Quantum physics
At the atomic scale light and matter behave in ways classical physics cannot explain. This unit shows that electromagnetic radiation comes in packets called photons, each with energy and momentum , and uses this idea to explain the photoelectric effect, the Compton effect, pair production and pair annihilation. It then shows that particles such as electrons behave as waves (), which makes the electron microscope possible, introduces Heisenberg's uncertainty principle, and explains line spectra with discrete energy levels ().
Learning outcomes (19)
- State that electromagnetic radiation has a particulate nature.
- Explain and apply the photonic model of light to solve problems (use ), and use the electronvolt (eV) as a unit of energy.
- Explain that a photon has momentum given by , and connect this with the idea that light can exert a force.
- Describe that photoelectrons may be emitted from a metal surface when it is illuminated by electromagnetic radiation.
- Describe and use the terms threshold frequency and threshold wavelength.
- Explain photoelectric emission in terms of photon energy and work function energy.
- State and apply .
- Explain why the maximum kinetic energy of photoelectrons is independent of intensity, whereas the photoelectric current is proportional to intensity.
- Juxtapose the evidence for light as a wave and as a particle: the photoelectric effect provides evidence for a particulate nature of electromagnetic radiation, while interference and diffraction provide evidence for a wave nature.
- Analyse qualitatively the evidence provided by electron diffraction for the wave nature of particles.
- Explain and apply the de Broglie wavelength to solve problems (use ).
- State that there are discrete electron energy levels in isolated atoms (for example atomic hydrogen).
- Explain the appearance and formation of emission and absorption line spectra.
- Use to solve problems.
- Describe the Compton effect qualitatively.
- Explain the phenomena of pair production and pair annihilation.
- Explain how electron microscopes achieve very high resolution.
- State and explain Heisenberg's uncertainty principle qualitatively.
- Use the uncertainty principle to explain why empirical measurements must necessarily have uncertainty in them.
What quantum physics is about
Newton's mechanics and Maxwell's theory of electricity and magnetism (together called classical physics) describe everyday objects very well. Around 1900, experiments on atoms and light gave results that classical physics could not explain at all: the radiation given out by hot bodies, the photoelectric effect and the line spectra of gases.
Quantum physics is the branch of physics that studies the behaviour of matter and energy at the atomic and subatomic level. At this scale the rules of classical physics are replaced by new ones that at first look strange. This unit builds four of those ideas.
| Quantum idea | What it says | Evidence in this unit |
|---|---|---|
| Light comes in packets | Light energy is carried by photons, each of energy and momentum . | Photoelectric effect, Compton effect, pair production |
| Matter behaves as waves | A moving particle has a wavelength . | Electron diffraction, electron microscope |
| Limits to measurement | Position and momentum cannot both be known exactly. | Heisenberg's uncertainty principle |
| Atoms have fixed energies | Electrons in atoms occupy only certain energy levels. | Emission and absorption line spectra |
21.1Electromagnetic waves and the spectrum
In 1864 the British physicist James Clerk Maxwell predicted that an accelerated electric charge produces linked electric and magnetic disturbances that travel through space. A changing electric field produces a magnetic field, and a changing magnetic field produces an electric field, so the two fields keep each other going. This travelling pair of fields is an electromagnetic (EM) wave.
- Electromagnetic wave
- A wave made of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of travel. It is produced by accelerated charges and travels through vacuum at .
Properties of electromagnetic waves
- The electric and magnetic fields are perpendicular to each other and to the direction of propagation, so EM waves are transverse waves.
- The direction of propagation is the direction of the vector product .
- The two fields oscillate in phase and with the same frequency.
- All EM waves travel through vacuum with the same speed . They need no medium, which is why sunlight reaches Earth through empty space. They can also travel through some materials, more slowly.
- EM waves carry energy and momentum from one point to another.
Frequency , wavelength and speed are related by the wave equation:
: speed of light (), : frequency in hertz (Hz), : wavelength in metres.
The electromagnetic spectrum
Visible light from the glowing filament of a bulb is one example of an EM wave. Radio waves, microwaves, infrared, ultraviolet, X-rays and gamma rays are the same kind of wave. They differ only in wavelength (and so in frequency). Arranged in order of wavelength they form the electromagnetic spectrum.
| Region | Typical wavelength | Typical source or use |
|---|---|---|
| Radio waves | Radio and TV broadcasting | |
| Microwaves | Radar, mobile phones, microwave ovens | |
| Infrared | Heat radiation, TV remote controls | |
| Visible light | to | Light seen by the eye |
| Ultraviolet | Sun, sterilising equipment | |
| X-rays | Medical imaging, crystal studies | |
| Gamma rays | Radioactive nuclei, cancer treatment |
Inside the visible band the colours in order of decreasing wavelength are red, orange, yellow, green, blue, indigo and violet. Red light has the longest wavelength and the lowest frequency; violet has the shortest wavelength and the highest frequency.
- Use for red light, with in metres.
- Repeat for violet light.
21.1Planck's quantum theory and the photonOften examined
Classical physics treated light energy as a smooth, continuous flow. In 1900 the German physicist Max Planck studied the radiation emitted by hot bodies (from their atoms and molecules). He found that the measured results at all wavelengths could be explained only if energy is emitted or absorbed in small, fixed packets. His idea is called Planck's quantum theory, and it gave birth to quantum physics.
Postulates of Planck's quantum theory
- Atoms and molecules emit or absorb energy only in discrete amounts (not continuously). The energy is , where is a whole number called the quantum number.
- The smallest amount of energy that can be emitted or absorbed as electromagnetic radiation is called a quantum (plural quanta). For radiation of frequency one quantum has energy .
- The energies of atoms and molecules are therefore quantized. The allowed energy states are called quantum states. An atom emits or absorbs energy only when it jumps from one quantum state to another.
(quantum number), : frequency of the radiation, : Planck's constant.
- Planck's constant (h)
- The constant that links the energy of a quantum to its frequency, (usually rounded to ). Its unit is joule second ().
- Quantum
- The smallest packet of energy that can be emitted or absorbed as electromagnetic radiation, .
- Photon
- A quantum of electromagnetic radiation (light). It travels at speed , has zero rest mass, energy and momentum .
Energy of a photon
In 1905 Einstein went further: light itself travels as a stream of these packets, now called photons. The energy of one photon depends only on the frequency of the light. Using :
Photon energy is directly proportional to frequency and inversely proportional to wavelength.
The electronvolt
Photon energies in joules are tiny numbers such as . Atomic and nuclear physics therefore uses a much smaller unit of energy, the electronvolt.
- Electronvolt (eV)
- The energy gained (or lost) by an electron when it moves through a potential difference of one volt. .
Derivation Relation between the electronvolt and the joule
Aim. Express in joules, the SI unit of energy.
- When a charge moves through a potential difference , the work done on it (and the kinetic energy it gains) is the charge times the p.d.
- For one electron the charge is . By definition of the electronvolt, the p.d. is .
- A coulomb times a volt is a joule, because .
- Larger multiples follow from the prefixes.
The electronvolt is a very small unit, the right size for single electrons, atoms and photons. To change eV into joules multiply by ; to change joules into eV divide by .
- photon energy
- frequency
- Change the energy into joules, because is in .
- Use and make the subject.
- Photon energy is inversely proportional to wavelength (), so write a ratio. and cancel.
- Substitute , and . The units of wavelength cancel, so there is no need to convert angstroms.
21.3Momentum of a photonOften examined
A photon has zero rest mass: a photon cannot be at rest, it always moves at speed . Even so it carries momentum. Maxwell's theory already showed that an electromagnetic wave carrying energy also carries momentum . In the photon picture every photon carries its own share of this momentum.
Derivation Momentum of a photon
Aim. Show that a photon of energy has momentum .
- The photon moves at the speed of light and has zero rest mass.
- Its energy is (Planck).
- By Einstein's mass-energy relation, energy is equivalent to a mass . A photon moving at speed therefore behaves as if it had this mass (it is not a rest mass).
- Momentum is mass times velocity, and the photon's velocity is .
- Put in the photon energy .
- Use , so .
A photon's momentum grows with its frequency: a short-wavelength X-ray photon carries far more momentum than a photon of visible light. Rewriting as shows that the energy of a photon is directly proportional to its momentum.
Light can exert a force
When light falls on a surface, the photons are absorbed or reflected, so their momentum changes. By Newton's second law, a rate of change of momentum is a force. So a beam of light pushes on any surface it hits. The push per unit area is called radiation pressure.
Derivation Force exerted by a beam of light
Aim. Find the force on a surface when a light beam of power falls on it at right angles.
- Each photon has energy , so the number of photons arriving per second is the power divided by the energy of one photon.
- If the surface absorbs the light, each photon gives up all its momentum .
- Force equals the momentum delivered per second: number of photons per second times momentum per photon.
- If the surface reflects the light straight back, each photon's momentum reverses from to , a change of , so the force doubles.
Because is so large the force is tiny: a beam absorbed by a surface pushes with only . It is still real and measurable.
- Energy of one photon.
- Photons per second = power / energy per photon.
- Force for complete absorption.
21.2Photoelectric effect: experiment and observationsOften examined
The word photoelectric joins two words: photo (light) and electric (electron). The photoelectric effect is an interaction between light (photons) and the electrons of a metal.
- Photoelectric effect
- The emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency falls on it.
- Photoelectrons
- The electrons emitted from a metal surface because of light falling on it.
- Photoelectric current
- The current produced when the photoelectrons are collected and flow round a circuit.
Experimental set-up
- An evacuated glass tube contains two metal plates. The emitter E (the metal surface under study) is joined to the negative terminal of a battery, and the collector C (also called the anode or detector) is kept at a positive potential.
- The tube is evacuated so that the emitted electrons do not collide with air molecules and the metal surface stays clean.
- When the tube is kept in the dark the ammeter reads zero: there is no current in the circuit.
- When light of suitable frequency falls on E, the ammeter shows a current at once. Electrons are emitted from the negative plate, cross the gap and are collected by the positive plate, completing the circuit.
