nuclear fusion. Its fuel is hydrogen.

Nuclear fusion: A process in which two or more light atomic nuclei come close together and join to form a heavier nucleus, releasing a large amount of energy.

Fusion happens naturally in stars, where the very high temperature and pressure make nuclei collide and fuse. The core of the Sun is the site of the fusion reactions that power its output. Scientists are developing fusion as a possible clean and sustainable energy source. A simple fusion reaction, deuterium with tritium (Figure 22.11), is:

\ce{^{2}_{1}H + ^{3}_{1}H -> ^{4}_{2}He + ^{1}_{0}n} + Q

Figure 22.11 (redrawn): fusion of deuterium and tritium. The helium nucleus and the neutron share the 17.6\,\mathrm{MeV} released.

Why fusion releases energy

Light nuclei have a low binding energy per nucleon, so they are less stable than nuclei in the middle of the binding energy curve (Figure 22.12). When they fuse, the product is more tightly bound: the binding energy per nucleon increases, the total mass decreases, and the mass difference is released as energy. The binding energy per nucleon rises to a maximum of about 8.8\,\mathrm{MeV} for \ce{^{56}_{26}Fe}, so the rising part of the curve favours fusion.

Energy released in deuterium-tritium fusion

Total mass before the reaction.
m_D+m_T=2.014102+3.016049=5.030151\,\mathrm{u}
Total mass after the reaction.
m_{\text{He}}+m_n=4.002603+1.008665=5.011268\,\mathrm{u}
Mass lost.
\Delta m=5.030151-5.011268=0.018883\,\mathrm{u}
Convert to energy with 1\,\mathrm{u}=931.5\,\mathrm{MeV}.
Q=0.018883\times931.5=17.6\,\mathrm{MeV}=2.8\times10^{-12}\,\mathrm{J}

Q is positive, so the reaction is exothermic: a single fusion of deuterium and tritium into helium and a neutron releases about 17.6\,\mathrm{MeV}.

Conditions for fusion

All nuclei are positive, so they repel each other. To come within the range of the strong nuclear force (about 10^{-15}\,\mathrm{m}) they must collide at very high speeds. This needs extremely high temperatures: about 1.5\times10^{7}\,\mathrm{K} in the core of the Sun and over 10^{8}\,\mathrm{K} in laboratory reactors. That is why fusion is also called a thermonuclear reaction. At such temperatures matter is a plasma, which must also be dense enough and held together long enough for many fusions to occur.

Practical applications of fusion

Energy of the Sun and stars: almost all the energy that reaches the Earth as sunlight comes from fusion.
Hydrogen bomb (thermonuclear weapon): an uncontrolled fusion reaction, started by the very high temperature of a fission bomb.
Fusion reactors (under research): devices called tokamaks, such as the international ITER project, hold a hot plasma of deuterium and tritium in strong magnetic fields, aiming to generate electricity.
Expected advantages: deuterium is plentiful in sea water, fusion produces no long-lived fission waste and no greenhouse gases, and it cannot run away like a fission chain reaction.
Difficulties: reaching and holding temperatures above 10^{8}\,\mathrm{K}; so far no reactor has produced more useful energy than it uses.

22.5.3Nuclear fissionOften examined

In 1938 Otto Hahn and Fritz Strassmann bombarded uranium with neutrons and found barium, an element with about half the mass of uranium, among the products. The uranium nucleus had split. This startling discovery is nuclear fission. (The book numbers this section 22.5.3; there is no 22.5.2.)

Nuclear fission: A process in which the nucleus of a heavy atom splits into two (or more) smaller nuclei of comparable mass, releasing a few neutrons and a large amount of energy.

Fission is usually started by bombarding the nucleus with a neutron. When uranium-235 absorbs a slow neutron it becomes uranium-236 in a highly excited, unstable state, which splits almost at once:

\ce{^{1}_{0}n + ^{235}_{92}U -> ^{236}_{92}U^{*} -> ^{141}_{56}Ba + ^{92}_{36}Kr + 3^{1}_{0}n} + Q

Figure 22.13 (redrawn): fission of uranium-235 by a slow neutron.

Fission does not give the same products every time. More than 100 different pairs of fragments are possible, and 2 or 3 neutrons are released (about 2.5 on average). Two other examples are:

\ce{^{1}_{0}n + ^{235}_{92}U -> ^{236}_{92}U^{*} -> ^{132}_{50}Sn + ^{101}_{42}Mo + 3^{1}_{0}n} + Q

\ce{^{1}_{0}n + ^{235}_{92}U -> ^{236}_{92}U^{*} -> ^{139}_{56}Ba + ^{95}_{36}Kr + 2^{1}_{0}n} + Q

Why fission releases energy

Heavy nuclei have a low binding energy per nucleon (about 7.6\,\mathrm{MeV} for uranium) because of the large electrostatic repulsion between their many protons. The fragments are medium-sized nuclei with about 8.5\,\mathrm{MeV} per nucleon, so they are more stable. The total binding energy increases by roughly 236\times0.9\approx200\,\mathrm{MeV}, and the products have correspondingly less mass than the reactants.

Most of this energy (about 165\,\mathrm{MeV}) appears as the kinetic energy of the two fragments, which fly apart because they repel each other. The rest is carried by the neutrons, by gamma rays and by the later radioactive decay of the fragments.

Fission energy compared with chemical energy

A single fission of uranium-235 releases about 200\,\mathrm{MeV}. A chemical reaction releases only a few electronvolts per reaction: burning one molecule of methane (\ce{CH4}) releases about 9.2\,\mathrm{eV}. So one fission releases about 2\times10^{7} times more energy than one chemical reaction.

Energy from 1 kg of uranium-235 compared with coal

Estimate the energy released by the fission of 1\,\mathrm{kg} of uranium-235 (about 200\,\mathrm{MeV} per fission) and compare it with the energy from burning 1\,\mathrm{kg} of coal (about 3\times10^{7}\,\mathrm{J}).

Number of nuclei in 1\,\mathrm{kg}: 1000\,\mathrm{g} divided by the molar mass 235\,\mathrm{g\,mol^{-1}} gives the moles; multiply by Avogadro's number.
N=\dfrac{1000}{235}\times6.02\times10^{23}=2.56\times10^{24}
Energy per fission in joules.
200\,\mathrm{MeV}=200\times1.6\times10^{-13}\,\mathrm{J}=3.2\times10^{-11}\,\mathrm{J}
Total energy.
E=(2.56\times10^{24})(3.2\times10^{-11})=8.2\times10^{13}\,\mathrm{J}
Compare with coal.
\dfrac{8.2\times10^{13}}{3\times10^{7}}\approx2.7\times10^{6}

E\approx8\times10^{13}\,\mathrm{J},\ \text{about 2.5 to 3 million times the energy from 1 kg of coal}

This huge difference is why nuclear power plants are so useful. A large nuclear power station uses about 25 tonnes of uranium fuel a year, while a coal station of the same power burns about 3 million tonnes of coal.

Example 22.7: energy released in a fission reaction

Find the amount of energy released by the fission reaction \ce{^{1}_{0}n + ^{235}_{92}U -> ^{141}_{56}Ba + ^{92}_{36}Kr + 3^{1}_{0}n} + Q if the masses of uranium, barium and krypton are 235.12142\,\mathrm{u}, 140.883\,\mathrm{u} and 91.9262\,\mathrm{u} respectively.

Given

m_U = 235.12142\,\mathrm{u}
m_{Ba} = 140.883\,\mathrm{u}
m_{Kr} = 91.9262\,\mathrm{u}
m_n = 1.008665\,\mathrm{u}

To find

Total mass on the left.
m_n+m_U=1.008665+235.12142=236.130085\,\mathrm{u}
Total mass on the right (three neutrons).
m_{Ba}+m_{Kr}+3m_n=140.883+91.9262+3(1.008665)=235.835195\,\mathrm{u}
Mass lost.
\Delta m=236.130085-235.835195=0.29489\,\mathrm{u}
Energy released.
Q=0.29489\times931.5=274.7\,\mathrm{MeV}

Q is positive, so energy is released in the fission.

Q\approx275\,\mathrm{MeV}\ \text{(with the book's data)}

Assignment 22.7

Find the energy released by \ce{^{1}_{0}n + ^{235}_{92}U -> ^{140}_{54}Xe + ^{94}_{38}Sr + 2^{1}_{0}n} + Q if the masses of uranium, xenon, strontium and the neutron are 235.12142\,\mathrm{u}, 139.9055\,\mathrm{u}, 93.9064\,\mathrm{u} and 1.008665\,\mathrm{u}.

Total mass on the left.
1.008665+235.12142=236.130085\,\mathrm{u}
Total mass on the right (two neutrons).
139.9055+93.9064+2(1.008665)=235.82923\,\mathrm{u}
Mass lost and energy released.
\Delta m=0.300855\,\mathrm{u},\qquad Q=0.300855\times931.5=280\,\mathrm{MeV}=4.5\times10^{-11}\,\mathrm{J}

Q\approx280\,\mathrm{MeV}\approx4.5\times10^{-11}\,\mathrm{J}

22.5.3Fission chain reaction and critical massOften examined

Each fission of uranium-235 releases 2 or 3 neutrons. If these neutrons strike other U-235 nuclei they cause more fissions, which release still more neutrons, and so on. The number of fissions can multiply very quickly: 1, 3, 9, 27, ...

Fission chain reaction: A succession of fission reactions in which, every time, at least one neutron released by a fission produces a further fission, so that the reaction keeps itself going.

Figure 22.15 (redrawn): a fission chain reaction in uranium-235. Neutrons from each fission cause further fissions.

Not every neutron causes another fission. The neutrons released by fission are fast (about 1 to 2\,\mathrm{MeV}), while U-235 is split most easily by slow neutrons, so fast neutrons often pass through without causing fission. Neutrons near the surface of a small piece of uranium escape before meeting another nucleus, and some are absorbed without fission. In a larger piece more neutrons are captured inside. So a chain reaction needs a minimum amount of fissile material.

Critical mass: The minimum mass of a fissile material that can sustain a fission chain reaction.
Critical volume: The volume occupied by the critical mass of a material.

Less than the critical mass (subcritical): too many neutrons are lost, and the chain reaction soon dies out.
Equal to the critical mass (critical): on average exactly one neutron from each fission causes another fission. The rate of fission stays constant and energy is produced steadily, as in a nuclear reactor.
More than the critical mass (supercritical): more than one neutron per fission causes further fissions. The number of fissions grows rapidly and uncontrollably and a huge amount of energy is released in a fraction of a second, as in a nuclear bomb.

The possibility of an uncontrolled chain reaction is what makes nuclear weapons so dangerous, and it is why nuclear reactors follow strict safety rules. In a reactor the chain reaction is controlled so that it stays exactly critical (Section 22.6).

22.6Nuclear reactorOften examined

Fission releases a huge amount of energy, so in a power station it must be carried out under controlled conditions. A nuclear reactor does this. The energy is used to produce electricity, and reactors are also used to produce radioisotopes.

Nuclear reactor: A device used to initiate and control a nuclear fission chain reaction, so that energy is released at a steady rate.

A reactor consists of the core, fuel rods, control rods (safety rods), moderator, coolant (heat exchanger) and shielding. Figure 22.16 shows a pressurised water reactor, the most common kind of water-moderated power reactor.

Figure 22.16 (redrawn): a pressurised water reactor. Water in the primary loop is both the moderator and the coolant; its heat makes steam in a separate loop, which drives the turbine.

Core

The core is the heart of the reactor. It contains the fuel, the moderator and the control rods. The controlled fission chain reaction takes place here and the heat is generated here. The core sits inside a thick steel pressure vessel.

Fuel rods

The fuel is usually uranium-235, sometimes plutonium-239. It is made into small pellets of uranium oxide, stacked in long metal tubes about 1\,\mathrm{cm} in diameter. Most reactors use uranium enriched to about 3% uranium-235. Many fuel rods are placed in the core with moderator between them.

Control rods (safety rods)

Control rods are made of a material that absorbs neutrons strongly, such as boron or cadmium. Pushing the rods further into the core absorbs more neutrons, so fewer fissions occur; pulling them out allows more fissions. They are adjusted so that, on average, only one neutron from each fission goes on to cause another fission. The rate of fission, and so the power output, then stays constant. They stop the fuel from reacting all at once. In an emergency the rods drop fully into the core and the chain reaction stops.

Moderator

A moderator is a material that slows down neutrons. Neutrons released by fission are fast, with energies of a few MeV, but uranium-235 is split far more easily by slow (thermal) neutrons of about 0.025\,\mathrm{eV}. The fast neutrons collide again and again with the light nuclei of the moderator, lose energy in each collision and become thermal neutrons, which are much more likely to cause fission. Water and graphite are the common moderators (Figure 22.17).