Effect of applied voltage on photoelectric current
Keep the frequency of the light fixed and plot the photoelectric current I against the potential V of the collector, for three different intensities. The graph shows these features:
- For positive V the current rises and then becomes constant. This maximum value is the saturation current: every electron emitted per second is now collected by the anode.
- The saturation current is larger for brighter light. Brighter light of the same frequency ejects more electrons per second, but it does not give each electron more energy.
- A small current flows even at V=0, because the electrons leave the metal with some kinetic energy.
- When V is made negative (battery reversed), the collector repels the electrons. Only electrons with kinetic energy greater than e|V| reach it, so the current falls.
- At a certain negative potential -V_0 the current becomes zero: even the fastest electrons are turned back. V_0 is the stopping potential. It is the same for all three intensities.
- Stopping potential (V_0)
- The minimum negative (retarding) potential of the collector, relative to the emitter, at which the photoelectric current becomes zero. It measures the maximum kinetic energy of the photoelectrons.
- Saturation current
- The largest photoelectric current, reached when all the photoelectrons emitted per second are collected.
Derivation Maximum kinetic energy and stopping potential
Aim. Show that K.E_{\max}=eV_0.
- e=1.6\times10^{-19}\,\mathrm{C} is the charge of an electron.
- The electrons move in a vacuum, so no energy is lost in collisions.
- The fastest photoelectron leaves the emitter with kinetic energy K.E_{\max}=\tfrac12mv_{\max}^{2} and moves towards the negative collector.K.E_{\max}=\tfrac12mv_{\max}^{2}
- Moving against a retarding p.d. V_0, the electron (charge e) does work eV_0 against the electric force. This work equals the electric potential energy it gains.\text{work done against field}=eV_0
- At the stopping potential this fastest electron just comes to rest at the collector, so all its kinetic energy has turned into potential energy (conservation of energy).\tfrac12mv_{\max}^{2}=eV_0
Measuring the stopping potential with a voltmeter gives the maximum kinetic energy directly. A stopping potential of 2\,\mathrm{V} means K.E_{\max}=2\,\mathrm{eV}=3.2\times10^{-19}\,\mathrm{J}.
Effect of intensity of incident radiation on photoelectric current
Keep the frequency and the potential difference fixed and change only the intensity. The photoelectric current increases linearly with intensity. The current is proportional to the number of photoelectrons emitted per second, so:
Effect of frequency of incident radiation on the kinetic energy of photoelectrons
Now keep the intensity fixed and change the frequency. The maximum kinetic energy of the photoelectrons depends on the frequency. Even very dim light of high frequency releases electrons with large kinetic energy. For every metal there is a threshold frequency f_0: below it no electrons are emitted, however intense the light is. Classical physics has no explanation for this frequency dependence.
- Below f_0: no emission at all, even with very bright light.
- Above f_0: K.E_{\max} increases linearly with f. The graph is a straight line whose slope is Planck's constant h for every metal.
- Extending the line backwards (dashed) meets the energy axis at -\Phi, where \Phi is the work function of the metal.
- Threshold frequency (f_0)
- The lowest frequency of electromagnetic radiation that can produce the photoelectric effect in a given metal.
- Threshold wavelength (\lambda_0)
- The longest wavelength of radiation that can eject electrons from a given metal, \lambda_0=c/f_0. No photoemission occurs if the wavelength is longer than this.
- Work function (\Phi)
- The minimum energy needed to remove an electron from the surface of a metal (the book also calls it the threshold energy). It is measured in joules or electronvolts, and \Phi=hf_0=hc/\lambda_0.
| Metal | Work function \Phi | Threshold wavelength \lambda_0=hc/\Phi | Light needed |
|---|---|---|---|
| Potassium | 2.1\,\mathrm{eV} | 590\,\mathrm{nm} | visible (yellow or shorter) |
| Sodium | 2.3\,\mathrm{eV} | 540\,\mathrm{nm} | visible (green or shorter) |
| Zinc | 4.3\,\mathrm{eV} | 290\,\mathrm{nm} | ultraviolet |
| Copper | 4.7\,\mathrm{eV} | 265\,\mathrm{nm} | ultraviolet |
Why classical wave theory fails
| Observation | Classical wave theory predicts | Experiment shows |
|---|---|---|
| Threshold frequency | Light of any frequency should eject electrons if it is bright enough, because energy keeps building up. | No electrons below f_0, however bright the light. |
| Kinetic energy and intensity | Brighter light (larger amplitude) should give faster electrons. | K.E_{\max} does not depend on intensity. |
| Kinetic energy and frequency | Frequency should not matter. | K.E_{\max} increases linearly with frequency. |
| Time lag | With dim light an electron needs time to collect enough energy (seconds or even minutes). | Emission starts at once (within about 10^{-9}\,\mathrm{s}), even in very dim light. |
21.2Photon theory of the photoelectric effectOften examined
In 1905 Albert Einstein reinterpreted Planck's theory: light itself exists as discrete packets of energy called photons, each of energy E=hf. Photoelectric emission is the result of an interaction between one photon of the incident radiation and one electron in the metal. The photon gives all its energy to the electron and disappears.
A part of this energy (the work function \Phi) is used by the electron to break away from the metal. The rest appears as the kinetic energy of the electron. Electrons from just below the surface lose a little extra energy on the way out, so only electrons from the surface leave with the maximum kinetic energy.
Derivation Einstein's photoelectric equation
Aim. Relate the photon energy hf to the work function \Phi and the maximum kinetic energy of the photoelectrons.
- Light is a stream of photons, each of energy hf.
- Each photon is absorbed completely by a single electron (one photon, one electron).
- The minimum energy needed to remove an electron from the surface is the work function \Phi.
- One photon of frequency f brings energy hf and gives all of it to one electron.E=hf
- Energy is conserved: the photon energy pays the work function and the remainder becomes kinetic energy. For an electron at the surface the kinetic energy is the maximum.hf=\Phi+K.E_{\max}\qquad(21.3)
- At the threshold frequency f_0 the electron just escapes, with zero kinetic energy. Put K.E_{\max}=0 and f=f_0.hf_0=\Phi
- Substitute \Phi=hf_0 back into (21.3).K.E_{\max}=hf-hf_0=h(f-f_0)
- Using K.E_{\max}=\tfrac12mv_{\max}^{2}=eV_0 gives other useful forms.hf=\Phi+\tfrac12mv_{\max}^{2}=\Phi+eV_0
- Divide by e to see how the stopping potential depends on frequency.V_0=\dfrac{h}{e}\,f-\dfrac{\Phi}{e}
The equation is a straight line: K.E_{\max} (or V_0) against f has slope h (or h/e), cuts the frequency axis at f_0 and, extended back, cuts the energy axis at -\Phi (or -\Phi/e).
The equation gives three cases:
- If hf<\Phi: there is no photoelectric effect.
- If hf=\Phi: the photoelectric effect just occurs, but the ejected electron has zero kinetic energy.
- If hf>\Phi: electrons are emitted with kinetic energy up to K.E_{\max}=hf-\Phi.
How the photon theory explains the observations
| Observation | Explanation with photons |
|---|---|
| Threshold frequency | A photon with hf<\Phi cannot free an electron. An electron cannot add up the energy of two photons, so a brighter beam (more photons) still gives no emission. |
| K.E_{\max} independent of intensity | Intensity is the number of photons per second. Each photon still has the same energy hf, so K.E_{\max}=hf-\Phi does not change. |
| Current proportional to intensity | More photons per second free more electrons per second, so the current increases in proportion. |
| K.E_{\max} increases with frequency | K.E_{\max}=hf-\Phi is a straight line in f with slope h. |
| No time lag | The energy arrives in one packet, so an electron can leave as soon as it absorbs a photon. |
Stopping potential against frequency
Since V_0=\dfrac{h}{e}f-\dfrac{\Phi}{e}, a graph of stopping potential against frequency is a straight line. It lets us find both the work function and Planck's constant.
- Slope =h/e, so h=e\times\text{slope}. The slope is the same for every metal.
- Frequency intercept =f_0, so \Phi=hf_0.
- Intercept on the stopping potential axis (line extended back to f=0) =-\Phi/e, so \Phi=e\times|\text{intercept}|.
Uses of the photoelectric effect
- Photocells (photoelectric cells) convert light energy into electric current. They switch on street lights at dusk, open automatic doors, operate burglar alarms and count objects on a conveyor belt.
- Photoconductive devices, such as light dependent resistors, whose resistance falls when light releases charge carriers.
- Solar cells, which convert sunlight directly into electrical energy.
- Light meters in cameras and the image sensors of digital cameras.
- \Phi = 4.2\,\mathrm{eV}=4.2\times1.6\times10^{-19}=6.72\times10^{-19}\,\mathrm{J} work function
- \lambda = 2000\,\text{\AA}=2000\times10^{-10}\,\mathrm{m}=2\times10^{-7}\,\mathrm{m}
- K.E_{\max}
- Einstein's photoelectric equation, with f=c/\lambda.K.E_{\max}=hf-\Phi=\dfrac{hc}{\lambda}-\Phi
- Energy of one photon.\dfrac{hc}{\lambda}=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{2\times10^{-7}}=9.945\times10^{-19}\,\mathrm{J}
- Subtract the work function.K.E_{\max}=9.945\times10^{-19}-6.72\times10^{-19}=3.23\times10^{-19}\,\mathrm{J}
- Change into electronvolts.K.E_{\max}=\dfrac{3.23\times10^{-19}}{1.6\times10^{-19}}=2.02\,\mathrm{eV}
- Photon energy in joules.E=\dfrac{hc}{\lambda}=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{3000\times10^{-10}}=6.63\times10^{-19}\,\mathrm{J}
- Change into electronvolts so it can be compared with \Phi directly.E=\dfrac{6.63\times10^{-19}}{1.6\times10^{-19}}=4.14\,\mathrm{eV}
- Einstein's equation gives the maximum kinetic energy.K.E_{\max}=E-\Phi=4.14-1.6=2.54\,\mathrm{eV}
- K.E_{\max}=eV_0, so the stopping potential in volts equals K.E_{\max} in electronvolts.V_0=\dfrac{K.E_{\max}}{e}=\dfrac{2.54\,\mathrm{eV}}{e}=2.54\,\mathrm{V}
- At threshold the whole photon energy is used as work function.f_0=\dfrac{\Phi}{h}=\dfrac{2.3\times1.6\times10^{-19}}{6.63\times10^{-34}}=5.6\times10^{14}\,\mathrm{Hz}
- Threshold wavelength from \lambda_0=c/f_0.\lambda_0=\dfrac{3\times10^{8}}{5.55\times10^{14}}=5.4\times10^{-7}\,\mathrm{m}=540\,\mathrm{nm}
- Compare: 700\,\mathrm{nm} is longer than \lambda_0, so each red photon has less energy than \Phi.700\,\mathrm{nm}>540\,\mathrm{nm}\ \Rightarrow\ \text{no emission}
21.3Compton's effectOften examined
In 1923 the American physicist Arthur H. Compton directed a beam of X-rays of wavelength \lambda at a block of graphite. He found that the scattered X-rays had a slightly longer wavelength \lambda' than the incident X-rays. The increase in wavelength (the loss of energy) depended on the angle through which the X-rays were scattered.
- Compton effect
- The scattering of X-ray (or gamma-ray) photons by electrons, in which the photons transfer energy and momentum to the electrons, so the wavelength of the scattered radiation increases.
- Compton shift (\Delta\lambda)
- The change in wavelength between the scattered and the incident X-ray, \Delta\lambda=\lambda'-\lambda.
Classical wave theory cannot explain this. A wave makes an electron oscillate at the wave's own frequency, so the electron should re-radiate at the same frequency and wavelength. Compton explained the result by treating the X-ray as a stream of photons.
Explanation with photons
Compton assumed that if a photon behaves like a particle, its collision with an electron is like a collision between two billiard balls: both energy and momentum must be conserved. The incident photon (energy hf, momentum h/\lambda) hits a free electron at rest and gives part of its energy and momentum to it. The electron recoils. The scattered photon is left with less energy, so its frequency is lower and its wavelength is longer.
Conservation of energy:
hf and hf': energies of the incident and scattered photons, K.E: kinetic energy given to the recoiling electron.
Conservation of momentum (a vector equation): the momentum \vec p of the incident photon equals the vector sum of the momenta of the scattered photon \vec p\,' and the recoil electron \vec p_e. The momentum of each photon is p=E/c=h/\lambda (Eq. 21.6).
Take the incident direction as the x-axis. Resolving momenta into components:
From these equations Compton derived the shift in wavelength:
m_0: rest mass of the electron, c: speed of light, \theta: scattering angle of the photon.
Derivation Compton shift formula (full working)
Aim. Derive \lambda'-\lambda=\dfrac{h}{m_0c}(1-\cos\theta) from conservation of energy and momentum.
- The electron is free and at rest before the collision (outer electrons in graphite are very loosely bound compared with the X-ray photon's energy).
- Energy and momentum are both conserved in the collision.
- Write photon momenta as p=h/\lambda and p'=h/\lambda', so photon energies are pc and p'c. Energy conservation, with E_e the total energy of the electron after the collision, is:pc+m_0c^{2}=p'c+E_e\ \Rightarrow\ E_e=(p-p')c+m_0c^{2}
- Momentum conservation gives \vec p_e=\vec p-\vec p\,'. Squaring (or combining 21.7 and 21.8), the angle between \vec p and \vec p\,' is \theta:p_e^{2}=p^{2}+p'^{2}-2pp'\cos\theta
- The recoil electron can be fast, so use the relativistic energy-momentum relation for it.E_e^{2}=p_e^{2}c^{2}+m_0^{2}c^{4}
- Square the energy equation of step 1 and set it equal to the relation of step 3, using step 2 for p_e^{2}.(p-p')^{2}c^{2}+2(p-p')m_0c^{3}+m_0^{2}c^{4}=(p^{2}+p'^{2}-2pp'\cos\theta)c^{2}+m_0^{2}c^{4}
- Expand (p-p')^{2}=p^{2}-2pp'+p'^{2} and cancel the terms that appear on both sides, then divide by 2c^{2}.(p-p')\,m_0c=pp'(1-\cos\theta)
- Divide both sides by pp'm_0c. Note \dfrac{p-p'}{pp'}=\dfrac{1}{p'}-\dfrac{1}{p}.\dfrac{1}{p'}-\dfrac{1}{p}=\dfrac{1-\cos\theta}{m_0c}
- Put back 1/p'=\lambda'/h and 1/p=\lambda/h, then multiply by h.\lambda'-\lambda=\dfrac{h}{m_0c}(1-\cos\theta)
The shift depends only on the scattering angle \theta. It does not depend on the incident wavelength or on the scattering material. The book quotes this result without working; the steps are given here for completeness.
- Compton wavelength (\lambda_c)
- The constant \lambda_c=\dfrac{h}{m_0c}=2.43\times10^{-12}\,\mathrm{m}=0.00243\,\mathrm{nm}. It has the dimension of a wavelength and equals the Compton shift for scattering at 90^\circ.
Features of the Compton shift
- The shift depends only on the scattering angle \theta: \Delta\lambda=0 at \theta=0^\circ, \Delta\lambda=\lambda_c at 90^\circ and \Delta\lambda=2\lambda_c=4.86\times10^{-12}\,\mathrm{m} (the maximum) at 180^\circ.
- It does not depend on the wavelength of the incident photon or on the nature of the scattering material.
- Since (1-\cos\theta) is never negative, \lambda'\ge\lambda: the scattered photon never gains energy from an electron at rest. (The book says (1-\cos\theta) is always positive; it is zero only at \theta=0, where there is no scattering.)
Physical significance of the Compton effect
- It is strong evidence for the particle nature of electromagnetic radiation: an X-ray photon collides with an electron like a particle.
- It confirms that a photon carries momentum p=h/\lambda as well as energy E=hf.
- It shows that the laws of conservation of energy and momentum hold in each individual photon-electron collision.
- Together with the photoelectric effect it describes how photons interact with electrons: in the photoelectric effect the photon is completely absorbed; in the Compton effect it is scattered with less energy.
- \lambda = 0.20\,\mathrm{nm}
- \theta = 45^\circ
- m_0 = 9.11\times10^{-31}\,\mathrm{kg}
- \Delta\lambda=\lambda'-\lambda
- Compton shift formula.\lambda'-\lambda=\dfrac{h}{m_0c}(1-\cos\theta)
- Substitute. The incident wavelength is not needed, because the shift does not depend on it.\lambda'-\lambda=\dfrac{6.63\times10^{-34}}{(9.11\times10^{-31})(3\times10^{8})}(1-\cos45^\circ)=2.43\times10^{-12}\times0.293
- Evaluate.\lambda'-\lambda=7.11\times10^{-13}\,\mathrm{m}
- Compton shift at \theta=60^\circ, where \cos60^\circ=0.5.\Delta\lambda=\dfrac{h}{m_0c}(1-\cos60^\circ)=2.43\times10^{-12}\times0.5=1.21\times10^{-12}\,\mathrm{m}=1.21\,\mathrm{pm}
- The scattered wavelength is the incident wavelength plus the shift.\lambda'=\lambda+\Delta\lambda=71+1.21=72.2\,\mathrm{pm}
21.4Pair productionOften examined
Every particle has an antiparticle with the same mass but opposite charge. The antiparticle of the electron is the positron (e^+): mass m_0=9.11\times10^{-31}\,\mathrm{kg}, charge +e.
- Pair production
- The creation of a particle and its antiparticle from energy. When a high-energy photon passes near a heavy nucleus, it disappears and an electron-positron pair is created. The photon is totally absorbed.
Pair production is direct evidence that energy can be converted into mass, as Einstein's relation E=mc^{2} says. Three quantities are conserved in the process: charge, momentum and energy.
Conservation laws in pair production
Conservation of charge. The photon has zero charge. The electron (-e) and positron (+e) together also have zero charge. So a photon can never produce a single electron, a single positron, two electrons or two positrons: charge would not be conserved.
Conservation of momentum. The photon's momentum hf/c must equal the total momentum of the products. Here m v_{e^+} and m v_{e^-} are the momenta of the positron and electron (vector sum), and the nucleus takes a small recoil:
Conservation of energy. The photon energy becomes the total energy of the electron and positron:
The total energy of a particle is its kinetic energy plus its rest mass energy m_0c^{2}, so
Derivation Minimum photon energy for pair production
Aim. Find the threshold (minimum) energy a photon needs to create an electron-positron pair.
- Rearrange Eq. (21.12 b): the photon energy provides two rest mass energies plus the kinetic energies.hf=2m_0c^{2}+K.E_{\text{electron}}+K.E_{\text{positron}}
- The photon energy is least when the pair is created with (almost) zero kinetic energy.hf_{\min}=2m_0c^{2}
- Rest mass energy of one electron (or positron).m_0c^{2}=(9.11\times10^{-31})(3\times10^{8})^{2}=8.2\times10^{-14}\,\mathrm{J}=0.511\,\mathrm{MeV}
- Double it for the pair.hf_{\min}=2\times0.511\,\mathrm{MeV}=1.02\,\mathrm{MeV}=1.64\times10^{-13}\,\mathrm{J}
- The largest wavelength that can do this follows from E=hc/\lambda.\lambda_{\max}=\dfrac{hc}{hf_{\min}}=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{1.64\times10^{-13}}=1.21\times10^{-12}\,\mathrm{m}
Only gamma-ray photons (wavelength about 10^{-12}\,\mathrm{m} or less) can produce a pair. If the photon energy is more than 1.02\,\mathrm{MeV}, the extra energy appears as kinetic energy of the electron and positron.
Conditions for pair production
- The photon energy must be at least 2m_0c^{2}=1.02\,\mathrm{MeV} (a gamma-ray photon).
- A heavy nucleus (or another particle) must be nearby to take up momentum.
- Charge, energy and momentum must all be conserved, so a particle and its antiparticle (electron and positron) are always created together.