Coolant (heat exchanger)

The coolant is a fluid, such as water or helium gas, that circulates through the core and carries away the heat produced by fission. It keeps the fuel rods and moderator from overheating; without it the core could melt. In the heat exchanger (steam generator) the hot coolant boils water in a second loop. The steam drives a turbine, which turns an electric generator. The steam is then cooled back to water in the condenser and pumped back.

Shielding

Fission produces intense radiation (neutrons and gamma rays) and radioactive products that are harmful to humans and the environment. The reactor vessel is surrounded by a thick layer of concrete (and steel), which absorbs this radiation and protects workers and the surroundings.

Principal components of a water-moderated power reactor.
Component	Typical material	Function
Fuel rods	uranium enriched to about 3% U-235 (or Pu-239)	undergo fission and produce heat
Moderator	ordinary water, heavy water or graphite	slows fast neutrons down to thermal energies
Control and safety rods	boron or cadmium	absorb neutrons to control or stop the chain reaction
Coolant and heat exchanger	water, carbon dioxide or helium	remove heat from the core and make steam for the turbine
Shielding	thick concrete and steel	stops radiation escaping from the reactor

How a reactor starts and controls the chain reaction

With the control rods fully inserted, the reactor is shut down: the rods absorb so many neutrons that no chain reaction occurs.
A few free neutrons are always present, from spontaneous fission of uranium or from a small neutron source placed in the core.
The control rods are slowly withdrawn. More neutrons survive, are slowed down by the moderator, and cause fission in U-235.
When exactly one neutron per fission causes a further fission, the reactor is critical and the power is steady. The rods are then held in place and moved slightly to raise or lower the power.

\text{nuclear energy}\;\to\;\text{heat}\;\to\;\text{steam}\;\to\;\text{kinetic energy of turbine}\;\to\;\text{electrical energy}

22.6Enrichment of uraniumOften examined

Natural uranium is a mixture of isotopes: about 99.3% uranium-238 and 0.7% uranium-235. Only uranium-235 is fissile: it absorbs a slow neutron, undergoes fission and can sustain a chain reaction. Uranium-238 is not directly fissile. It usually captures neutrons without splitting, and it can be slowly converted into plutonium-239, which is fissile.

Why uranium fuel needs to be enriched

With only 0.7% U-235, most neutrons in an ordinary water-moderated reactor are absorbed by U-238 and by the water before they meet a U-235 nucleus. Fewer than one neutron per fission then causes a further fission, so the chain reaction cannot be sustained. Raising the proportion of U-235 to about 3 to 5% makes enough neutrons cause fission. The reactor can then reach and keep criticality, operate efficiently and use its fuel for longer.

Enrichment of uranium: The process of increasing the percentage of the fissile isotope uranium-235 in uranium, by separating it from uranium-238.
Depleted uranium: The left-over uranium (the 'tails') that contains less than 0.7% uranium-235.

Methods of enrichment

Because U-235 and U-238 behave in the same way chemically, they are separated using the small difference in their masses. The uranium is first converted into uranium hexafluoride (\ce{UF6}), which is a gas at moderate temperatures. The molecular masses of its two forms are:

M\left(\ce{^{235}UF6}\right)=235+6(19)=349,\qquad M\left(\ce{^{238}UF6}\right)=238+6(19)=352

Gaseous diffusion

At the same temperature all gas molecules have the same average kinetic energy, \tfrac{1}{2}mv^{2}. So the lighter \ce{^{235}UF6} molecules move slightly faster and pass through the tiny holes of a porous barrier slightly more often than the heavier \ce{^{238}UF6} molecules. The gas that has passed through the barrier is slightly richer in U-235.

\dfrac{v_{235}}{v_{238}}=\sqrt{\dfrac{M_{238}}{M_{235}}}=\sqrt{\dfrac{352}{349}}=1.0043

Each stage enriches the gas only a little, so the process is repeated through hundreds or thousands of barriers in a row (a cascade) until the desired level of enrichment is reached.

Figure 22.18 (redrawn): gaseous diffusion. Lighter uranium-235 hexafluoride molecules pass through the porous barrier slightly more often.

Gas centrifuge

This is the most common method today. \ce{UF6} gas is fed into a tall cylinder (rotor) that a motor spins at very high speed. The heavier \ce{^{238}UF6} molecules are thrown towards the wall, while the lighter \ce{^{235}UF6} molecules stay nearer the axis. Slightly enriched gas is drawn off near the axis and depleted gas near the wall. Many centrifuges are connected in cascades.

Figure 22.19 (redrawn): gas centrifuge.

Laser enrichment

A laser tuned to a frequency that only uranium-235 atoms (or molecules) absorb excites or ionises them selectively, so that they can be separated from uranium-238.

22.7Energy in annihilation reactionsOften examined

Every particle has an antiparticle with the same mass but opposite charge. The antiparticle of the electron is the positron. When a particle meets its antiparticle, both are destroyed and their mass is converted completely into energy, usually as gamma ray photons. This is annihilation. It is the basis of PET scans in medical imaging.

Antiparticle: A particle with the same mass as a given particle but opposite charge (and other opposite properties), for example the positron e^{+} for the electron e^{-} and the antiproton for the proton.
Annihilation: The process in which a particle collides with its antiparticle and both are destroyed, their total energy (rest mass energy plus kinetic energy) appearing as photons or other particles.

\ce{e- + e+ -> \gamma + \gamma}\qquad(22.2)

Electron-positron annihilation: two gamma photons of 0.511\,\mathrm{MeV} move off in opposite directions.

Conservation laws in annihilation

Charge: the total charge before is (-e)+(+e)=0, and photons carry no charge.
Mass-energy: the total energy before (rest mass energy 2m_0c^{2} plus any kinetic energy) equals the total energy of the photons, as Einstein's equation E=mc^{2} describes.
Momentum: when the electron and positron are almost at rest, the total momentum is zero. A single photon always carries momentum (p=E/c), so one photon alone cannot be produced. Two photons of equal energy moving in opposite directions have zero total momentum, so momentum is conserved.

Derivation Energy of each photon in electron-positron annihilation

Aim. Find the energy of each gamma photon produced when an electron and a positron annihilate at rest.

Assumptions.

The electron and the positron are at rest (their kinetic energy is negligible).
Rest mass of each particle: m_0=9.1\times10^{-31}\,\mathrm{kg}.

Energy is conserved: the rest mass energy of the two particles becomes the energy of the two photons.
m_0c^{2}+m_0c^{2}=2hf
Rest mass energy of one electron (the positron has the same mass).
m_0c^{2}=(9.1\times10^{-31})(3\times10^{8})^{2}=8.19\times10^{-14}\,\mathrm{J}
Convert to electronvolts.
m_0c^{2}=\dfrac{8.19\times10^{-14}}{1.6\times10^{-19}}\,\mathrm{eV}=5.12\times10^{5}\,\mathrm{eV}\approx0.51\,\mathrm{MeV}
Each photon carries hf=m_0c^{2}; the total is twice this.
hf=0.51\,\mathrm{MeV},\qquad 2hf=1.02\,\mathrm{MeV}

Resulthf=m_0c^{2}\approx0.511\,\mathrm{MeV}\ \text{per photon},\qquad E_{\text{total}}\approx1.02\,\mathrm{MeV}

1.02\,\mathrm{MeV} is the minimum energy produced by electron-positron annihilation, and each photon has 0.511\,\mathrm{MeV} (511\,\mathrm{keV}), the energy that PET detectors look for. Its wavelength is \lambda=hc/E=2.4\times10^{-12}\,\mathrm{m}, in the gamma ray region.

If the electron and positron are moving, their kinetic energies are also converted, and the photons carry more energy:

K.E_{\text{electron}}+K.E_{\text{positron}}+2m_0c^{2}=2hf\qquad(22.3)

Example 22.8: annihilation of moving particles

Find the amount of energy released by the annihilation of an electron and a positron initially moving with kinetic energy of 0.12\,\mathrm{MeV} each. The rest mass energy of the electron and the positron is 0.51\,\mathrm{MeV}.

Given

m_0c^{2} = 0.51\,\mathrm{MeV} each particle
K.E_{e^-}=K.E_{e^+} = 0.12\,\mathrm{MeV}

To find

energy of each photon hf

Use equation (22.3) (the book calls it equation 22.15, a misprint).
K.E_{e^-}+K.E_{e^+}+2m_0c^{2}=2hf
Substitute.
0.12+0.12+2(0.51)=2hf\;\Rightarrow\;1.26\,\mathrm{MeV}=2hf
Divide by 2.
hf=0.63\,\mathrm{MeV}

hf=0.63\,\mathrm{MeV}\ \text{per photon}\ (1.26\,\mathrm{MeV}\ \text{in total})

Proton-antiproton annihilation

A proton and an antiproton can also annihilate. Each is about 1836 times heavier than an electron, so much more energy is released. They produce gamma photons, or other particles such as pions:

\ce{p+ + p- -> \gamma + \gamma}\qquad\text{or}\qquad\ce{p+ + p- -> \pi+ + \pi- + \pi^0}

Assignment 22.8: proton-antiproton annihilation

Find the energy of the gamma ray photons released by the reaction \ce{p+ + p- -> \gamma + \gamma}, where the mass of the proton and of the antiproton is 1.007276\,\mathrm{u}.

Rest mass energy of one proton, with 1\,\mathrm{u}=931.5\,\mathrm{MeV}.
m_pc^{2}=1.007276\times931.5=938.3\,\mathrm{MeV}
Energy conservation with the particles at rest: the two rest energies become two photons of equal energy.
2m_pc^{2}=2hf\;\Rightarrow\;hf=m_pc^{2}=938.3\,\mathrm{MeV}
In joules.
hf=938.3\times1.6\times10^{-13}=1.50\times10^{-10}\,\mathrm{J}

hf\approx938\,\mathrm{MeV}\approx1.50\times10^{-10}\,\mathrm{J}\ \text{per photon}

22.8.1Medical uses of radiation and medical tracersOften examined

Radiation has wide uses in medical diagnosis (finding what is wrong) and in treatment. The most common are medical X-rays, gamma ray therapy, medical tracers and PET scanning.

Medical tracer: A substance containing radioactive nuclei that can be introduced into the body and is then absorbed by the tissue being studied, so that its radiation shows how that tissue or organ is working.

The tracer is introduced by injection into the blood stream, by swallowing (ingestion) or by breathing in (inhalation), depending on the study. It collects in particular tissues because of its chemistry; for example, iodine collects in the thyroid gland. The radioactive nuclei emit gamma rays, which pass out of the body and are detected by imaging equipment such as a gamma camera, a PET scanner or a SPECT (single photon emission computed tomography) scanner. The images help doctors diagnose and monitor cancer, heart disease and neurological disorders (Figure 22.20).

Properties of a good tracer

It emits gamma rays (or positrons, for PET), which leave the body and can be detected. Alpha emitters are not used for imaging: alpha particles cannot leave the body and they damage tissue badly.
Its half-life is short (minutes to days): long enough to finish the scan, short enough to keep the patient's radiation dose low.
It is chemically suitable to be absorbed by the organ being studied.
It is not poisonous and decays into a harmless product.

Table 22.4: medical tracers and their uses.
Tracer	Use	Half-life
Technetium-99 (as Tc-99m)	bone scans, cardiac stress tests and thyroid	6 hours
Fluorodeoxyglucose (fluorine-18)	metabolic activity and cancerous tumours	110 minutes
Iodine-131	therapy for hyperthyroidism and thyroid cancer	8 days
Iodine-123	thyroid gland, brain and heart scans	13 hours
Gallium-67	infections, tumours, lymphoma and lung cancer	78 hours
Thallium-201	myocardial perfusion imaging, heart function and coronary artery disease	73 hours
Radium-223	bone metastases and prostate cancer; reduces cancer spread in bone	11.4 days
Carbon-11	neurodegenerative diseases and some cancers	20 minutes
Oxygen-15	blood flow and metabolism in tissues	2 minutes

Radiation in treatment (radiotherapy)

High-energy gamma rays from cobalt-60 (or X-rays from an accelerator) are aimed at a tumour from several directions, so that the tumour receives a large dose while the healthy tissue around it receives less. Radioactive isotopes can also be put inside the body: iodine-131 treats thyroid cancer and hyperthyroidism, and beta emitters such as lutetium-177 and yttrium-90 destroy cancer cells (radionuclide therapy). Gamma rays are also used to sterilise medical instruments.

22.8.2PET scannerOften examined

Positron emission tomography (PET) is a powerful imaging technique that shows metabolic and physiological processes inside the body, for example which tissues are using the most glucose.