- Energy balance: the photon supplies both rest mass energies and the kinetic energies.K.E_{\text{total}}=hf-2m_0c^{2}=3.0-1.02=1.98\,\mathrm{MeV}
- Equal sharing.K.E_{\text{each}}=\dfrac{1.98}{2}=0.99\,\mathrm{MeV}
21.5Pair annihilationOften examined
Pair annihilation is the reverse of pair production. In pair production energy (a photon) is converted into mass (an electron-positron pair). In annihilation mass (an electron-positron pair) is converted into energy (photons).
- Pair annihilation
- The process in which an electron and a positron combine, both disappear, and their mass is converted into two gamma-ray photons.
Why two photons and not one
Just before annihilation the electron and positron are nearly at rest, so their total momentum is (almost) zero. A single photon always moves at speed c and carries momentum hf/c, which is never zero. One photon therefore cannot conserve momentum. Two photons of equal energy moving in opposite directions have equal and opposite momenta, so their total momentum is zero, just as before. Charge is also conserved: (+e)+(-e)=0, and photons carry no charge.
Derivation Energy of each annihilation photon
Aim. Find the energy and wavelength of each photon produced when an electron and a positron at rest annihilate.
- The total energy before annihilation is the rest mass energy of two particles (kinetic energy is negligible).E_{\text{before}}=2m_0c^{2}
- Two photons share this energy equally (they have equal momenta, so equal energies).2hf=2m_0c^{2}\ \Rightarrow\ hf=m_0c^{2}
- Substitute the rest mass energy of the electron.hf=0.511\,\mathrm{MeV}=8.2\times10^{-14}\,\mathrm{J}
- Wavelength of each photon.\lambda=\dfrac{hc}{m_0c^{2}}=\dfrac{h}{m_0c}=2.43\times10^{-12}\,\mathrm{m}
- If the particles are moving, their kinetic energies are also given to the photons.2m_0c^{2}+K.E_{\text{electron}}+K.E_{\text{positron}}=\text{energy of two gamma photons}\qquad(21.13)
The photons are gamma rays. Their wavelength happens to equal the Compton wavelength h/m_0c. Annihilation shows mass being turned completely into energy, E=mc^{2}.
| Feature | Pair production | Pair annihilation |
|---|---|---|
| Conversion | Energy into mass | Mass into energy |
| Before | One gamma photon (hf\ge1.02\,\mathrm{MeV}) | An electron and a positron |
| After | An electron and a positron | Two gamma photons (0.511\,\mathrm{MeV} each when the pair is at rest) |
| Momentum | A heavy nucleus takes the extra momentum | Photons move in opposite directions |
| Energy equation | hf=2m_0c^{2}+K.E_{e^-}+K.E_{e^+} | 2m_0c^{2}+K.E_{e^-}+K.E_{e^+}=\text{energy of two photons} |
Another example of annihilation is proton-antiproton annihilation. Besides confirming the photon model of electromagnetic radiation, pair production and pair annihilation clearly illustrate Einstein's idea of mass and rest energy.
21.6Wave-particle duality and de Broglie wavesOften examined
Interference, diffraction and polarization are the basic properties of all electromagnetic waves. Light shows all three, so light can be considered a wave. But the photoelectric effect can be explained only if light is made of small energy packets (particles) called photons. So light displays a dual nature: it has particle characteristics as well as wave characteristics.
- Wave-particle duality
- The property of light (and of matter) of behaving as a wave in some experiments and as a stream of particles in others.
| Evidence for the wave nature | Evidence for the particle nature |
|---|---|
| Interference (Young's double slit experiment) | Photoelectric effect |
| Diffraction (single slit, diffraction grating) | Compton effect (photon momentum) |
| Polarization (light is a transverse wave) | Pair production and pair annihilation |
| Speed c, wavelength \lambda and frequency f with c=f\lambda | Line spectra (atoms emit photons of definite energy) |
The two pictures are joined by two equations. The particle properties (energy E and momentum p) are fixed by the wave properties (frequency f and wavelength \lambda):
de Broglie's hypothesis
In 1924 the French physicist Louis de Broglie argued that nature should be symmetric. If light, which shows pure wave properties, can also have particle-like properties, then particles such as electrons can also behave as waves. Particles of matter obey a wave equation just as photons do. Compton's work showed that a photon of wavelength \lambda has momentum p=h/\lambda; de Broglie proposed the same relation for every moving particle.
Derivation de Broglie wavelength
Aim. Find the wavelength associated with a particle of mass m moving with speed v.
- For a photon, the momentum and wavelength are related by Compton's result.p=\dfrac{h}{\lambda}
- de Broglie assumed the same relation holds for a material particle, so its wavelength is Planck's constant divided by its momentum.\lambda=\dfrac{h}{p}\qquad(21.14)
- For a particle of mass m and speed v (much less than c) the momentum is p=mv.\lambda=\dfrac{h}{mv}
Every moving particle has a wavelength. The greater its momentum, the shorter its wavelength. Heavy everyday objects have wavelengths far too small to observe, but electrons have wavelengths comparable to the spacing of atoms, so their wave nature can be seen.
- de Broglie wavelength
- The wavelength of the matter wave associated with a moving particle, \lambda=h/p=h/mv.
- Matter waves
- Waves associated with moving material particles such as electrons, protons and neutrons.
Macroscopic objects and electrons compared
- (a) Use \lambda=h/mv.\lambda=\dfrac{6.63\times10^{-34}}{(0.5)(10)}=1.3\times10^{-34}\,\mathrm{m}
- This is far smaller than even a nucleus (10^{-15}\,\mathrm{m}), so the wave nature of the object can never be observed.
- (b) Same formula for the electron.\lambda=\dfrac{6.63\times10^{-34}}{(9.1\times10^{-31})(10^{6})}=7.3\times10^{-10}\,\mathrm{m}
- This is comparable to the spacing between atoms in a crystal (about 10^{-10}\,\mathrm{m}), which is just right for diffraction and interference. So the wavelengths of very small particles are observable.
- m = 5.0\,\mathrm{mg}=5.0\times10^{-6}\,\mathrm{kg}
- v = 8.0\,\mathrm{m\,s^{-1}}
- \lambda
- de Broglie relation.\lambda=\dfrac{h}{mv}
- Substitute (1\,\mathrm{mg}=10^{-6}\,\mathrm{kg}).\lambda=\dfrac{6.63\times10^{-34}}{(5.0\times10^{-6})(8.0)}=1.66\times10^{-29}\,\mathrm{m}
- Convert the mass: 20\,\mathrm{g}=0.020\,\mathrm{kg}. Use \lambda=h/mv.\lambda=\dfrac{6.63\times10^{-34}}{(0.020)(1500)}=2.2\times10^{-35}\,\mathrm{m}
- No. To observe a wave we need diffraction or interference by an obstacle or slit about as small as the wavelength. Nothing in nature is this small (a nucleus is about 10^{-15}\,\mathrm{m}), so the bullet behaves purely as a particle.
Wavelength of an electron accelerated through a potential difference
Derivation de Broglie wavelength of an accelerated electron
Aim. Find the wavelength of an electron accelerated from rest through a potential difference V.
- The electron starts from rest.
- Its speed stays well below c, so p=mv can be used (true up to a few tens of kilovolts).
- The electron gains kinetic energy equal to the work done by the field.\tfrac12mv^{2}=eV
- Write kinetic energy in terms of momentum: \tfrac12mv^{2}=\dfrac{p^{2}}{2m}.\dfrac{p^{2}}{2m}=eV\ \Rightarrow\ p=\sqrt{2meV}
- Put this momentum into the de Broglie relation.\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2meV}}
- Insert h, m=9.11\times10^{-31}\,\mathrm{kg} and e=1.6\times10^{-19}\,\mathrm{C} to get a handy form (V in volts).\lambda\approx\dfrac{1.23\times10^{-9}}{\sqrt{V}}\,\mathrm{m}
A larger accelerating voltage gives a shorter wavelength. For V=100\,\mathrm{V}, \lambda=1.23\times10^{-10}\,\mathrm{m} (about the size of an atom); for V=10\,\mathrm{kV}, \lambda=1.2\times10^{-11}\,\mathrm{m}=0.12\,\text{\AA}. This is the key to the electron microscope.
21.6Electron diffraction: evidence for matter waves
If electrons are waves they must show diffraction, just like light passing through a grating. A diffraction grating works best when its spacing is comparable to the wavelength. The atoms in a crystal are about 10^{-10}\,\mathrm{m} apart, and electrons accelerated through a few hundred to a few thousand volts have wavelengths of about 10^{-10} to 10^{-11}\,\mathrm{m}, comparable to this spacing. So a thin crystal acts as a natural grating for electrons.
What the experiment shows
- A pattern of bright and dark concentric rings appears on the screen, similar to the pattern made when X-rays pass through a powdered crystal. Rings of this kind are produced by constructive and destructive interference, which only waves can do.
- When the accelerating voltage is increased, the electrons move faster, their wavelength \lambda=h/\sqrt{2meV} becomes shorter, and the rings become smaller (less diffraction).
- The wavelength calculated from the size of the rings agrees with de Broglie's formula \lambda=h/mv.
- A magnet brought near the tube shifts the whole pattern. So the pattern is made by charged electrons, not by X-rays or light produced in the tube.
21.7Electron microscopeOften examined
The electron microscope is a device based on the wave nature of electrons. Instead of light, it uses a beam of accelerated electrons as its source of illumination. It lets us observe objects in detail down to the nanometre scale, with remarkably high resolution. It is commonly used to study microscopic organisms such as viruses and bacteria, and the structure of metals and crystals.
Why the wavelength decides the resolution
Resolution is the ability of a microscope to show two very close points as separate. Because of diffraction, any microscope can only show details that are about as large as, or larger than, the wavelength of the radiation it uses. Shorter wavelength means finer detail and higher resolution.
- Visible light has wavelengths from 4000\,\text{\AA} to 7000\,\text{\AA} (400 to 700\,\mathrm{nm}). The best optical microscope cannot resolve details smaller than about 200\,\mathrm{nm}.
- Electrons accelerated through a p.d. V have wavelength \lambda=h/\sqrt{2meV}. Through 10\,000\,\mathrm{V} (10\,\mathrm{keV}) the wavelength is about 0.12\,\text{\AA} (0.012\,\mathrm{nm}), tens of thousands of times shorter than visible light. At 100\,\mathrm{kV} it is about 0.004\,\mathrm{nm}.