How a PET scan works

(i) The patient is injected with a tracer containing a positron-emitting radionuclide, such as carbon-11 or fluorine-18 (in fluorodeoxyglucose). The tracer travels through the blood stream and collects in the tissues or organs being studied.
(ii) The radionuclide undergoes beta-plus decay and emits a positron, the antiparticle of the electron (same mass, opposite charge).
\ce{^{11}_{6}C -> ^{11}_{5}B + ^{0}_{+1}e} + \nu
(iii) The positron travels a short distance (about a millimetre) in the tissue and meets an electron.
(iv) The electron and the positron annihilate, producing a pair of gamma ray photons (Figure 22.21).
(v) The two photons have 511\,\mathrm{keV} each and travel in opposite directions, 180° apart, to conserve momentum. They pass out of the body.
(vi) The patient lies inside a ring of gamma ray detectors (Figure 22.22). When the photons strike the detectors they are converted into electrical signals.
(vii) The scanner registers two photons that arrive at opposite detectors at almost the same instant (in coincidence, within a few nanoseconds). The annihilation must have happened on the straight line joining those two detectors. The computer records the arrival times and positions of many such pairs and builds up a 3-D image of the tracer concentration (Figure 22.23).

Figures 22.21 and 22.22 (redrawn): the two annihilation photons travel in opposite directions and strike opposite detectors in the ring.

Locating the annihilation point from the arrival times

If the annihilation happens exactly midway between the two detectors, both photons arrive together. If it happens a distance x from the midpoint, one photon travels 2x further than the other and arrives later by \Delta t=2x/c. Measuring \Delta t therefore locates the point along the line:

\begin{aligned}x&=\dfrac{c\,\Delta t}{2}\\ \text{e.g. }\Delta t&=0.40\,\mathrm{ns}:\\ x&=\dfrac{(3\times10^{8})(0.40\times10^{-9})}{2}=0.06\,\mathrm{m}=6\,\mathrm{cm}\end{aligned}

Tumours use glucose faster than normal tissue, so fluorodeoxyglucose collects in them and they show up as bright regions in the image.

Uses and hazards of radiation

Radioactivity and nuclear radiation have many uses beyond medicine, but ionising radiation can also damage living cells. Safe use depends on knowing both.

Uses

Medicine: diagnosis with tracers and PET, treatment of cancer with cobalt-60 gamma rays or with radionuclides, and sterilisation of instruments.
Industry: thickness gauges (the amount of beta or gamma radiation passing through a sheet shows its thickness; cesium-137 is used in such devices), gamma radiography to find flaws in welds and castings (non-destructive testing).
Dating: carbon-14 dates once-living material up to about 50000 years old.
Energy: nuclear power stations generate electricity from fission.

Hazards

Radiation ionises the atoms of living cells. It can break DNA molecules and kill or damage cells. Large doses cause radiation burns and radiation sickness (nausea, hair loss, damage to blood-forming organs). Smaller doses increase the risk of cancer years later (somatic effects). Damage to reproductive cells can cause mutations that are passed on to children (genetic effects).

Outside the body, gamma rays and beta particles are the main danger because they can reach the organs. Inside the body (if a source is swallowed or breathed in), alpha emitters are the most dangerous, because all their energy is deposited in a small volume of tissue.

Safety precautions

Nuclear energy: benefits and drawbacks

Nuclear power compared with other energy sources.
Benefits	Drawbacks
A very large amount of energy from a small mass of fuel (1\,\mathrm{kg} of U-235 gives about 2.5 million times the energy of 1\,\mathrm{kg} of coal).	Radioactive waste stays dangerous for thousands of years and must be stored safely.
No carbon dioxide or smoke during operation, so it adds little to global warming and air pollution.	Risk of serious accidents that spread radioactive material (Chernobyl 1986, Fukushima 2011).
Steady, reliable electricity day and night, independent of the weather.	High cost and long time to build; expensive to dismantle safely at the end of its life.
Fuel is easy to transport and store, and reserves last a long time.	Uranium is not renewable, enrichment technology can be misused for weapons, and waste heat warms rivers and seas (thermal pollution).

Formula sheet

Mass number

A=Z+N

Z: atomic number (protons), N: neutron number

no unit (a count)

Atomic mass unit

1\,\mathrm{u}=1.66\times10^{-27}\,\mathrm{kg}\;\equiv\;931.5\,\mathrm{MeV}

one twelfth of the mass of a carbon-12 atom

Units: \mathrm{kg} or \mathrm{MeV}

Electronvolt conversion

1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J},\qquad 1\,\mathrm{MeV}=1.6\times10^{-13}\,\mathrm{J}

energy gained by an electron through 1\,\mathrm{V}

Unit: \mathrm{J}

Mass defect (22.1)

\Delta m=\left[(A-Z)\,m_n+Z\,m_p\right]-m_{\text{nucleus}}

m_p=1.007276\,\mathrm{u}, m_n=1.008665\,\mathrm{u}, m_{\text{nucleus}}: mass of the nucleus (atomic mass minus Z m_e)

Units: \mathrm{u} or \mathrm{kg}

Binding energy (22.2)

E_b=\Delta m\,c^{2}

\Delta m: mass defect in kg, c=3\times10^{8}\,\mathrm{m\,s^{-1}}

Unit: \mathrm{J}

Binding energy in MeV

E_b\,(\mathrm{MeV})=\Delta m\,(\mathrm{u})\times931.5

\Delta m in atomic mass units

Unit: \mathrm{MeV}

Binding energy per nucleon (22.3)

f=\dfrac{E_b}{A}=\dfrac{\left[(A-Z)m_n+Zm_p-m_{\text{nucleus}}\right]c^{2}}{A}

called packing fraction in the book; A: mass number

Unit: \mathrm{MeV} per nucleon

Alpha decay

\ce{^{A}_{Z}X -> ^{A-4}_{Z-2}Y + ^{4}_{2}He} + Q

Z falls by 2, A falls by 4

Beta-minus decay

\ce{^{A}_{Z}X -> ^{A}_{Z+1}Y + ^{0}_{-1}e} + \bar{\nu}

a neutron becomes a proton; Z rises by 1

Beta-plus decay

\ce{^{A}_{Z}X -> ^{A}_{Z-1}Y + ^{0}_{+1}e} + \nu

a proton becomes a neutron; Z falls by 1

Gamma decay

\ce{^{A}_{Z}X^{*} -> ^{A}_{Z}X + \gamma},\qquad hf=E_2-E_1

X^{*}: excited nucleus; Z and A unchanged

Q value of a decay or reaction

Q=\left(m_{\text{before}}-m_{\text{after}}\right)c^{2}=\Delta m\,(\mathrm{u})\times931.5\,\mathrm{MeV}

Q>0: energy released (exothermic); Q<0: energy absorbed

Units: \mathrm{MeV} or \mathrm{J}

Rate law

\dfrac{\Delta N}{\Delta t}=-\lambda N

N: undecayed nuclei, \lambda: decay constant; minus sign: N decreases

Unit: nuclei per second

Decay constant

\lambda=\dfrac{1}{N}\times\dfrac{\Delta N}{\Delta t}

fraction of nuclei decaying per unit time

Unit: \mathrm{s^{-1}}

Activity

A=\lambda N

A: decays per second

Units: becquerel, 1\,\mathrm{Bq}=1\,\mathrm{s^{-1}}

Curie

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

older unit of activity

Unit: \mathrm{Ci}

Exponential decay law

N=N_0\,e^{-\lambda t},\qquad x=x_0\,e^{-\lambda t}

x: undecayed nuclei, activity, mass of isotope or count rate; N_0: value at t=0

Unit: same as x

Decay in half-lives

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

n: number of half-lives

Half-life and decay constant

T_{1/2}=\dfrac{\ln2}{\lambda}=\dfrac{0.693}{\lambda}

T_{1/2}: half-life

Unit: \mathrm{s} (if \lambda in \mathrm{s^{-1}})

Straight-line form

\ln N=\ln N_0-\lambda t

graph of \ln N against t: gradient -\lambda, intercept \ln N_0

Annihilation at rest

2m_0c^{2}=2hf\;\Rightarrow\;hf=m_0c^{2}=0.511\,\mathrm{MeV}

m_0=9.1\times10^{-31}\,\mathrm{kg}: rest mass of electron or positron

Units: \mathrm{MeV} or \mathrm{J}

Annihilation of moving particles (22.3)

K.E_{\text{electron}}+K.E_{\text{positron}}+2m_0c^{2}=2hf

hf: energy of each of the two photons

Unit: \mathrm{MeV}

Gaseous diffusion

\dfrac{v_{235}}{v_{238}}=\sqrt{\dfrac{352}{349}}=1.0043

speeds of \ce{^{235}UF6} and \ce{^{238}UF6} molecules at the same temperature

no unit

PET: position along the line

x=\dfrac{c\,\Delta t}{2}

\Delta t: difference in arrival times of the two photons, x: distance from the midpoint

Unit: \mathrm{m}

Key terms

Nucleon: A proton or a neutron.
Atomic number Z: The number of protons in a nucleus.
Mass number A: The number of nucleons (protons plus neutrons) in a nucleus.
Isotopes: Nuclides of the same element with different numbers of neutrons.
Atomic mass unit (u): One twelfth of the mass of a carbon-12 atom, 1.66\times10^{-27}\,\mathrm{kg}, equivalent to 931.5\,\mathrm{MeV}.
Mass defect: The difference between the sum of the masses of the constituent nucleons and the mass of the nucleus.
Binding energy: The minimum energy required to break a nucleus into its separate nucleons, E_b=\Delta mc^{2}.
Binding energy per nucleon: Total binding energy divided by the mass number; a measure of stability (called packing fraction in the book).
Radioactivity: The natural (spontaneous) emission of radiation from unstable nuclei.
Parent and daughter nucleus: The nucleus that decays, and the nucleus formed by the decay.
Alpha decay: Emission of a helium-4 nucleus; Z decreases by 2 and A by 4.
Beta decay: Emission of an electron (\beta^-) or a positron (\beta^+) from a nucleus; A is unchanged and Z changes by 1.
Gamma decay: Emission of a high-energy photon by an excited nucleus; Z and A are unchanged.
Electron capture: A nucleus captures an inner electron, which combines with a proton to form a neutron; Z decreases by 1.
Positron: The antiparticle of the electron: same mass, charge +e.
Neutrino and antineutrino: Uncharged, almost massless particles emitted in beta decay.
Count rate: The number of decays detected per unit time.
Background radiation: Low-level radiation always present, from rocks, radon, cosmic rays, food and medical sources.
Half-life: The time taken for half of the radioactive nuclei in a sample to decay.
Rate law: The rate of decay is directly proportional to the number of undecayed nuclei present.
Decay constant: The fraction of radioactive nuclei that decay per unit time; unit \mathrm{s^{-1}}.
Activity: The number of disintegrations per unit time, A=\lambda N.
Becquerel: SI unit of activity: one disintegration per second.
Curie: Unit of activity equal to 3.70\times10^{10} disintegrations per second.
Nuclear reaction: A process in which nuclei interact and are changed, often releasing or absorbing energy.
Q value: The energy released (Q>0) or absorbed (Q<0) in a nuclear reaction, equal to the change in rest-mass energy.
Thermal neutron: A slow neutron with kinetic energy of about 0.025\,\mathrm{eV}.
Nuclear fusion: Joining of light nuclei to form a heavier nucleus, releasing a large amount of energy.
Nuclear fission: Splitting of a heavy nucleus into two smaller nuclei, releasing neutrons and a large amount of energy.
Fission chain reaction: A succession of fissions in which at least one neutron from each fission causes a further fission.
Critical mass: The minimum mass of fissile material that can sustain a chain reaction.
Critical volume: The volume occupied by the critical mass of a material.
Nuclear reactor: A device used to initiate and control a fission chain reaction.
Control rods: Rods of boron or cadmium that absorb neutrons to control the rate of fission (also called safety rods in the book).
Moderator: A material (water or graphite) that slows fast neutrons down to thermal energies.
Coolant: A fluid (water, carbon dioxide or helium) that carries heat away from the core to the heat exchanger.
Shielding: A thick layer of concrete and steel around the reactor that stops radiation escaping.
Enrichment of uranium: Increasing the percentage of fissile U-235 in uranium (from 0.7% to 3 to 5%) by separating it from U-238.
Gaseous diffusion: Enrichment method in which \ce{UF6} gas is passed through porous barriers; lighter \ce{^{235}UF6} passes slightly faster.
Gas centrifuge: Enrichment method using rapidly spinning cylinders that push heavier \ce{^{238}UF6} towards the wall.
Antiparticle: A particle with the same mass as another particle but opposite charge.
Annihilation: The destruction of a particle and its antiparticle, their energy appearing as photons.
Medical tracer: A substance containing radioactive nuclei that is introduced into the body and absorbed by the tissue being studied.
PET scanner: Positron emission tomography: imaging that detects pairs of 511\,\mathrm{keV} annihilation photons to map a tracer in the body.

Multiple choice questionsp. 168

11 mark

The process by which a heavy nucleus splits into two smaller nuclei is called:

AFissionAnswer
BFusion
CAlpha decay
DBeta decay

Why

Fission is the splitting of a heavy nucleus, such as uranium-235, into two smaller nuclei of comparable size. Fusion is the opposite (light nuclei join), and alpha or beta decay emits only a small particle, not two comparable nuclei.

21 mark

Radiation with highest ionizing power is:

AGamma rays
BX-rays
CAlpha particleAnswer
DBeta particle

Why

Alpha particles have the largest charge (+2e) and mass and move slowly, so they ionise about 10^{4} atoms per cm of air, far more than beta particles. Gamma rays and X-rays are uncharged photons and ionise the least.

31 mark

Which of the following particles has almost the same mass as a proton but carries no charge:

ANeutronAnswer
BProton
CElectron
DBeta particle

Why

The neutron has mass 1.008665\,\mathrm{u}, almost equal to the proton's 1.007276\,\mathrm{u}, and it is uncharged. The electron and the beta particle are about 1840 times lighter and are charged.

41 mark

The charge on an alpha particle is:

A-1
B-2
C+1
D+2Answer

Why

An alpha particle is a helium nucleus containing 2 protons and 2 neutrons, so its charge is +2e.

51 mark

In the nucleus of uranium-235, the number of neutrons is:

A92
B143Answer
C235
D134

Why

Uranium has Z=92, so N=A-Z=235-92=143. The value 92 is the number of protons and 235 is the total number of nucleons.

61 mark

The half-life of radium is 1590 years. In how many years shall the Earth lose all its radium due to radioactive decay?

A795
B1590
C3180
DInfiniteAnswer

Why

In every half-life only half of the remaining nuclei decay, so the amount left is N_0(1/2)^{n}, which is never exactly zero. In principle it takes an infinite time for all the radium to decay. (1590 years is only the time for half of it to decay, and 3180 years leaves one quarter.)

71 mark

The energy released in nuclear reactor is produced by:

AFissionAnswer
BFusion
CCoal
DGas

Why

Nuclear reactors release energy by the controlled fission of uranium-235. Fusion has not yet been achieved in a power reactor, and coal and gas release chemical energy in ordinary power stations.

81 mark

C-14 has a half-life 5730 years. The number of nuclei in a sample will drop to 1/8 of initial quantity in ______ years:

A1.44\times10^{4}
B1.72\times10^{4}Answer
C2.58\times10^{4}
D2.85\times10^{4}

Why

\dfrac{1}{8}=\left(\dfrac{1}{2}\right)^{3}, so three half-lives are needed: t=3\times5730=17190\,\text{years}\approx1.72\times10^{4}\,\text{years}.

Short questionsp. 168–169

Each answer is written for 3 marks: the key statement first, then the reason, then an equation or example.

13 marks

What are some of the potential benefits and drawbacks of using nuclear energy as a source of electricity compared to other forms of energy?

Solution

A nuclear power station produces electricity from the heat released by the controlled fission of uranium-235.

Benefits:

A huge amount of energy from very little fuel: one fission gives about 200\,\mathrm{MeV}, so 1\,\mathrm{kg} of U-235 gives about 2.5 million times the energy of 1\,\mathrm{kg} of coal.
No carbon dioxide, smoke or acid gases during operation, so little contribution to global warming and air pollution.
Steady, reliable power day and night, independent of the weather (unlike solar and wind).

Drawbacks:

Radioactive waste that stays dangerous for thousands of years and needs safe long-term storage.
Risk of serious accidents that release radiation (Chernobyl, Fukushima), and possible misuse of the technology for weapons.
High cost and long construction time; uranium is non-renewable.

23 marks

What happens to the atomic number and mass number of nucleus that (a) emits electron (b) undergoes electron capture (c) emits \alpha particle?

Solution

(a) emits an electron

This is beta-minus decay. A neutron in the nucleus changes into a proton, emitting an electron and an antineutrino. The atomic number increases by 1; the mass number does not change, because the number of nucleons stays the same.

\ce{^{A}_{Z}X -> ^{A}_{Z+1}Y + ^{0}_{-1}e} + \bar{\nu}\qquad\text{e.g.}\quad\ce{^{14}_{6}C -> ^{14}_{7}N + ^{0}_{-1}e} + \bar{\nu}

(b) undergoes electron capture

The nucleus captures one of its own inner (K-shell) electrons, which combines with a proton to form a neutron, and a neutrino is emitted. The atomic number decreases by 1; the mass number does not change.

\ce{^{A}_{Z}X + ^{0}_{-1}e -> ^{A}_{Z-1}Y} + \nu\qquad\text{e.g.}\quad\ce{^{7}_{4}Be + ^{0}_{-1}e -> ^{7}_{3}Li} + \nu

The nucleus loses 2 protons and 2 neutrons (a helium-4 nucleus). The atomic number decreases by 2 and the mass number decreases by 4.

\ce{^{A}_{Z}X -> ^{A-4}_{Z-2}Y + ^{4}_{2}He}\qquad\text{e.g.}\quad\ce{^{238}_{92}U -> ^{234}_{90}Th + ^{4}_{2}He}

33 marks

Why does the low energy alpha particle not make physical contact with the nucleus when an alpha particle is headed directly towards the nucleus of an atom?

Solution

The alpha particle (+2e) and the nucleus (+Ze) are both positive, so they repel each other with a Coulomb force that grows rapidly as the distance decreases (F\propto1/r^{2}). As the alpha particle approaches, its kinetic energy is converted into electric potential energy. It stops at the distance of closest approach r_0, where all its kinetic energy has become potential energy, and is then pushed back along its path.

K.E=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{(2e)(Ze)}{r_0}\qquad\Longrightarrow\qquad r_0=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{2Ze^{2}}{K.E}

For a low-energy alpha particle r_0 is much larger than the radius of the nucleus, so it never comes within the short range of the attractive strong nuclear force (about 10^{-15}\,\mathrm{m}). For example, a 5\,\mathrm{MeV} alpha particle heading for a gold nucleus (Z=79) turns back at r_0\approx4.6\times10^{-14}\,\mathrm{m}, about six times the radius of the gold nucleus.

43 marks

An alpha particle has twice the charge of a beta particle. Why does the former deflect less than the latter when passing between electrically charged plates, assuming they both have the same speed?

Solution

Deflection depends on the acceleration, not on the charge alone. Between the plates the electric force is F=qE, so the sideways acceleration is a=qE/m. In the time t=L/v spent between plates of length L, the sideways deflection is:

y=\dfrac{1}{2}at^{2}=\dfrac{1}{2}\left(\dfrac{q}{m}\right)E\,\dfrac{L^{2}}{v^{2}}\;\propto\;\dfrac{q}{m}\qquad(\text{same }v)

The alpha particle has twice the charge but about 7300 times the mass of a beta particle, so its charge-to-mass ratio is about 3600 times smaller:

\begin{aligned}\dfrac{q_\alpha}{m_\alpha}&=\dfrac{2(1.6\times10^{-19})}{6.64\times10^{-27}}=4.8\times10^{7}\,\mathrm{C\,kg^{-1}}\\ \dfrac{q_\beta}{m_\beta}&=\dfrac{1.6\times10^{-19}}{9.11\times10^{-31}}=1.76\times10^{11}\,\mathrm{C\,kg^{-1}}\end{aligned}

Its greater inertia means the same kind of force gives it a far smaller acceleration, so the alpha particle is deflected much less (and in the opposite direction, because its charge is positive).

Alpha bends slightly towards the negative plate; beta bends strongly towards the positive plate; gamma is not deflected.

52026 Annual I, Set B, Q.6 OR (a)3 marks

Why uranium fuel needs to be enriched before use?

Solution

Natural uranium contains only about 0.7% uranium-235, the isotope that undergoes fission with slow neutrons. The other 99.3% is uranium-238, which is not fissile and mostly absorbs neutrons without splitting. In a reactor moderated by ordinary water, so many neutrons are absorbed by U-238 and by the water that fewer than one neutron per fission causes another fission, and the chain reaction cannot be sustained.

Enrichment raises the proportion of U-235 to about 3 to 5%. Then enough neutrons cause fission to keep the chain reaction going steadily (the reactor can reach criticality), the reactor works efficiently, the core can be smaller and the fuel lasts longer.

63 marks

If U-238 undergoes alpha decay, what is the resulting nucleus?

Solution

In alpha decay the nucleus emits a helium nucleus (\ce{^{4}_{2}He}), so its atomic number falls by 2 and its mass number falls by 4. For uranium-238 (Z=92, A=238): the new atomic number is 92-2=90 and the new mass number is 238-4=234. The element with Z=90 is thorium.

\ce{^{238}_{92}U -> ^{234}_{90}Th + ^{4}_{2}He} + Q

The resulting nucleus is thorium-234, with 90 protons and 234-90=144 neutrons. It is itself radioactive (a beta emitter). The energy released in this decay is about 4.3\,\mathrm{MeV}, carried mainly by the alpha particle.

\ce{^{234}_{90}Th}\ \text{(thorium-234)}

73 marks

How do chain reactions occur in nuclear fission?

Solution

When a slow neutron is absorbed by a uranium-235 nucleus, it forms unstable U-236, which splits into two medium-sized nuclei and releases 2 or 3 fast neutrons and about 200\,\mathrm{MeV} of energy:

\ce{^{1}_{0}n + ^{235}_{92}U -> ^{141}_{56}Ba + ^{92}_{36}Kr + 3^{1}_{0}n} + Q

These neutrons, once slowed down, can strike other U-235 nuclei and cause further fissions, each releasing more neutrons. The number of fissions multiplies (1, 3, 9, 27, ...). This self-sustaining series of fissions, in which at least one neutron from each fission causes another fission, is the chain reaction. It continues only if the fissile material is at least the critical mass; in a smaller piece too many neutrons escape.

Neutrons from each fission cause further fissions.

83 marks

How does the molecular weight difference between U-235 and U-238 hexafluoride molecules enable enrichment? What role do porous barriers play in the gaseous diffusion process?

Solution

Uranium is converted into uranium hexafluoride gas. The two kinds of molecule have molecular masses 349 (\ce{^{235}UF6}) and 352 (\ce{^{238}UF6}). At the same temperature both have the same average kinetic energy \tfrac{1}{2}mv^{2}, so the lighter molecules move slightly faster:

\dfrac{v_{235}}{v_{238}}=\sqrt{\dfrac{352}{349}}=1.0043

Role of porous barriers: a porous barrier contains millions of tiny holes. The faster, lighter \ce{^{235}UF6} molecules reach the holes and pass through slightly more often than the heavier ones, so the gas that passes through is slightly richer in U-235 and the gas left behind is depleted. The separation per barrier is tiny (about 0.4%), so the gas is passed through a cascade of hundreds or thousands of barriers to reach 3 to 5% U-235.

Comprehensive questionsp. 169

Full answers with headings that follow a typical 6 to 8 mark split: definition, explanation, equation or derivation, diagram and applications.

12026 Annual I, Set B, Q.2 (xi) OR6 marks

Explain the terms mass defect and binding energy.

Solution

Mass defect

The mass of a nucleus is always less than the total mass of the separate protons and neutrons that make it. This difference is the mass defect.

Mass defect: The difference between the sum of the masses of the constituent nucleons of a nucleus and the actual mass of the nucleus.

\Delta m=\left[Z\,m_p+(A-Z)\,m_n\right]-m_{\text{nucleus}}

Z: number of protons, A-Z: number of neutrons, m_p=1.007276\,\mathrm{u}, m_n=1.008665\,\mathrm{u}.

Mass defect of deuterium: the bound atom has 0.002388\,\mathrm{u} less mass than its separate particles.

Reason. When nucleons come together, the strong nuclear force binds them and energy is released. According to Einstein, energy has mass (E=mc^{2}), so the energy that leaves carries away mass. The bound nucleus is therefore lighter than its parts.

Binding energy

Binding energy: The minimum energy required to break a nucleus into its separate protons and neutrons. It equals the energy released when the nucleus is formed from them.

The binding energy is the energy equivalent of the mass defect:

E_b=\Delta m\,c^{2}

\Delta m in kg gives E_b in joules. With \Delta m in u: E_b(\mathrm{MeV})=\Delta m(\mathrm{u})\times931.5, because 1\,\mathrm{u}\equiv931.5\,\mathrm{MeV}.

Example: helium-4

The helium-4 nucleus (alpha particle, 4.001506\,\mathrm{u}) has 2 protons and 2 neutrons.
\Delta m=2(1.007276)+2(1.008665)-4.001506=0.030376\,\mathrm{u}
Binding energy.
E_b=0.030376\times931.5=28.3\,\mathrm{MeV}=4.5\times10^{-12}\,\mathrm{J}
Binding energy per nucleon.
\dfrac{E_b}{A}=\dfrac{28.3}{4}=7.07\,\mathrm{MeV}

Binding energy per nucleon and stability

Dividing the binding energy by the number of nucleons gives the binding energy per nucleon, the average energy needed to remove one nucleon. The larger it is, the more stable the nucleus. It is highest (about 8.8\,\mathrm{MeV}) for iron-56, the most stable nucleus, and lower for very light and very heavy nuclei. That is why energy is released when light nuclei fuse or heavy nuclei split: both processes form more tightly bound nuclei, so mass is lost and energy is released.