- In practice the magnetic lenses are not perfect, so the resolution is limited to about 0.1 to 0.2\,\mathrm{nm}. This is still about a thousand times finer than an optical microscope, enough to see individual large molecules and rows of atoms.
Construction and working
- Electron gun: a tungsten filament is heated and emits electrons. A high potential on the anode (tens of kilovolts) accelerates them into a narrow, fast beam, like a beam of light.
- Condenser lens: a magnetic coil that focuses the electron beam onto the specimen.
- Specimen (sample) stage: the sample must be very thin, so that electrons can pass through it, and carefully prepared. It is placed in the vacuum chamber.
- Objective lens: forms the first magnified image of the specimen.
- Intermediate and projector lenses: magnify the image further and project it downwards.
- Fluorescent screen or detector: at the very bottom. Electrons striking it make it glow, forming a visible image. A photograph of it is called an electron micrograph.
- Vacuum system: the whole column is evacuated so that the electrons are not scattered or stopped by air molecules.
The electron beam passes through a series of coil-shaped electromagnets that replace the glass lenses of an optical microscope. The magnetic field of a coil bends the paths of the electrons and brings them to a focus. Increasing the current in a coil makes its field stronger, so the magnetic lens becomes stronger (shorter focal length). The electron gun and condenser lens are in the upper part; the sample stage and objective lens are in the middle; the intermediate and projector lenses are in the lower part; the screen is at the bottom.
| Feature | Optical microscope | Electron microscope |
|---|---|---|
| Illumination | Visible light | Beam of fast electrons |
| Wavelength | 400 to 700\,\mathrm{nm} | about 0.01\,\mathrm{nm} or less (depends on V) |
| Lenses | Glass lenses | Magnetic lenses (current-carrying coils) |
| Best resolution | about 200\,\mathrm{nm} | about 0.1 to 0.2\,\mathrm{nm} |
| Useful magnification | up to about 1500 times | up to about a million times |
| Medium | Works in air | Needs a vacuum |
| Specimen | Can be living | Thin, dry, non-living, in vacuum |
| Image | Seen directly by the eye, in colour | Formed on a fluorescent screen or detector (black and white) |
21.8Heisenberg's uncertainty principleOften examined
On the microscopic level we cannot measure any property of a particle without interacting with it in some way, for example by bouncing a photon off it. This interaction introduces an unavoidable uncertainty into the result. From Planck's point of view, E=hc/\lambda: a photon of short wavelength has a large energy and a large momentum, so it disturbs whatever it hits.
The German physicist and Nobel laureate Werner Heisenberg formulated the uncertainty principle in 1927. It is a consequence of the dual (wave and particle) nature of matter.
- Heisenberg's uncertainty principle
- It is impossible to find both the position and the momentum of a particle accurately at the same time. The product of the uncertainties in position and momentum, measured at the same instant, is approximately equal to \hbar (or greater).
\Delta x: uncertainty in position, \Delta p: uncertainty in momentum, \hbar (read as h-cut or h-bar): the reduced Planck constant.
If the position were known exactly (\Delta x=0), the momentum would be completely unknown (\Delta p=\infty), and the other way round. The more precisely we fix one, the less precisely we can know the other.
The wave picture
In quantum mechanics a particle is described by a wave. Position means where the wave is concentrated; the uncertainty in position is how far the wave is spread out. Momentum is fixed by the wavelength (p=h/\lambda); the uncertainty in momentum is how unclear the wavelength is. A wave cannot be both short (definite position) and have one exact wavelength (definite momentum).
Why measuring position disturbs momentum
If we want accuracy in position we must use photons of shorter wavelength, because the best resolution we can get is about the wavelength of the radiation used. But short-wavelength radiation means high-frequency, high-energy photons with large momentum h/\lambda. When these collide with the electron they give it a large and unpredictable kick. If we use longer wavelength (less energetic) photons, the kick is gentler but the position becomes blurred.
Derivation Rough estimate of the uncertainty product
Aim. Show that locating an electron with light of wavelength \lambda gives \Delta x\,\Delta p\approx h.
- A photon of wavelength \lambda can locate the electron only to within about one wavelength.\Delta x\approx\lambda
- The photon has momentum h/\lambda. After it bounces off, the electron's momentum is uncertain by up to about this much, because the direction of the scattered photon is not known exactly.\Delta p\approx\dfrac{h}{\lambda}
- Multiply. The wavelength cancels, so the product cannot be made smaller by choosing a different \lambda.\Delta x\,\Delta p\approx\lambda\times\dfrac{h}{\lambda}=h
This rough argument gives the order of magnitude. A careful treatment gives \Delta x\,\Delta p\ge\hbar (the book's form); the exact modern form is \Delta x\,\Delta p\ge\hbar/2. All versions say the same thing: the product has a lower limit of the order of Planck's constant.
Uncertainty in energy and time
Another kind of uncertainty concerns the energy of a quantum state and the time for which it lasts. If \Delta E is the uncertainty in energy and \Delta t is the uncertainty in the time to which that energy refers, then:
Derivation Energy-time uncertainty
Aim. Show how a limited time of observation makes the energy uncertain.
- To measure the frequency of a wave we count the number of waves in a time interval \Delta t. The count is uncertain by about one wave, so the uncertainty in frequency is about one wave divided by the time.\Delta f\ \ge\ \dfrac{1}{\Delta t}
- The energy of a photon is E=hf, so an uncertainty in frequency means an uncertainty in energy.\Delta E=h\,\Delta f
- Combine the two lines.\Delta E\ \ge\ \dfrac{h}{\Delta t}\ \Rightarrow\ \Delta E\,\Delta t\ \ge\ h\ \text{(order of magnitude)}
A system that stays in a state for a very long time (a metastable state) can have a well-defined energy. A state that lasts only a short time has a correspondingly large uncertainty in energy. As with \Delta x\,\Delta p, the rough argument gives h and the book writes \hbar; they differ only by the factor 2\pi.
Why every measurement has an uncertainty
- These uncertainties are built into the physical world. They have nothing to do with poor instruments or the skill of the observer, and better apparatus cannot remove them.
- Every measurement needs an interaction (for example a photon hitting the object), and every interaction disturbs the object being measured.
- Because Planck's constant is so small, the uncertainties are far too small to notice for everyday objects. For electrons inside atoms they are very important.
- \Delta x = 10^{-10}\,\mathrm{m}
- \Delta p
- Uncertainty principle.\Delta x\,\Delta p\ \ge\ \dfrac{h}{2\pi}\ \Rightarrow\ \Delta p\ \ge\ \dfrac{h}{2\pi\,\Delta x}
- Substitute.\Delta p=\dfrac{6.63\times10^{-34}}{2(3.14)(10^{-10})}=1.06\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}
- Momentum of the ball.p=mv=0.1\times30=3\,\mathrm{kg\,m\,s^{-1}}
- Uncertainty in momentum is 10^{-6} of this.\Delta p=10^{-6}\times3=3\times10^{-6}\,\mathrm{kg\,m\,s^{-1}}
- Uncertainty principle, \Delta x\,\Delta p\ge\hbar.\Delta x\ \ge\ \dfrac{\hbar}{\Delta p}=\dfrac{1.054\times10^{-34}}{3\times10^{-6}}=3.5\times10^{-29}\,\mathrm{m}
- This is far smaller than a nucleus, so for a ball the uncertainty is completely negligible. The uncertainty principle matters only for tiny particles.
21.9Discrete energy levels in an atom and its spectraOften examined
The electrons in an atom can occupy only certain allowed orbits (also called shells). Each orbit has a definite energy, called an energy level: E_1, E_2, E_3 and so on. The energy of the atom is therefore quantized, as Planck's theory requires. An electron can move from one allowed level to another only by gaining or losing a definite amount of energy.
- Energy level
- One of the definite, allowed energies that an electron in an atom can have.
- Ground state
- The lowest energy level of an atom, E_1. Higher levels are called excited states.
- Excitation
- The transition of an electron from a lower to a higher energy level. It needs energy, for example the absorption of a photon.
- De-excitation
- The transition of an electron from a higher to a lower energy level. Energy is released as a photon.
An electron moves to a higher or lower level by absorbing or emitting electromagnetic radiation of frequency f. The frequency depends only on the energy difference between the two levels:
E_n: higher energy level, E_p: lower energy level, f: frequency and \lambda: wavelength of the photon absorbed or emitted.
Obtaining a spectrum
A spectrum is produced by using a prism or a diffraction grating to separate the different wavelengths in a beam of light. Photons emitted in such transitions are passed through a narrow slit and a prism (or grating); each wavelength is bent through a different angle and forms a separate line on the detector.
A line spectrum contains a discrete set of wavelengths. It is a characteristic feature of an element, unique to it.
Emission line spectrum
An emission spectrum of an element consists of a series of separate bright lines of definite frequencies (or wavelengths) on a dark background. It is produced by the excited atoms of the element in a vapour or gas at low pressure, for example in a discharge tube. The photons are emitted as electrons fall from higher to lower excited states or to the ground state. Each line corresponds to one particular transition.
Since each element has a unique set of orbital electrons and energy levels, its emission line spectrum is also unique. The spectrum can therefore be used to identify the element, like a fingerprint. Different coloured lines show which elements a substance is made of, because each element radiates its own particular energies.