26 marks

What is radioactivity? Explain.

Solution

Definition

Radioactivity: The natural (spontaneous) emission of radiation from unstable nuclei.

It was discovered by Henri Becquerel in 1896 in uranium salts and studied by Marie Curie, who found radioactivity in uranium and thorium and showed that the radiation makes the surrounding air conducting. An unstable nucleus (with too many neutrons, too many protons, or simply too large) changes into a more stable one by emitting radiation. The original nucleus is the parent and the new one is the daughter.

Types of radioactive decay

Alpha decay: the nucleus emits a helium nucleus; Z falls by 2 and A by 4.

\ce{^{A}_{Z}X -> ^{A-4}_{Z-2}Y + ^{4}_{2}He}\qquad\text{e.g.}\quad\ce{^{238}_{92}U -> ^{234}_{90}Th + ^{4}_{2}He}

Beta decay: a neutron becomes a proton and emits an electron (\beta^-, Z rises by 1), or a proton becomes a neutron and emits a positron (\beta^+, Z falls by 1). A does not change.

\ce{^{A}_{Z}X -> ^{A}_{Z+1}Y + ^{0}_{-1}e} + \bar{\nu}\qquad\text{e.g.}\quad\ce{^{14}_{6}C -> ^{14}_{7}N + ^{0}_{-1}e} + \bar{\nu}

Gamma decay: an excited nucleus drops to a lower energy state and emits a high-energy photon; Z and A do not change.

\ce{^{60}_{27}Co^{*} -> ^{60}_{27}Co + \gamma}

Properties of the radiations

	Alpha	Beta	Gamma
Charge	+2e	-e or +e	0
Ionising power	highest	medium	lowest
Penetration	stopped by paper	stopped by a few mm of aluminium	reduced by several cm of lead

Penetrating power of alpha, beta and gamma radiation.

Characteristics of radioactivity

It is spontaneous: it happens by itself and is not affected by temperature, pressure, chemical combination or electric and magnetic fields, because it is a property of the nucleus.
It is random: we cannot predict when a particular nucleus will decay. Fluctuations in the count rate show this.
Each isotope decays with its own fixed half-life, and the number of undecayed nuclei falls exponentially, N=N_0e^{-\lambda t}.
Mass number and charge are conserved in every decay, and energy is released (Q>0).
The radiations ionise matter; this is used to detect them, and it makes them harmful to living cells.

36 marks

How can you find energy from a nuclear decay?

Solution

Principle

In every spontaneous decay the total mass of the products (daughter nucleus plus emitted particle) is less than the mass of the parent nucleus. The lost mass \Delta m is converted into energy according to Einstein's mass-energy equation. This energy is called the Q value of the decay.

Q=\Delta m\,c^{2}=\left[m_{\text{parent}}-\left(m_{\text{daughter}}+m_{\text{particle}}\right)\right]c^{2}

Q\,(\mathrm{MeV})=\Delta m\,(\mathrm{u})\times931.5\qquad\text{or}\qquad Q\,(\mathrm{J})=\Delta m\,(\mathrm{kg})\times(3\times10^{8})^{2}

Method

Write the balanced decay equation (check A and Z on both sides).
Add the masses on each side, using atomic masses consistently so that the electrons balance.
Subtract: \Delta m= mass before - mass after.
Convert \Delta m into energy. A positive Q means energy is released; a spontaneous decay always has Q>0.

Example 1: alpha decay of uranium-238

\ce{^{238}_{92}U -> ^{234}_{90}Th + ^{4}_{2}He} + Q

Atomic masses: U-238 =238.05079\,\mathrm{u}, Th-234 =234.04360\,\mathrm{u}, He-4 =4.00260\,\mathrm{u}.
\Delta m=238.05079-(234.04360+4.00260)=0.00459\,\mathrm{u}
Energy released.
Q=0.00459\times931.5=4.28\,\mathrm{MeV}

Example 2: beta decay of carbon-14

With atomic masses the electron of the beta particle is already included in the nitrogen atom, so only the two atomic masses are needed.
\Delta m=M(\ce{^{14}C})-M(\ce{^{14}N})=14.003242-14.003074=0.000168\,\mathrm{u}
Energy released.
Q=0.000168\times931.5=0.16\,\mathrm{MeV}

Using binding energies

The Q value also equals the gain in total binding energy: Q=E_b(\text{products})-E_b(\text{parent}). For the uranium decay: Q=1777.7+28.3-1801.7=4.3\,\mathrm{MeV}, the same answer.

Where the energy goes

The energy appears mostly as kinetic energy of the emitted particle. Because momentum is conserved, the light alpha particle takes about 98% of Q and the heavy daughter recoils slowly. In beta decay the energy is shared between the beta particle and the antineutrino (or neutrino), so beta particles have a range of energies. If the daughter is left excited, part of the energy is emitted later as gamma rays; in gamma decay the whole energy difference E_2-E_1 is carried by the photon.

46 marks

Describe the importance of radiations in medical field?

Solution

Radiation is used in medicine both to diagnose disease (see inside the body) and to treat it (destroy diseased cells).

Diagnosis

X-rays pass through soft tissue but are absorbed by bone, giving images of bones and teeth; CT scans combine many X-ray images.
Medical tracers: a substance containing radioactive nuclei is injected, swallowed or inhaled and collects in the organ being studied. Its gamma rays are detected outside the body by a gamma camera or SPECT scanner. Examples: technetium-99m (half-life 6 h) for bone scans and cardiac stress tests, iodine-123 for the thyroid, thallium-201 for heart function, gallium-67 for infections and tumours.
PET scanning: a positron-emitting tracer such as fluorine-18 (in fluorodeoxyglucose) or carbon-11 collects in active tissue. Each positron annihilates with an electron to give two 511\,\mathrm{keV} gamma photons in opposite directions, which a ring of detectors records to build a 3-D image. Tumours, which use glucose quickly, show up clearly.

Treatment

Radiotherapy: high-energy gamma rays from cobalt-60 (or X-rays from an accelerator) are aimed at a tumour from several directions, so the tumour gets a large dose and healthy tissue much less.
Internal radionuclide therapy: iodine-131 collects in the thyroid and destroys cancerous or overactive thyroid tissue; beta emitters such as lutetium-177 and yttrium-90 are carried to cancer cells; radium-223 treats cancer that has spread to the bones.
Sterilisation: gamma rays kill bacteria on syringes, dressings and surgical instruments after packing, without heating them.

Choosing a radioisotope

Purpose	Radiation needed	Half-life
Imaging (tracer)	gamma (or positrons for PET), so that it leaves the body	short: minutes to hours
Destroying tissue inside the body	beta or alpha, which deposit energy locally	days
External therapy source	penetrating gamma	long (cobalt-60: 5.27 years)

Because radiation can also damage healthy cells and DNA, doctors use the smallest dose that does the job, choose isotopes with short half-lives, and protect staff with shielding, distance and dosimeters.

56 marks

What is nuclear fusion? Explain in detail. Also discuss some practical applications of this reaction.

Solution

Definition

Nuclear fusion: A process in which two or more light nuclei come close together and join to form a heavier nucleus, releasing a large amount of energy.

Example

\ce{^{2}_{1}H + ^{3}_{1}H -> ^{4}_{2}He + ^{1}_{0}n} + 17.6\,\mathrm{MeV}

Fusion of deuterium and tritium.

Energy released

Mass before the reaction.
2.014102+3.016049=5.030151\,\mathrm{u}
Mass after the reaction.
4.002603+1.008665=5.011268\,\mathrm{u}
Mass lost and energy released.
\Delta m=0.018883\,\mathrm{u},\qquad Q=0.018883\times931.5=17.6\,\mathrm{MeV}

Why energy is released

Light nuclei have a low binding energy per nucleon (1.1\,\mathrm{MeV} for deuterium, 2.8\,\mathrm{MeV} for tritium). The product, helium-4, has 7.07\,\mathrm{MeV} per nucleon, so it is much more tightly bound. The total binding energy increases, the total mass decreases, and the mass difference appears as energy. Fusion is favoured on the rising left part of the binding energy curve, up to iron-56.

Conditions for fusion

Nuclei are positively charged and repel each other strongly. To come within the range of the strong nuclear force (10^{-15}\,\mathrm{m}) they must collide at very high speeds, which needs temperatures of about 10^{7} to 10^{8}\,\mathrm{K}. Fusion is therefore called a thermonuclear reaction. At such temperatures matter is a plasma, which must be dense and confined long enough for many fusions to occur.

Fusion in the Sun and stars

In the core of the Sun (about 1.5\times10^{7}\,\mathrm{K}) hydrogen is converted into helium by the proton-proton chain. Overall, \ce{4^{1}_{1}H -> ^{4}_{2}He + 2^{0}_{+1}e + 2\nu}+26.7\,\mathrm{MeV}. This is the source of sunlight and so of almost all energy on Earth.

Practical applications

Energy of the Sun and stars, on which all life depends.
Hydrogen (thermonuclear) bomb: uncontrolled fusion started by the heat of a fission bomb.
Fusion reactors (research): tokamaks such as ITER hold a deuterium-tritium plasma in strong magnetic fields, aiming to generate electricity. Deuterium is plentiful in sea water, and fusion produces no long-lived fission waste and no carbon dioxide.
Neutron sources: small fusion devices produce neutrons for research and industry.

The main difficulty is producing and confining a plasma above 10^{8}\,\mathrm{K}; so far no fusion reactor gives out more useful energy than it uses.

66 marks

What is nuclear fission? Explain in detail.

Solution

Definition

Nuclear fission: A process in which a heavy nucleus splits into two (or more) smaller nuclei of comparable mass, releasing a few neutrons and a large amount of energy.

Fission was discovered in 1938 by Otto Hahn and Fritz Strassmann, who found barium among the products when uranium was bombarded with neutrons.

Process

A slow neutron is absorbed by a uranium-235 nucleus, forming uranium-236 in a highly excited, unstable state. It oscillates, becomes elongated and splits into two medium-sized fragments, releasing 2 or 3 fast neutrons and about 200\,\mathrm{MeV} of energy:

\ce{^{1}_{0}n + ^{235}_{92}U -> ^{236}_{92}U^{*} -> ^{141}_{56}Ba + ^{92}_{36}Kr + 3^{1}_{0}n} + Q

Fission of uranium-235 by a slow neutron.

The fragments are not always the same. Another example is \ce{^{1}_{0}n + ^{235}_{92}U -> ^{139}_{56}Ba + ^{95}_{36}Kr + 2^{1}_{0}n} + Q. The fragments have too many neutrons, so they are radioactive and decay further.

Why energy is released

Heavy nuclei have a low binding energy per nucleon (about 7.6\,\mathrm{MeV} for uranium) because of the strong electrostatic repulsion between their many protons. The fragments have about 8.5\,\mathrm{MeV} per nucleon. The gain of about 0.9\,\mathrm{MeV} for each of the 236 nucleons gives roughly 200\,\mathrm{MeV} per fission, and the products have correspondingly less mass (Q=\Delta mc^{2}). About 165\,\mathrm{MeV} appears as the kinetic energy of the fragments; the rest is carried by neutrons and gamma rays and released later by the radioactive fragments.

Chain reaction and critical mass

The neutrons released can cause further fissions in other U-235 nuclei, giving a chain reaction. It is self-sustaining only if at least one neutron per fission causes another fission, which needs at least the critical mass of fuel. With exactly one (a controlled chain reaction) energy is released steadily, as in a nuclear reactor, where moderators slow the neutrons and control rods absorb the extra ones. With more than one (uncontrolled) the reaction grows explosively, as in an atomic bomb.

Comparison with chemical energy

One fission gives about 200\,\mathrm{MeV}, while burning one methane molecule gives about 9.2\,\mathrm{eV}. The fission of 1\,\mathrm{kg} of U-235 releases about 8\times10^{13}\,\mathrm{J}, about 2.5 million times the energy from burning 1\,\mathrm{kg} of coal. This is why fission is used in nuclear power plants.

76 marks

Discuss the function of the principle components of water moderated power reactor: such as core, fuel, rods, moderator, control rods, heat exchange, safety rods and shielding.

Solution

A nuclear reactor is a device used to initiate and control a fission chain reaction, so that the energy of fission is released steadily and can be used to generate electricity. In a water-moderated (pressurised water) reactor, ordinary water is both the moderator and the coolant.

A pressurised water reactor.

Core

The heart of the reactor, inside a thick steel pressure vessel. It holds the fuel rods, the moderator and the control rods. The controlled chain reaction happens here and the heat is produced here.

Fuel and fuel rods

The fuel is uranium enriched to about 3% uranium-235 (sometimes plutonium-239), made into small oxide pellets and sealed in long metal tubes about 1\,\mathrm{cm} in diameter. Fission of U-235 in the rods releases the energy, mostly as kinetic energy of the fragments, which becomes heat.

Moderator

Fission gives fast neutrons (a few MeV), but U-235 is split far more easily by slow (thermal) neutrons of about 0.025\,\mathrm{eV}. The moderator (water or graphite) contains light nuclei; the neutrons lose energy in repeated collisions with them and are slowed to thermal speeds, so the chain reaction can continue.

Control rods

Rods of boron or cadmium, which absorb neutrons strongly. They can be moved into or out of the core. Inserting them absorbs more neutrons and slows the reaction; withdrawing them speeds it up. They are set so that exactly one neutron from each fission causes another fission, which keeps the power constant.

Safety rods

Rods of the same neutron-absorbing material, held above the core. In an emergency they drop fully into the core, absorb most of the neutrons and stop the chain reaction at once (the book also uses 'safety rods' as another name for control rods).

Coolant and heat exchanger

Water under high pressure is pumped through the core, carries away the heat and stops the fuel and core from melting. In the heat exchanger (steam generator) this hot water boils water in a separate loop. The steam drives a turbine connected to an electric generator; it is then condensed and pumped back.

Shielding

A thick layer of concrete (and steel) around the reactor absorbs the neutrons and gamma rays and protects the workers and the environment from radiation.

\text{nuclear energy}\to\text{heat}\to\text{steam}\to\text{turbine}\to\text{electricity}

82026 Annual I, Set D, Q.6 OR (a)6 marks

Explain the exponential nature of radioactive decay.

Solution

Meaning

Radioactive decay is random for a single nucleus, but a large sample obeys a simple rule: in equal intervals of time, the same fraction of the remaining nuclei decays. For example, in each half-life half of the remaining nuclei decay: N_0\to N_0/2\to N_0/4\to N_0/8. A quantity that falls by the same fraction in equal times decreases exponentially.

Rate law

The rate of decay is directly proportional to the number of undecayed nuclei present: \Delta N/\Delta t=-\lambda N, where the decay constant \lambda is the fraction of nuclei decaying per second (unit \mathrm{s^{-1}}).

Derivation

Derivation Exponential decay law

Aim. Show that N=N_0e^{-\lambda t}.

Assumptions.

The sample contains a very large number of nuclei, so N changes smoothly.
The decay constant \lambda is constant for the isotope.

Rate law: the rate of decay is proportional to the number of undecayed nuclei; the minus sign shows N decreases.
\dfrac{\Delta N}{\Delta t}=-\lambda N
For a very small interval the ratio becomes a derivative.
\dfrac{dN}{dt}=-\lambda N
Separate the variables.
\dfrac{dN}{N}=-\lambda\,dt
Integrate both sides.
\ln N=-\lambda t+C\qquad(\text{i})
Initial condition: at t=0, N=N_0, so the constant is C=\ln N_0.
\ln N_0=0+C
Substitute C in (i) and combine the logarithms.
\ln N-\ln N_0=-\lambda t\;\Rightarrow\;\ln\dfrac{N}{N_0}=-\lambda t
Take the exponential of both sides.
\dfrac{N}{N_0}=e^{-\lambda t}

ResultN=N_0\,e^{-\lambda t}

The number of undecayed nuclei decreases exponentially with time and never becomes exactly zero.

Graph

Exponential decay curve

The curve falls steeply at first and then more and more slowly, approaching zero but never reaching it. Equal time intervals (T_{1/2}) halve the number each time.

Activity and count rate

Since activity A=\lambda N, it follows the same law, A=A_0e^{-\lambda t}, and so does the count rate of a detector. In general x=x_0e^{-\lambda t}.

Half-life from the law

Putting N=N_0/2 at t=T_{1/2} gives e^{\lambda T_{1/2}}=2, so T_{1/2}=\ln2/\lambda=0.693/\lambda. The half-life is constant, which is exactly what exponential decay means.

Graphical form

Taking logarithms, \ln N=\ln N_0-\lambda t. A graph of \ln N against t is a straight line of gradient -\lambda; this is used to find \lambda and T_{1/2} from experimental readings.

96 marks

Explain annihilation reaction in detail.

Solution

Definition

Every particle has an antiparticle with the same mass but opposite charge; the positron (e^{+}) is the antiparticle of the electron (e^{-}). Annihilation is the process in which a particle and its antiparticle meet and are destroyed, their total energy (rest mass energy plus kinetic energy) appearing as photons or other particles.

\ce{e- + e+ -> \gamma + \gamma}

Electron-positron annihilation produces two gamma photons moving in opposite directions.

Conservation laws

Charge: (-e)+(+e)=0 before; photons are uncharged, so charge is conserved.
Mass-energy: the rest mass energy and kinetic energy of the pair become the energy of the photons, E=mc^{2}.
Momentum: for a pair almost at rest, the total momentum is zero. One photon would carry momentum E/c, so at least two photons are produced, with equal energies and opposite directions, making the total momentum zero.

Energy of the photons

For particles at rest, the rest energy of the pair equals the energy of the two photons.
2m_0c^{2}=2hf\;\Rightarrow\;hf=m_0c^{2}
Rest energy of an electron (m_0=9.1\times10^{-31}\,\mathrm{kg}).
m_0c^{2}=(9.1\times10^{-31})(3\times10^{8})^{2}=8.19\times10^{-14}\,\mathrm{J}=0.51\,\mathrm{MeV}
So each photon has about 0.51\,\mathrm{MeV} and the total is 1.02\,\mathrm{MeV}, the minimum energy of this annihilation.
hf=0.51\,\mathrm{MeV},\qquad 2hf=1.02\,\mathrm{MeV}

If the particles are moving, their kinetic energies add to the photon energies: K.E_{e^-}+K.E_{e^+}+2m_0c^{2}=2hf. For example, with 0.12\,\mathrm{MeV} each, 2hf=0.12+0.12+1.02=1.26\,\mathrm{MeV} and each photon has 0.63\,\mathrm{MeV}.

Other annihilations

A proton and an antiproton annihilate to give gamma photons (\ce{p+ + p- -> \gamma + \gamma}), each of about 938\,\mathrm{MeV}, or particles such as pions (\ce{p+ + p- -> \pi+ + \pi- + \pi^0}).

Applications and the reverse process

Annihilation is the basis of PET scanning: positrons from a tracer annihilate with electrons in the body, and the pairs of 511\,\mathrm{keV} photons are detected to form an image. The reverse process is pair production: a photon of at least 1.02\,\mathrm{MeV} near a nucleus becomes an electron-positron pair. Together they show that mass and energy are inter-convertible.

106 marks

Illustrate how PET scanning works.

Solution

Positron emission tomography (PET) is an imaging technique that shows the metabolic and physiological activity of tissues inside the body. It uses a positron-emitting tracer and detects the gamma photons produced when positrons annihilate with electrons.

Steps

The patient is injected with a tracer containing a positron-emitting radionuclide, such as carbon-11 or fluorine-18 (in fluorodeoxyglucose, FDG). The tracer travels in the blood and collects in the tissues being studied; tumours take up FDG quickly.
The radionuclide decays by beta-plus decay and emits a positron, the antiparticle of the electron: \ce{^{11}_{6}C -> ^{11}_{5}B + ^{0}_{+1}e} + \nu.
The positron travels only about a millimetre in the tissue before it meets an electron.
The electron and positron annihilate, producing two gamma photons.
The two photons, each of 511\,\mathrm{keV}, move in exactly opposite directions (180° apart) to conserve momentum, and leave the body.
A ring of gamma ray detectors around the patient converts the photons that strike it into electrical signals.
The electronics accept only two photons that arrive at opposite detectors at the same time (in coincidence). The computer records their arrival times and positions and builds up a 3-D image of the tracer concentration.

Detectors in a PET scanner: the two photons from one annihilation strike opposite detectors in the ring.

How the position is found

The annihilation must lie on the straight line joining the two detectors that fired together (the line of response). Many thousands of such lines cross at the places where the tracer is concentrated. The difference in arrival times, \Delta t, fixes the point along each line: x=c\Delta t/2 from the midpoint. For example, \Delta t=0.40\,\mathrm{ns} gives x=6\,\mathrm{cm}.

Uses

PET images show how actively tissues use glucose or oxygen, so they reveal cancers and their spread, heart muscle that is not receiving enough blood, and brain disorders. Because carbon-11 and fluorine-18 have short half-lives (20 and 110 minutes), the patient's dose is small, but the tracers must be produced near the hospital.

Numerical problemsp. 169

Data are converted to SI units first. Where the book's printed answer is wrong or cannot be reproduced, the note explains why.

13 marks

If an isotope has a half-life of 10 days and there are initially 5000 nuclei, how many will remain after 30 days?

Solution

Given

T_{1/2} = 10\,\text{days}
N_0 = 5000
t = 30\,\text{days}

To find

undecayed nuclei N after 30 days

Number of half-lives in 30 days. Both times are in days, so no conversion is needed.
n=\dfrac{t}{T_{1/2}}=\dfrac{30}{10}=3
The number halves in each half-life.
N=N_0\left(\dfrac{1}{2}\right)^{n}=5000\left(\dfrac{1}{2}\right)^{3}=\dfrac{5000}{8}
Evaluate.
N=625

N=625\ \text{nuclei}

Book answer. 625

23 marks

If a nucleus of carbon-14 has a decay constant of 0.00012\,\mathrm{s^{-1}}, what is its half-life?

Solution

Given

\lambda = 0.00012\,\mathrm{s^{-1}}=1.2\times10^{-4}\,\mathrm{s^{-1}}

To find

half-life T_{1/2}

Use the relation between half-life and decay constant.
T_{1/2}=\dfrac{0.693}{\lambda}
Substitute.
T_{1/2}=\dfrac{0.693}{1.2\times10^{-4}\,\mathrm{s^{-1}}}=5775\,\mathrm{s}
In hours, for a feel of the size.
T_{1/2}=\dfrac{5775}{3600}\,\mathrm{h}\approx1.6\,\mathrm{h}

T_{1/2}=5775\,\mathrm{s}\approx5.8\times10^{3}\,\mathrm{s}

Book answer. 5775\,\mathrm{s}

Note. The data are not realistic for carbon-14, whose actual half-life is 5730 years (\lambda=3.8\times10^{-12}\,\mathrm{s^{-1}}), but the method is what the question tests.

33 marks

What is energy released when a nucleus of U-235 undergoes fission and releases two neutrons?

Solution

The question gives no masses, so choose a fission of U-235 that releases two neutrons and use accurate atomic masses. The book's own example of such a fission (Section 22.5.3) is:

\ce{^{1}_{0}n + ^{235}_{92}U -> ^{139}_{56}Ba + ^{95}_{36}Kr + 2^{1}_{0}n} + Q

Given

m_U = 235.04393\,\mathrm{u}
m_{Ba} = 138.90884\,\mathrm{u}
m_{Kr} = 94.93971\,\mathrm{u}
m_n = 1.008665\,\mathrm{u}

To find

energy released Q in joules

Total mass before the reaction.
m_n+m_U=1.008665+235.04393=236.052595\,\mathrm{u}
Total mass after the reaction (two neutrons).
m_{Ba}+m_{Kr}+2m_n=138.90884+94.93971+2(1.008665)=235.86588\,\mathrm{u}
Mass lost.
\Delta m=236.052595-235.86588=0.18672\,\mathrm{u}
Energy in MeV.
Q=0.18672\times931.5=173.9\,\mathrm{MeV}
Energy in joules, with 1\,\mathrm{MeV}=1.6\times10^{-13}\,\mathrm{J}.
Q=173.9\times1.6\times10^{-13}=2.78\times10^{-11}\,\mathrm{J}

Q\approx174\,\mathrm{MeV}\approx2.8\times10^{-11}\,\mathrm{J}

Book answer. 1.793\times10^{-11}\,\mathrm{J}

Note. The book prints no data, and its answer (1.793\times10^{-11}\,\mathrm{J}=112\,\mathrm{MeV}) cannot be obtained from any standard two-neutron fission of U-235: with accurate masses such fissions release about 170 to 185 MeV.

43 marks

Calculate the number of nuclei of C-14 that remain undecayed after 40110 years if the initial atoms were 10000. (The half-life of C-14 is approximately 5730 years.)

Solution

Given

N_0 = 10000
t = 40110\,\text{years}
T_{1/2} = 5730\,\text{years}

To find

Number of half-lives.
n=\dfrac{t}{T_{1/2}}=\dfrac{40110}{5730}=7
Seven halvings.
N=N_0\left(\dfrac{1}{2}\right)^{7}=\dfrac{10000}{128}=78.1
The same result from the exponential law, with \lambda=0.693/5730=1.209\times10^{-4}\,\text{year}^{-1}.
N=10000\,e^{-(1.209\times10^{-4})(40110)}=10000\,e^{-4.85}=78

N\approx78\ \text{nuclei}

Book answer. 4670

Note. The book's answer is wrong for the printed data. 40110 years is exactly 7 half-lives, so only 10000/128\approx78 nuclei remain. The printed answer 4670 would be correct for a time of about 6300 years (10000\times2^{-6300/5730}=4667).