Absorption line spectrum
An absorption spectrum of an element consists of a series of separate dark lines of definite frequencies (or wavelengths) on a continuous coloured background. White light from a source such as a tungsten filament lamp (containing all wavelengths) is passed through the cool vapour of the element and then through the slit and grating. The atoms absorb only those photons whose energy exactly matches a jump between their levels. These wavelengths are removed from the beam, leaving dark lines. The excited atoms soon return to the ground state and re-emit the same photons, but in all directions, so very few of them continue in the original direction.
| Feature | Emission line spectrum | Absorption line spectrum |
|---|---|---|
| Appearance | Bright coloured lines on a dark background | Dark lines on a continuous coloured background |
| Produced by | Hot, excited gas or vapour at low pressure | White light passed through cool gas or vapour |
| Electron transitions | Downwards (de-excitation), photons emitted | Upwards (excitation), photons absorbed |
| Line positions | Characteristic of the element | Same wavelengths as that element's emission lines |
- (a) Energy difference between the two levels.\Delta E=E_3-E_2=-1.51-(-3.40)=1.89\,\mathrm{eV}=1.89\times1.6\times10^{-19}=3.02\times10^{-19}\,\mathrm{J}
- Frequency from hf=\Delta E.f=\dfrac{\Delta E}{h}=\dfrac{3.02\times10^{-19}}{6.63\times10^{-34}}=4.56\times10^{14}\,\mathrm{Hz}
- Wavelength from \lambda=c/f. This is the red line of hydrogen (measured value 656\,\mathrm{nm}; the small difference comes from rounding the constants).\lambda=\dfrac{3\times10^{8}}{4.56\times10^{14}}=6.58\times10^{-7}\,\mathrm{m}\approx658\,\mathrm{nm}
- (b) Excitation needs exactly the energy difference; the photon must be absorbed whole.hf=E_2-E_1=-3.40-(-13.6)=10.2\,\mathrm{eV}\ \ (\text{ultraviolet, }\lambda\approx122\,\mathrm{nm})
Formula sheet
Key terms
- Quantum physics
- The branch of physics that studies the behaviour of matter and energy at the atomic and subatomic level.
- Electromagnetic wave
- A transverse wave of oscillating electric and magnetic fields, perpendicular to each other and to the direction of travel, moving through vacuum at c=3\times10^{8}\,\mathrm{m\,s^{-1}}.
- Quantum
- The smallest amount of energy that can be emitted or absorbed as electromagnetic radiation, E=hf.
- Photon
- A quantum (packet) of electromagnetic radiation with energy E=hf, momentum p=h/\lambda and zero rest mass.
- Planck's constant
- h=6.63\times10^{-34}\,\mathrm{J\,s}, the constant linking photon energy and frequency.
- Quantum number
- The whole number n in E=nhf that labels the allowed (quantized) energy states.
- Electronvolt
- The energy gained by an electron moving through a potential difference of one volt; 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}.
- Radiation pressure
- The force per unit area exerted by light on a surface because the photons' momentum changes.
- Photoelectric effect
- The emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency falls on it.
- Photoelectrons
- Electrons emitted from a metal surface by the action of light.
- Saturation current
- The maximum photoelectric current, reached when all emitted photoelectrons are collected.
- Stopping potential
- The minimum retarding potential at which the photoelectric current becomes zero; eV_0=K.E_{\max}.
- Threshold frequency
- The lowest frequency of radiation that can eject electrons from a given metal, f_0=\Phi/h.
- Threshold wavelength
- The longest wavelength of radiation that can eject electrons from a given metal, \lambda_0=c/f_0.
- Work function (threshold energy)
- The minimum energy needed by an electron to break free from a metal surface, \Phi=hf_0.
- Einstein's photoelectric equation
- hf=\Phi+K.E_{\max}: the photon energy equals the work function plus the maximum kinetic energy of the photoelectron.
- Photocell
- A device based on the photoelectric effect that converts light energy into an electric current.
- Compton effect
- The increase in wavelength of X-rays or gamma rays scattered by electrons, because the photons give some energy and momentum to the electrons.
- Compton shift
- The change in wavelength \Delta\lambda=\lambda'-\lambda=\dfrac{h}{m_0c}(1-\cos\theta) of a scattered photon.
- Compton wavelength
- h/m_0c=2.43\times10^{-12}\,\mathrm{m}, the Compton shift at a scattering angle of 90^\circ.
- Antiparticle
- A particle with the same mass as another particle but opposite charge; the positron is the antiparticle of the electron.
- Positron
- The antiparticle of the electron: mass 9.11\times10^{-31}\,\mathrm{kg}, charge +e.
- Rest mass energy
- The energy m_0c^{2} equivalent to the rest mass of a particle; 0.511\,\mathrm{MeV} for an electron.
- Pair production
- The creation of an electron and a positron from a photon of energy at least 1.02\,\mathrm{MeV} near a heavy nucleus (energy converted into mass).
- Pair annihilation
- The combination of an electron and a positron into two gamma-ray photons moving in opposite directions (mass converted into energy).
- Wave-particle duality
- The property of light and matter of behaving as waves in some experiments and as particles in others.
- de Broglie wavelength
- The wavelength of the wave associated with a moving particle, \lambda=h/p=h/mv.
- Matter waves
- Waves associated with moving material particles such as electrons.
- Electron diffraction
- The spreading of an electron beam into a ring or spot pattern when it passes through a thin crystal; evidence that electrons behave as waves.
- Electron microscope
- A microscope that uses a beam of fast electrons, focused by magnetic lenses, instead of light; its very short electron wavelength gives very high resolution.
- Resolution
- The ability of an instrument to show two close points as separate; it is limited by the wavelength of the radiation used.
- Electron micrograph
- A photograph of the image formed in an electron microscope.
- Heisenberg's uncertainty principle
- The position and momentum of a particle cannot both be measured exactly at the same time; \Delta x\,\Delta p\ge\hbar.
- Reduced Planck constant
- \hbar=h/2\pi=1.054\times10^{-34}\,\mathrm{J\,s}, read as h-cut or h-bar.
- Metastable state
- An excited state in which an atom stays for a comparatively long time, so its energy is well defined.
- Energy level
- One of the definite allowed energies of an electron in an atom.
- Ground state
- The lowest energy level of an atom.
- Excitation
- The jump of an electron to a higher energy level by absorbing energy.
- De-excitation
- The fall of an electron to a lower energy level, emitting a photon of energy hf=E_n-E_p.
- Line spectrum
- A spectrum containing only a discrete set of wavelengths, characteristic of an element.
- Emission line spectrum
- Bright lines of definite wavelengths on a dark background, produced by excited atoms of a gas at low pressure.
- Absorption line spectrum
- Dark lines of definite wavelengths on a continuous background, produced when white light passes through a cool gas.
Multiple choice questionsp. 138–139
- AInfrared
- BVioletAnswer
- CRed
- DBlue
Photon energy is E=hf, so the region with the highest frequency has the highest energy. Of these, violet light has the highest frequency (shortest wavelength, about 400\,\mathrm{nm}). Infrared has an even lower frequency than red, and blue lies below violet.
- AA photon has zero mass and zero momentum.
- BA photon has finite mass and a finite value of momentum.
- CA photon has zero mass but finite value of momentum.Answer
- DA photon has finite mass but zero momentum.
A photon has zero rest mass, but it carries momentum p=E/c=h/\lambda. The Compton effect proves that photons have momentum, so option A is wrong; B and D are wrong because a photon has no rest mass.
- AIt varies with the frequency of light.Answer
- BIt varies with the light intensity.
- CIt varies with the speed of light.
- DIt varies irregularly.
By Einstein's equation K.E_{\max}=hf-\Phi, the kinetic energy depends on the frequency f. Intensity changes only the number of electrons emitted per second, not their energy.
- AConverts light energy into electricity.Answer
- BConverts electricity into light energy.
- CStores light energy.
- DStores electricity.
In a photocell, light falling on the cathode ejects photoelectrons, which produce an electric current. So it converts light energy into electrical energy; it does not store energy.
- AMass
- BMomentumAnswer
- CWave properties
- DSpeed rates
The Compton effect is explained only if the photon behaves as a particle carrying momentum p=h/\lambda that it shares with the electron in a collision. It shows the particle nature, not the wave properties.
- Adecreases frequencyAnswer
- Bdecreases wavelength
- Cincreases energy
- Dincreases momentum
The photon gives part of its energy to the electron, so its energy hf' is smaller and its frequency decreases. Its wavelength increases and its momentum h/\lambda' decreases, so B, C and D are wrong.
- A12.4\times10^{1}\,\mathrm{eV}
- B12.4\times10^{2}\,\mathrm{eV}
- C12.4\times10^{3}\,\mathrm{eV}Answer
- D12.4\times10^{4}\,\mathrm{eV}
So the energy is 12.4\,\mathrm{keV} (1.24\times10^{4}\,\mathrm{eV}). A photon of this wavelength is usually called an X-ray photon, but the calculation is the same.
Note. Book misprint: all four options are printed as 12.4\times10^{4}\,\mathrm{eV}. The options here are restored with different powers of ten so that the question can be answered; the correct energy is 12.4\times10^{3}\,\mathrm{eV}.
- Aless than h/4\pi
- Bgreater than h/4\pi
- Czero
- DinfiniteAnswer
From \Delta x\,\Delta p\ge\hbar, \Delta p\ge\hbar/\Delta x. As \Delta x\to0, \Delta p\to\infty: knowing the position exactly makes the momentum completely unknown. Option B is true in general, but the exact consequence of \Delta x=0 is an infinite uncertainty.
- APair production
- BPair annihilationAnswer
- CNuclear fission
- DNuclear fusion
When a particle meets its antiparticle (for example an electron and a positron) both disappear and their mass becomes gamma-ray photons. This is pair annihilation. Pair production is the reverse process.
- APhotonAnswer
- BElectron
- CProton
- DNeutron
An electron and a positron annihilate into two gamma-ray photons moving in opposite directions. Electrons are consumed, not produced; protons and neutrons are not involved.
Short questionsp. 139
A photon moves with the speed of light c. By Einstein's mass-energy relation its energy E is equivalent to a mass m=E/c^{2} (not a rest mass). Its momentum is mass times velocity:
Since c=3\times10^{8}\,\mathrm{m\,s^{-1}} is a constant, E\propto p: doubling the momentum of a photon doubles its energy. The same result follows from E=hf and p=h/\lambda:
Hence the energy of a photon is directly proportional to its momentum, and the constant of proportionality is the speed of light.
One electronvolt is the energy gained by an electron moving through a p.d. of one volt: 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}. The prefix mega means 10^{6}, so 1\,\mathrm{MeV}=10^{6}\,\mathrm{eV}=1.6\times10^{-13}\,\mathrm{J}.