53 marks

Calculate the mass defect and binding energy per nucleon for the silver (Ag) nucleus if its atomic mass is 107.905949\,\mathrm{u}.

Solution

Given

M_{\text{atom}} = 107.905949\,\mathrm{u}
Z = 47 silver
A = 108 the mass is close to 108 u
m_p,\ m_n,\ m_e = 1.007276,\ 1.008665,\ 0.000549\,\mathrm{u}

To find

\Delta m
E_b/A

The nucleus is silver-108: Z=47 protons and A-Z=108-47=61 neutrons. The given mass is an atomic mass, so remove the 47 electrons to get the nuclear mass.
m_{\text{nucleus}}=107.905949-47(0.000549)=107.880146\,\mathrm{u}
Total mass of the separate nucleons.
Zm_p+(A-Z)m_n=47(1.007276)+61(1.008665)=47.341972+61.528565=108.870537\,\mathrm{u}
Mass defect.
\Delta m=108.870537-107.880146=0.99039\,\mathrm{u}=0.99039\times1.66\times10^{-27}\,\mathrm{kg}=1.644\times10^{-27}\,\mathrm{kg}
Binding energy.
E_b=0.99039\times931.5=922.5\,\mathrm{MeV}=1.476\times10^{-10}\,\mathrm{J}
Binding energy per nucleon.
\dfrac{E_b}{A}=\dfrac{922.5}{108}=8.54\,\mathrm{MeV}=8.54\times1.6\times10^{-13}\,\mathrm{J}=1.37\times10^{-12}\,\mathrm{J}

\begin{aligned}\Delta m&\approx0.990\,\mathrm{u}\ (1.64\times10^{-27}\,\mathrm{kg})\\ \dfrac{E_b}{A}&\approx8.5\,\mathrm{MeV}=1.37\times10^{-12}\,\mathrm{J\ per\ nucleon}\end{aligned}

Book answer. 41.031\times10^{-12}\,\mathrm{J/nucleon}

Note. The book's answer is impossible: 41.0\times10^{-12}\,\mathrm{J} is about 256\,\mathrm{MeV} per nucleon, while every nucleus has between 1 and 9 MeV per nucleon. Silver lies near the top of the binding energy curve, so about 8.5\,\mathrm{MeV} is expected.

63 marks

Calculate the energy released in the following reaction:

\ce{^{1}_{0}n + ^{238}_{92}U -> ^{96}_{38}Sr + ^{140}_{54}Xe + 3^{1}_{0}n} + Q

Solution

The question gives no masses, so use accurate atomic masses:

Given

m_n = 1.008665\,\mathrm{u}
m_U = 238.05078\,\mathrm{u} U-238
m_{Sr} = 95.92172\,\mathrm{u} Sr-96
m_{Xe} = 139.92164\,\mathrm{u} Xe-140

To find

Check the equation: upper numbers 1+238=96+140+3=239; lower numbers 92=38+54.
Total mass before the reaction.
m_n+m_U=1.008665+238.05078=239.059445\,\mathrm{u}
Total mass after the reaction (three neutrons).
m_{Sr}+m_{Xe}+3m_n=95.92172+139.92164+3(1.008665)=238.869355\,\mathrm{u}
Mass lost.
\Delta m=239.059445-238.869355=0.19009\,\mathrm{u}
Energy released.
Q=0.19009\times931.5=177.1\,\mathrm{MeV}=177.1\times1.6\times10^{-13}\,\mathrm{J}=2.83\times10^{-11}\,\mathrm{J}

Q is positive: energy is released, as in every fission of a heavy nucleus, because the fragments (about 8.5\,\mathrm{MeV} per nucleon) are more tightly bound than uranium (7.6\,\mathrm{MeV} per nucleon).

Q\approx177\,\mathrm{MeV}\approx2.83\times10^{-11}\,\mathrm{J}\ \text{(released)}

Book answer. -6.474\times10^{-11}\,\mathrm{J}

Note. The book's answer cannot be right: it is negative (energy absorbed) and far too large (-405\,\mathrm{MeV}). With accurate masses the reaction releases about 2.8\times10^{-11}\,\mathrm{J}. (Uranium-238 needs a fast neutron to start fission, but that is because of the energy barrier to splitting, not because Q is negative.)

FBISE 2026 questions

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

2026 Annual I, Set B, MCQ 131 mark

Which of the following is an example of nuclear fusion?

AUranium decay
BAtomic bomb
CEnergy production in the SunAnswer
DRadioactive decay

Why

In the Sun, hydrogen nuclei fuse to form helium, releasing energy: this is fusion. Uranium decay and radioactive decay are spontaneous emissions from unstable nuclei, and an atomic bomb works by uncontrolled fission.

2026 Annual I, Set B, Q.2 (ix)1.5+1.5 = 3 marks

How do control rods and moderator play an important role in the action of nuclear reactor?

Solution

Control rods (1.5 marks)

Control rods are made of a material that absorbs neutrons strongly, such as boron or cadmium. They can be moved into or out of the reactor core. Pushing them in absorbs more neutrons, so fewer fissions occur; pulling them out allows more fissions. They are set so that only one neutron from each fission goes on to cause another fission, which keeps the chain reaction steady and the power output constant. Dropped fully in, they stop the reaction in an emergency.

Moderator (1.5 marks)

The moderator (water or graphite) slows down the fast neutrons produced by fission (energies of a few MeV) to thermal neutrons of about 0.025\,\mathrm{eV}. Slow neutrons are far more likely to be captured by uranium-235 and cause fission, so the moderator keeps the chain reaction going. The neutrons lose energy by colliding many times with the light nuclei (hydrogen or carbon) of the moderator.

2026 Annual I, Set B, Q.2 (xi) OR1.5+1.5 = 3 marks

Briefly explain the terms: a. Mass Defect b. Binding Energy

Solution

(a) Mass defect (1.5 marks)

The mass of a nucleus is always less than the sum of the masses of its separate protons and neutrons. This difference is the mass defect: \Delta m=\left[Zm_p+(A-Z)m_n\right]-m_{\text{nucleus}}. For deuterium, \Delta m=(1.007276+1.008665)-2.013553=0.002388\,\mathrm{u}. The missing mass has been converted into the energy released when the nucleons bound together.

(b) Binding energy (1.5 marks)

The binding energy is the minimum energy required to break a nucleus into its separate nucleons; the same energy is released when the nucleus forms. It is the energy equivalent of the mass defect: E_b=\Delta mc^{2}, or E_b(\mathrm{MeV})=\Delta m(\mathrm{u})\times931.5. For deuterium, E_b=0.002388\times931.5=2.22\,\mathrm{MeV}. The greater the binding energy per nucleon, the more stable the nucleus.

2026 Annual I, Set B, Q.6 OR(2+1)+3 = 6 marks

a. What is meant by enrichment? Why Uranium fuel needs to enriched before use? b. Find decay constant and activity of an element if its 100 out of 700 nuclei decay in 5 min.

Solution

(a) What is meant by enrichment? Why does uranium fuel need to be enriched before use? (2+1 marks)

Enrichment (2 marks)

Enrichment is the process of increasing the percentage of the fissile isotope uranium-235 in uranium. Natural uranium contains only about 0.7% U-235 and 99.3% U-238. The two isotopes are chemically identical, so they are separated physically using their small difference in mass. The uranium is converted into uranium hexafluoride gas (\ce{UF6}, molecular masses 349 and 352) and passed through porous barriers (gaseous diffusion) or spun in gas centrifuges. The lighter \ce{^{235}UF6} molecules move slightly faster and are separated a little at each stage; many stages raise the U-235 content to 3 to 5%.

Why it is needed (1 mark)

With only 0.7% U-235, most neutrons are absorbed by U-238 and by the moderator before they can cause fission, so fewer than one neutron per fission produces another fission and the chain reaction cannot be sustained. Enriched fuel lets the reactor reach and keep criticality and run efficiently.

(b) Find the decay constant and activity if 100 out of 700 nuclei decay in 5 min. (3 marks)

Given

N = 700
\Delta N = 100
\Delta t = 5\,\mathrm{min}=300\,\mathrm{s}

To find

\lambda
A

Decay constant: the fraction of nuclei decaying per second.
\lambda=\dfrac{1}{N}\times\dfrac{\Delta N}{\Delta t}=\dfrac{1}{700}\times\dfrac{100}{300}=4.76\times10^{-4}\,\mathrm{s^{-1}}
Activity.
A=\lambda N=(4.76\times10^{-4})(700)=0.333\,\mathrm{Bq}
In curie (optional).
A=\dfrac{0.333}{3.70\times10^{10}}=9\times10^{-12}\,\mathrm{Ci}

\lambda=4.76\times10^{-4}\,\mathrm{s^{-1}},\qquad A=0.333\,\mathrm{Bq}

Note. Part (b) is the book's Example 22.5; part (a) matches Short Question 5 of the exercise.

2026 Annual I, Set D, MCQ 151 mark

In nuclear fission, a chain reaction occurs due to the release of:

APhotons
BNeutronsAnswer
CProtons
DElectrons

Why

Each fission of uranium-235 releases 2 or 3 neutrons. These neutrons cause further fissions, which release more neutrons, so the reaction keeps itself going. Photons, protons and electrons do not cause further fissions.

2026 Annual I, Set D, Q.2 (ix)1.5+1.5 = 3 marks

How is the variation of binding energy per nucleon an important factor in nuclear physics? (Discuss both high binding energy and low binding energy)

Solution

The binding energy per nucleon E_b/A is the average energy needed to remove one nucleon from a nucleus. Its variation with mass number (the binding energy curve) explains the stability of nuclei and the release of nuclear energy.

High binding energy per nucleon (1.5 marks)

Medium nuclei (A\approx50 to 100) have the highest values, with a peak of about 8.8\,\mathrm{MeV} for iron-56. Their nucleons are most tightly bound, so these nuclei are the most stable; they undergo neither fission nor fusion, and no energy can be obtained from them.

Low binding energy per nucleon (1.5 marks)

Very light nuclei (such as \ce{^{2}H}, 1.1\,\mathrm{MeV}) and very heavy nuclei (such as U-235, 7.6\,\mathrm{MeV}) have lower values and are less stable. They can gain stability by moving towards the peak, and energy is released when they do: light nuclei by fusion (hydrogen into helium in the Sun) and heavy nuclei by fission (U-235 into two medium nuclei, about 200\,\mathrm{MeV}). This is the basis of nuclear power and of stellar energy.

Binding energy per nucleon (in \mathrm{MeV}) against mass number

Sketch of the binding energy curve: low at both ends, highest near iron-56.

2026 Annual I, Set D, Q.2 (xi) OR1.5+1.5 = 3 marks

Briefly explain the function of control rods and moderators in nuclear reactor.

Solution

Function of control rods (1.5 marks)

Control rods, made of neutron-absorbing boron or cadmium, control the rate of the chain reaction by absorbing some of the neutrons. Inserting them further lowers the number of fissions and the power; withdrawing them raises it. In normal running they are adjusted so that exactly one neutron per fission causes a further fission, so the reaction neither dies out nor runs out of control. In an emergency they are dropped fully in to shut the reactor down.

Function of moderators (1.5 marks)

A moderator is a material that slows down fast neutrons. Fission neutrons are fast (about 2\,\mathrm{MeV}), but uranium-235 is split most easily by slow (thermal) neutrons of about 0.025\,\mathrm{eV}. In repeated collisions with the light nuclei of the moderator (water or graphite) the neutrons lose most of their energy and become thermal, which keeps the chain reaction going.

2026 Annual I, Set D, Q.6 OR3+3 = 6 marks

a. Derive and show that the decay of nucleus is exponential with time. b. If half-life of strontium (Sr. 91) is 9.70 hours. Find its decay constant.

Solution

(a) Derive and show that the decay of a nucleus is exponential with time. (3 marks)

The rate of decay of a radioactive sample is directly proportional to the number N of undecayed nuclei present (rate law). The constant of proportionality is the decay constant \lambda.

Derivation Exponential decay law

Aim. Show that N=N_0e^{-\lambda t}.

Assumptions.

The sample contains a very large number of nuclei, so N changes smoothly.
The decay constant \lambda is constant for the isotope.