- Change MeV into eV.800\,\mathrm{MeV}=800\times10^{6}\,\mathrm{eV}=8\times10^{8}\,\mathrm{eV}
- Multiply by 1.6\times10^{-19}\,\mathrm{J} per eV.E=8\times10^{8}\times1.6\times10^{-19}=1.28\times10^{-10}\,\mathrm{J}
So 800\,\mathrm{MeV} is 1.28\times10^{-10}\,\mathrm{J}. Energies of this size occur in nuclear and particle physics; even so, in joules they are very small numbers.
The two phenomena are the photoelectric effect and the Compton effect. In both, a photon behaves as a particle and interacts with a single electron.
| Photoelectric effect | Compton effect | |
|---|---|---|
| What happens to the photon | Completely absorbed by one electron | Scattered with less energy and longer wavelength |
| Electron | Bound electron is ejected from a metal surface | Free (loosely bound) electron recoils |
| Energy equation | hf=\Phi+K.E_{\max} | hf=hf'+K.E |
| Typical radiation | Visible or ultraviolet (a few eV) | X-rays or gamma rays (keV and above) |
| Shows | Energy of light comes in quanta | Photons carry momentum h/\lambda |
At still higher photon energies (above 1.02\,\mathrm{MeV}) a photon can also create an electron-positron pair near a nucleus (pair production).
Einstein assumed that one photon gives all its energy to one electron, so K.E_{\max}=hf-\Phi. The experimental results confirm that an electron does not absorb two or more photons:
- No emission below the threshold frequency. If an electron could collect the energy of several photons, very intense light with hf<\Phi would eventually eject electrons. It never does, however intense the light is.
- Maximum kinetic energy does not depend on intensity. With multi-photon absorption, brighter light (more photons) would give electrons energies such as 2hf-\Phi. In fact K.E_{\max} and the stopping potential stay the same for all intensities.
- Slope of the graph: the K.E_{\max} against f graph has a slope of exactly h, the energy per unit frequency of one photon; two-photon absorption would give a slope of 2h.
Hence each photoelectron is released by exactly one photon.
Einstein's equation eV_0=hf-\Phi can be written as a straight line of the form y=mx+c:
- Work function: the line cuts the frequency axis at the threshold frequency f_0 (where V_0=0), so \Phi=hf_0. Or extend the line back to f=0: it cuts the V_0 axis at -\Phi/e, so \Phi=e\times|\text{intercept}|.
- Planck's constant: yes. The slope of the line is h/e, so h=e\times\text{slope}=e\,\dfrac{\Delta V_0}{\Delta f}. The slope is the same for every metal.
The free electrons in a metal already have some thermal energy. Heating the photo-electrode (emitter) gives them more, which can change the results in these ways:
- The work function decreases slightly as the metal is heated, so the threshold frequency becomes a little lower and the stopping potential (and K.E_{\max}) for a given frequency becomes a little larger.
- At high temperature some electrons escape by thermionic emission even in the dark. This adds a dark current to the photoelectric current, and the cut-off at the stopping potential becomes less sharp.
- At ordinary temperatures the thermal energy of an electron (about 0.03\,\mathrm{eV}) is tiny compared with \Phi (2 to 5\,\mathrm{eV}), so the effect is very small.
So the basic laws (threshold frequency, K.E_{\max}=hf-\Phi, current proportional to intensity) still hold; only small shifts and an extra thermal current appear when the electrode is made very hot.
Classical wave theory treats light as a continuous wave whose energy depends on its intensity (amplitude). It cannot explain these observations:
- Threshold frequency: no electrons are emitted below f_0 however intense the light, whereas wave theory predicts emission at any frequency if the light is bright enough.
- Kinetic energy independent of intensity: K.E_{\max} does not increase with intensity, although a stronger wave should give electrons more energy.
- Kinetic energy depends on frequency: K.E_{\max} increases linearly with f, which wave theory cannot explain.
- No time lag: emission is instant (about 10^{-9}\,\mathrm{s}) even in dim light, whereas wave theory predicts a delay while the electron collects energy.
All four are explained by the photon theory, hf=\Phi+K.E_{\max}.
In the Compton effect X-ray photons scattered by electrons come out with a longer wavelength, \Delta\lambda=\dfrac{h}{m_0c}(1-\cos\theta). Its significance is:
- It gives strong evidence for the particle nature of electromagnetic radiation: the photon collides with an electron like a billiard ball.
- It confirms that a photon carries momentum p=h/\lambda as well as energy E=hf.
- It shows that energy and momentum are conserved in each individual photon-electron collision.
- Classical wave theory predicts no change in wavelength, so the effect supports quantum theory.
| Compton shift | Compton wavelength |
|---|---|
| The change in wavelength of a scattered photon: \Delta\lambda=\lambda'-\lambda. | The constant \lambda_c=\dfrac{h}{m_0c} that appears in the Compton equation. |
| Depends on the scattering angle: \Delta\lambda=\lambda_c(1-\cos\theta). | Fixed value for the electron: 2.43\times10^{-12}\,\mathrm{m}=0.00243\,\mathrm{nm}. |
| Varies from 0 (at \theta=0^\circ) to 2\lambda_c=4.86\,\mathrm{pm} (at 180^\circ). | Equals the Compton shift at \theta=90^\circ. |
| A measured change, different for each angle. | A constant with the dimension of length, depending only on h, m_0 and c. |
In short, the Compton wavelength is the scale factor, and the Compton shift is that factor multiplied by (1-\cos\theta).
The Compton shift is given by
Here h, m_0 and c are constants, so the only variable is the scattering angle \theta. The shift does not depend on the incident wavelength or on the scattering material. Changing \theta changes the shift:
| \theta | 0^\circ | 60^\circ | 90^\circ | 180^\circ |
|---|---|---|---|---|
| \Delta\lambda | 0 | 1.21\,\mathrm{pm} | 2.43\,\mathrm{pm} | 4.86\,\mathrm{pm} (maximum) |
| \theta | \Delta\lambda |
|---|---|
| 0^\circ | 0 |
| 60^\circ | 1.21\,\mathrm{pm} |
| 90^\circ | 2.43\,\mathrm{pm} |
| 180^\circ | 4.86\,\mathrm{pm} (maximum) |
So the scattering angle alone decides the size of the shift: it is zero for no deflection and largest when the photon bounces straight back.
In pair production a gamma-ray photon changes into an electron and a positron. The conditions are:
- Minimum energy: the photon energy must be at least the rest mass energy of the two particles, hf\ge2m_0c^{2}=2(0.511)=1.02\,\mathrm{MeV} (a gamma-ray photon, \lambda\le1.2\times10^{-12}\,\mathrm{m}). Any extra energy becomes kinetic energy of the pair.
- A heavy nucleus nearby: the process must happen close to a nucleus, which takes up part of the momentum, so that momentum can be conserved.
- Conservation laws: charge, energy and momentum must all be conserved. Since the photon has no charge, a particle and its antiparticle (electron and positron, total charge zero) must be created together.
In 1924 de Broglie suggested that every moving particle has a wave associated with it (a matter wave). The de Broglie wavelength is the wavelength of this wave. It equals Planck's constant divided by the momentum of the particle:
Greater momentum means a shorter wavelength. For a 0.5\,\mathrm{kg} ball at 10\,\mathrm{m\,s^{-1}}, \lambda=1.3\times10^{-34}\,\mathrm{m} (unobservable), but for an electron at 10^{6}\,\mathrm{m\,s^{-1}}, \lambda=7.3\times10^{-10}\,\mathrm{m}, similar to atomic spacing, so electron diffraction can be observed.
The photoelectric effect provides the clearest evidence. It shows a threshold frequency, a maximum kinetic energy that depends on frequency but not on intensity, and no time lag. These results can be explained only if radiation consists of particles (photons) of energy hf, each absorbed by one electron: hf=\Phi+K.E_{\max}.
The Compton effect (photons colliding with electrons and transferring momentum h/\lambda) and pair production (a single photon turning into an electron and a positron) give further evidence of the particle nature of electromagnetic radiation.
Comprehensive questionsp. 139–140
Definition
The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency falls on it. The emitted electrons are called photoelectrons and the current they produce is the photoelectric current.
Experimental arrangement
Light falls on the emitter E inside an evacuated tube. Photoelectrons move to the collector C and the ammeter records the current. The p.d. V between C and E can be changed and reversed.
Experimental results
- No electrons are emitted if the frequency is below a certain threshold frequency f_0, however intense the light.
- Above f_0, emission starts instantly, even in very dim light.
- The photoelectric current (number of electrons per second) is proportional to the intensity of light.
- The current becomes zero at a negative collector potential -V_0, the stopping potential, and K.E_{\max}=eV_0. V_0 is the same for all intensities, so K.E_{\max} does not depend on intensity.
- K.E_{\max} increases linearly with frequency.
Failure of classical wave theory
Wave theory says a more intense wave carries more energy, so brighter light of any frequency should eventually eject electrons, and should give them more kinetic energy after some delay. It cannot explain the threshold frequency, the independence of K.E_{\max} from intensity, its dependence on frequency, or the absence of a time lag.
Quantum (photon) explanation
In 1905 Einstein used Planck's quantum idea: light is a stream of photons, each of energy hf. A photon interacts with a single electron and is absorbed completely (one photon, one electron).
Derivation Einstein's photoelectric equation
Aim. Relate photon energy, work function and maximum kinetic energy.
- Light consists of photons. One photon of frequency f carries energy hf and gives all of it to one electron.E=hf
- Part of this energy, the work function \Phi, is used to pull the electron out of the metal. By conservation of energy the rest is the electron's kinetic energy, which is maximum for an electron from the surface.hf=\Phi+K.E_{\max}
- At the threshold frequency f_0 the electron just escapes with zero kinetic energy.hf_0=\Phi
- Substitute for \Phi and use K.E_{\max}=\tfrac12mv_{\max}^{2}=eV_0.K.E_{\max}=\tfrac12mv_{\max}^{2}=h(f-f_0)=eV_0
Photoemission happens only if hf\ge\Phi, and the maximum kinetic energy rises linearly with frequency with slope h.
How the equation explains the results
- Threshold frequency: if hf<\Phi one photon cannot free an electron, and an electron cannot combine the energy of two photons, so there is no emission however many photons arrive.