Rate law: the rate of decay is proportional to the number of undecayed nuclei; the minus sign shows N decreases.
\dfrac{\Delta N}{\Delta t}=-\lambda N
For a very small interval the ratio becomes a derivative.
\dfrac{dN}{dt}=-\lambda N
Separate the variables.
\dfrac{dN}{N}=-\lambda\,dt
Integrate both sides.
\ln N=-\lambda t+C\qquad(\text{i})
Initial condition: at t=0, N=N_0, so the constant is C=\ln N_0.
\ln N_0=0+C
Substitute C in (i) and combine the logarithms.
\ln N-\ln N_0=-\lambda t\;\Rightarrow\;\ln\dfrac{N}{N_0}=-\lambda t
Take the exponential of both sides.
\dfrac{N}{N_0}=e^{-\lambda t}

ResultN=N_0\,e^{-\lambda t}

The number of undecayed nuclei decreases exponentially with time and never becomes exactly zero.

This shows that the number of undecayed nuclei falls exponentially with time: in equal time intervals the same fraction decays (half in each half-life), and N never becomes exactly zero.

Exponential decay curve

N against t: the number halves in every half-life.

N=N_0e^{-\lambda t}

(b) The half-life of strontium-91 is 9.70 hours. Find its decay constant. (3 marks)

Given

T_{1/2} = 9.70\,\mathrm{h}=9.70\times3600\,\mathrm{s}=3.492\times10^{4}\,\mathrm{s}

To find

\lambda

Relation between decay constant and half-life.
\lambda=\dfrac{0.693}{T_{1/2}}
Substitute the half-life in seconds.
\lambda=\dfrac{0.693}{3.492\times10^{4}\,\mathrm{s}}=1.98\times10^{-5}\,\mathrm{s^{-1}}
Equivalent value per hour.
\lambda=\dfrac{0.693}{9.70\,\mathrm{h}}=0.0714\,\mathrm{h^{-1}}

\lambda\approx1.98\times10^{-5}\,\mathrm{s^{-1}}

Note. Part (a) matches Long Question 8 of the exercise; part (b) is the book's Assignment 22.6.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 8, old syllabus1 mark

The number of neutrons in {}_{56}\mathrm{Ba}^{138} are:

A192
B138
C82Answer
D56

Why

The nucleus has Z=56 protons and mass number A=138, so N=A-Z=138-56=82. (192 is A+Z, 138 is the number of nucleons and 56 the number of protons.)

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 11, old syllabus1 mark

The unit of decay constant is:

AMeter
BSecond
CPer meter
DPer secondAnswer

Why

The decay constant is the fraction of nuclei that decay per unit time, \lambda=\dfrac{1}{N}\dfrac{\Delta N}{\Delta t}, so its unit is \mathrm{s^{-1}} (per second). This also follows from \lambda=0.693/T_{1/2}.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 17, old syllabus1 mark

Various types of cancer are treated by:

AStrontium-90
BCobalt 60Answer
CCarbon 14
DNickel-63

Why

Cobalt-60 emits penetrating gamma rays (and has a convenient half-life of 5.27 years), so it is the standard source for radiotherapy of many cancers. Carbon-14 is used for dating, nickel-63 in small batteries and detectors, and strontium-90 only for a few surface treatments.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (vi) OR, old syllabus3 marks

How can nuclear reactor initiate a controlled fission reaction?

Solution

A reactor starts and keeps a controlled chain reaction, in which on average exactly one neutron from each fission causes another fission:

The core contains fuel rods of uranium enriched to about 3% U-235, surrounded by a moderator (water or graphite). A few free neutrons are always present, from spontaneous fission or a small neutron source.
The moderator slows these neutrons to thermal energies (about 0.025\,\mathrm{eV}), at which they readily cause fission of U-235. Each fission releases 2 or 3 more fast neutrons, which are slowed in turn.
Control rods of cadmium or boron absorb neutrons. They are slowly withdrawn until the reactor becomes critical (one neutron per fission causes a further fission), and are then adjusted to keep the rate of fission, and so the power, constant.
The coolant removes the heat, and safety rods drop in to stop the reaction at once in an emergency.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (viii), old syllabus3 marks

How Geiger-Muller counter detects and counts radiation?

Solution

A GM tube is a metal cylinder (cathode) with a thin central wire (anode) at about 400\,\mathrm{V} positive, filled with argon at low pressure and a little quenching gas, with a thin mica window at one end.

Geiger-Muller tube and counter.

Radiation entering the window ionises a few gas atoms.
The electrons are accelerated towards the wire by the strong field and ionise more atoms: an avalanche of electrons reaches the anode.
This gives a pulse of current, which makes a voltage pulse across the resistor R.
Each pulse is amplified and counted by a scaler or ratemeter: one pulse for each particle or photon, so the count rate measures the radiation.
The quenching gas stops the discharge so the tube is ready for the next particle (after a dead time of about 10^{-4}\,\mathrm{s}).

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (xii), old syllabus3 marks

The half-life of radioactive nucleus {}_{88}\mathrm{Ra}^{226} is 1.6\times10^{3} years. Determine the decay constant.

Solution

Given

T_{1/2} = 1.6\times10^{3}\,\text{years}=1.6\times10^{3}\times3.15\times10^{7}\,\mathrm{s}=5.05\times10^{10}\,\mathrm{s} 1 year = 365 days

To find

decay constant \lambda

Relation between decay constant and half-life (1 mark).
\lambda=\dfrac{0.693}{T_{1/2}}
Substitute the half-life in seconds (1 mark).
\lambda=\dfrac{0.693}{5.05\times10^{10}\,\mathrm{s}}
Evaluate, with the unit (1 mark).
\lambda=1.37\times10^{-11}\,\mathrm{s^{-1}}
In per year: \lambda=0.693/1600=4.33\times10^{-4}\,\text{year}^{-1}.

\lambda\approx1.37\times10^{-11}\,\mathrm{s^{-1}}\ (4.33\times10^{-4}\ \text{per year})

2026 Annual I, Old syllabus (Curriculum 2006), Q.5 OR, old syllabus3+4 = 7 marks

What is meant by 'half-life' of a radio active element? Show that T_{1/2}=\dfrac{0.693}{\lambda}.

Solution

Half-life (3 marks)

The half-life T_{1/2} of a radioactive element is the time taken for half of the radioactive nuclei present in a sample to decay.

Starting with N_0 undecayed nuclei, N_0/2 remain after one half-life, N_0/4 after two, N_0/8 after three, and so on: N=N_0(1/2)^{n} after n half-lives. The half-life is a fixed property of each isotope. It does not depend on the amount of the sample or on temperature, pressure or chemical state. Examples: radon-222, 3.82 days; cobalt-60, 5.27 years; carbon-14, 5730 years; uranium-238, 4.5 billion years.

Decay curve showing successive half-lives

The number of undecayed nuclei halves in each half-life.

Derivation of T_{1/2}=0.693/\lambda (4 marks)

Derivation Relation between half-life and decay constant

Aim. Show that T_{1/2}=\dfrac{0.693}{\lambda}.

Assumptions.

The sample obeys the exponential decay law N=N_0e^{-\lambda t}, which follows from the rate law dN/dt=-\lambda N.

Write the decay law.
N=N_0\,e^{-\lambda t}
At t=T_{1/2} half of the nuclei remain, N=N_0/2. Substitute.
\dfrac{N_0}{2}=N_0\,e^{-\lambda T_{1/2}}
Cancel N_0 and take the reciprocal of both sides.
\dfrac{1}{2}=e^{-\lambda T_{1/2}}\;\Longrightarrow\;2=e^{\lambda T_{1/2}}
Take the natural logarithm of both sides.
\ln2=\lambda T_{1/2}
Rearrange, using \ln2=0.693.
T_{1/2}=\dfrac{\ln2}{\lambda}=\dfrac{0.693}{\lambda}

ResultT_{1/2}=\dfrac{0.693}{\lambda}

The half-life is inversely proportional to the decay constant: a rapidly decaying isotope (large \lambda) has a short half-life.

Exam focus

Derive N=N_0e^{-\lambda t} and sketch the decay curve. (long)
Asked in FBISE 2026 (Set D, Q.6 OR (a)) and as Long Question 8; a standard derivation worth 3 to 6 marks.
Go to answer
Define half-life and show that T_{1/2}=0.693/\lambda. (long)
Asked in the 2026 old-syllabus paper (Q.5 OR, 7 marks) and used in almost every decay numerical.
Go to answer
Explain mass defect and binding energy, and calculate them for a given nucleus. (short)
FBISE 2026 Set B Q.2 (xi) OR and Long Question 1; numericals use 1\,\mathrm{u}=931.5\,\mathrm{MeV}.
Go to answer
Explain the importance of the variation of binding energy per nucleon (high and low values) and sketch the curve. (short)
FBISE 2026 Set D Q.2 (ix); the curve explains why both fission and fusion release energy.
Go to answer
Functions of control rods and the moderator; principal components of a water-moderated reactor. (short)
Asked twice in 2026 (Set B Q.2 (ix), Set D Q.2 (xi) OR); Long Question 7 covers every component.
Go to answer
What is enrichment, and why must uranium fuel be enriched? (short)
FBISE 2026 Set B Q.6 OR (a), and Short Questions 5 and 8 of the exercise.
Go to answer
Decay constant and activity: \lambda=\dfrac{1}{N}\dfrac{\Delta N}{\Delta t}, A=\lambda N, \lambda=0.693/T_{1/2}. (numerical)
Three 2026 numericals used these: Set B Q.6 OR (b), Set D Q.6 OR (b) and old Q.2 (xii).
Go to answer
Half-life numericals: how much is left after a given time. (numerical)
Numericals 1 and 4 and MCQ 8; use N=N_0(1/2)^{t/T_{1/2}} and check by halving.
Go to answer
Fission, the chain reaction and critical mass. (long)
2026 Set D MCQ 15, Short Question 7 and Long Question 6.
Go to answer
Nuclear fusion: conditions, energy released and applications. (long)
2026 Set B MCQ 13 and Long Question 5.
Go to answer
Energy released in a decay or reaction from Q=\Delta m\times931.5\,\mathrm{MeV}. (numerical)
Long Question 3 and Numericals 3 and 6; the mass bookkeeping (atomic or nuclear masses) is where marks are lost.
Go to answer
Annihilation (0.511\,\mathrm{MeV} photons) and how PET scanning works. (long)
Long Questions 9 and 10 and four SLOs; the photon energy is a likely short calculation.
Go to answer

22The nucleus: nucleons, atomic number and mass number

22.1Mass defectOften examined

22.2Binding energy and mass-energy equivalenceOften examined

Energy equivalent of one atomic mass unit

Derivation 1u is equivalent to 931.5MeV

22.2Binding energy per nucleonOften examined

22.2The binding energy curveOften examined

Features of the curve

Why the curve has this shape

Relevance to fission and fusion

22.3RadioactivityOften examined

Main features of radioactivity

22.3.1Alpha decayOften examined

22.3.2Beta decayOften examined

Beta-minus decay

Beta-plus decay

Electron capture

22.3.3Gamma decayOften examined

22.3Properties of alpha, beta and gamma radiationOften examined

Ionising power and penetrating power

22.3.4Spontaneous and random nature of decay

22.3.4Energy released in a nuclear decayOften examined

22.4Half-lifeOften examined

22.4Rate law, decay constant and activityOften examined

22.4.1Exponential nature of nuclear decayOften examined

Derivation Exponential decay law, N=N_0e^{-\lambda t}

The same law for activity and count rate

Solving decay problems analytically

Solving decay problems graphically

22.4.2Relation between decay constant and half-lifeOften examined

Derivation Relation between half-life and decay constant

22.5Nuclear reactions

Conservation laws in nuclear reactions

22.5.1Nuclear fusionOften examined

Why fusion releases energy

Energy released in deuterium-tritium fusion

Conditions for fusion

Practical applications of fusion

22.5.3Nuclear fissionOften examined

Why fission releases energy

Fission energy compared with chemical energy

22.5.3Fission chain reaction and critical massOften examined

22.6Nuclear reactorOften examined

Core

Fuel rods

Control rods (safety rods)

Moderator

Coolant (heat exchanger)

Shielding

How a reactor starts and controls the chain reaction

22.6Enrichment of uraniumOften examined

Why uranium fuel needs to be enriched

Methods of enrichment

Gaseous diffusion

Gas centrifuge

Laser enrichment

22.7Energy in annihilation reactionsOften examined

Conservation laws in annihilation

Derivation Energy of each photon in electron-positron annihilation

Proton-antiproton annihilation

22.8.1Medical uses of radiation and medical tracersOften examined

Properties of a good tracer

Radiation in treatment (radiotherapy)

22.8.2PET scannerOften examined

Locating the annihilation point from the arrival times

Uses and hazards of radiation

Uses

Hazards

Safety precautions

Nuclear energy: benefits and drawbacks

Formula sheet

Key terms

Multiple choice questionsp. 168

Short questionsp. 168–169

Comprehensive questionsp. 169

Mass defect

Binding energy

Example: helium-4

Binding energy per nucleon and stability

Definition

Derivation $1 u$ is equivalent to $931.5 MeV$