- No time lag: the energy arrives in one packet, so the electron leaves as soon as it absorbs a photon.
- Current proportional to intensity: a brighter beam contains more photons per second, which free more electrons per second.
- Kinetic energy independent of intensity: each photon still has the same energy hf, so K.E_{\max}=hf-\Phi is unchanged.
- Kinetic energy increases with frequency: K.E_{\max}=hf-\Phi is a straight line in f with slope h and intercept f_0.
So the photoelectric effect is explained completely by the photon (quantum) theory, and it is strong evidence that electromagnetic radiation has a particle nature.
Definition
When a beam of X-rays of wavelength \lambda is scattered by the electrons of a material (Compton used graphite in 1923), the scattered X-rays have a slightly longer wavelength \lambda'. The increase depends on the scattering angle. This is the Compton effect; the increase \Delta\lambda=\lambda'-\lambda is the Compton shift.
Diagram
Explanation
Classical theory predicts no change of wavelength. Compton treated the X-ray as photons, each of energy hf and momentum h/\lambda, colliding with free electrons like billiard balls. In each collision energy and momentum are conserved. The electron recoils and takes some energy, so the scattered photon has less energy: lower frequency f' and longer wavelength \lambda'.
Conservation of energy (K.E is the kinetic energy of the recoil electron):
Conservation of momentum along and perpendicular to the incident direction (p_e is the momentum of the electron):
Compton shift
Derivation Compton equation
Aim. Obtain \Delta\lambda from conservation of energy and momentum.
- With p=h/\lambda and p'=h/\lambda', energy conservation gives the electron's total energy after the collision.E_e=(p-p')c+m_0c^{2}
- Momentum conservation, squared (angle \theta between \vec p and \vec p\,'):p_e^{2}=p^{2}+p'^{2}-2pp'\cos\theta
- For the electron E_e^{2}=p_e^{2}c^{2}+m_0^{2}c^{4}. Substituting steps 1 and 2 and cancelling equal terms leaves:(p-p')\,m_0c=pp'(1-\cos\theta)
- Divide by pp'm_0c and use 1/p=\lambda/h, 1/p'=\lambda'/h.\lambda'-\lambda=\dfrac{h}{m_0c}(1-\cos\theta)
The shift depends only on the scattering angle.
The constant h/m_0c=2.43\times10^{-12}\,\mathrm{m} is the Compton wavelength.
Main features
- \Delta\lambda depends only on \theta: zero at 0^\circ, 2.43\,\mathrm{pm} at 90^\circ, maximum 4.86\,\mathrm{pm} at 180^\circ.
- It is independent of the incident wavelength and of the scattering material.
- The effect is noticeable only for X-rays and gamma rays; for visible light a shift of a few picometres is negligible.
Significance
- It proves the particle nature of electromagnetic radiation.
- It shows that photons carry momentum p=h/\lambda.
- It confirms conservation of energy and momentum in single photon-electron collisions.
Statement
Heisenberg's uncertainty principle (1927): it is impossible to measure accurately both the position and the momentum of a particle at the same time. The product of the uncertainties in position and momentum at the same instant is approximately equal to \hbar or greater:
Here \Delta x is the uncertainty in position and \Delta p the uncertainty in momentum. If \Delta x\to0 then \Delta p\to\infty, and the other way round.
Explanation: measurement disturbs the particle
To observe an electron we must bounce at least one photon off it. The position can be found only to within about one wavelength, \Delta x\approx\lambda. To locate it precisely we need a short wavelength, but a short-wavelength photon has a large momentum h/\lambda and gives the electron a large, unknown kick, \Delta p\approx h/\lambda. Using a longer wavelength gives a smaller kick but a blurred position. In every case
which agrees in order of magnitude with \Delta x\,\Delta p\ge\hbar.
Explanation: wave nature
A particle is described by a wave. Position means where the wave is concentrated; momentum is fixed by the wavelength, p=h/\lambda. A long wave has a definite wavelength (definite momentum) but no definite position. A short pulse has a definite position, but it is made by mixing many wavelengths, so its momentum is uncertain.
Energy and time
A similar relation holds for energy and time. If a state lasts for a time \Delta t, its energy is uncertain by \Delta E, where
A metastable state that lasts a long time has a sharply defined energy; a short-lived state has a large uncertainty in energy.
Examples
- Electron located within \Delta x=10^{-10}\,\mathrm{m} (size of an atom): \Delta p\ge\hbar/\Delta x=1.05\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}, which corresponds to a speed uncertainty of about 10^{6}\,\mathrm{m\,s^{-1}}. This is huge for an electron.
- Ball of mass 0.1\,\mathrm{kg} with \Delta p=3\times10^{-6}\,\mathrm{kg\,m\,s^{-1}}: \Delta x\ge3.5\times10^{-29}\,\mathrm{m}, far too small to notice.
Conclusion
These uncertainties are a property of nature, not of poor instruments or of the observer's skill. Every measurement needs an interaction that disturbs the object, so every empirical measurement must carry some uncertainty. Because h is so small, the effect matters only for atomic particles.
Definition
Pair production is the creation of a particle and its antiparticle from energy. When a high-energy gamma-ray photon passes close to a heavy nucleus, it disappears and an electron (e^-) and a positron (e^+) are created. The photon is completely absorbed.
Energy converted into mass
Before the event there is only electromagnetic energy (the photon has no rest mass). Afterwards there are two particles, each of rest mass m_0. Energy has been converted into mass, exactly as Einstein's mass-energy relation E=mc^{2} predicts: creating a rest mass m_0 needs energy m_0c^{2}.
Conservation laws
- Charge: the photon has zero charge; the electron and positron have charges -e and +e, total zero. So the photon can never produce a single electron, a single positron, or two particles of the same charge.
- Momentum: the photon's momentum hf/c must equal the total momentum of the products, \dfrac{hf}{c}=mv_{e^+}+mv_{e^-} (plus the recoil of the nucleus). A pair alone cannot carry all of the photon's momentum, so the heavy nucleus takes up the rest; this is why the process needs a nucleus.
- Energy: the photon energy equals the total energy (rest energy plus kinetic energy) of the two particles.
Minimum (threshold) energy
Derivation Threshold energy for pair production
Aim. Find the least photon energy that can create a pair.
- The photon energy is least when the pair has no kinetic energy.hf_{\min}=2m_0c^{2}
- Rest energy of an electron.m_0c^{2}=(9.11\times10^{-31})(3\times10^{8})^{2}=8.2\times10^{-14}\,\mathrm{J}=0.511\,\mathrm{MeV}
- Double it.hf_{\min}=1.02\,\mathrm{MeV}=1.64\times10^{-13}\,\mathrm{J}
- Corresponding largest wavelength.\lambda_{\max}=\dfrac{hc}{hf_{\min}}=1.21\times10^{-12}\,\mathrm{m}
Only gamma-ray photons can produce pairs; any energy above 1.02\,\mathrm{MeV} becomes kinetic energy of the electron and positron.
Example
A 3.0\,\mathrm{MeV} photon: 1.02\,\mathrm{MeV} becomes the rest mass of the pair and the remaining 1.98\,\mathrm{MeV} becomes their kinetic energy.
Pair production is therefore a practical demonstration of Einstein's equation E=mc^{2}: electromagnetic energy is turned into matter.
Note. The book prints Eisenstein for Einstein; the question refers to Einstein's mass-energy equation E=mc^{2}.
Definition
Pair annihilation is the process in which an electron and its antiparticle, the positron, combine and disappear, and their mass is converted into two gamma-ray photons. It is the reverse of pair production.
Mass converted into energy
Before annihilation there are two particles, each of rest mass m_0. Afterwards there is no rest mass at all, only electromagnetic energy. The whole mass 2m_0 has been converted into energy E=(2m_0)c^{2}, as Einstein's relation predicts.
Conservation laws
- Charge: (+e)+(-e)=0 before; photons have no charge, so charge is conserved.
- Momentum: the pair is nearly at rest, so its total momentum is about zero. One photon always has momentum hf/c\ne0, so a single photon cannot be produced. Two photons of equal energy moving in opposite directions have a total momentum of zero, so momentum is conserved.
- Energy: the total energy of the pair (rest energy plus kinetic energy) equals the energy of the two photons.
Energy of each photon
Derivation Photon energy in annihilation at rest
Aim. Find the energy and wavelength of each photon when the pair is at rest.
- Total energy available is the rest energy of the two particles.2hf=2m_0c^{2}
- Each photon gets half.hf=m_0c^{2}=0.511\,\mathrm{MeV}=8.2\times10^{-14}\,\mathrm{J}
- Wavelength of each photon.\lambda=\dfrac{h}{m_0c}=2.43\times10^{-12}\,\mathrm{m}
The photons are gamma rays. Their energy comes entirely from the mass of the particles.
Examples and uses
Proton-antiproton annihilation is another example. In medicine, PET scanners detect the pairs of 0.511\,\mathrm{MeV} photons produced when positrons from a tracer annihilate with electrons in the body.
Annihilation is therefore a practical demonstration of E=mc^{2} in the direction mass into energy, while pair production shows energy into mass. Together they confirm the photon model of radiation and Einstein's idea of rest mass energy.
Note. The book prints Eisenstein for Einstein.
Meaning
Dual nature of light means that light behaves as a wave in some experiments and as a stream of particles (photons) in others. Neither picture alone explains everything, so light has both wave and particle characteristics (wave-particle duality).
Evidence for the wave nature
- Interference: in Young's double slit experiment light from two slits produces bright and dark fringes. Only waves can add and cancel in this way.
- Diffraction: light spreads round the edges of narrow slits and a diffraction grating gives sharp maxima.
- Polarization: light can be polarized, which shows that it is a transverse wave.
Evidence for the particle nature
- Photoelectric effect: threshold frequency, instant emission and K.E_{\max}=hf-\Phi independent of intensity can be explained only by photons of energy hf, each absorbed by one electron.
- Compton effect: X-ray photons collide with electrons like particles and transfer momentum h/\lambda, so the scattered wavelength increases.
- Pair production and annihilation: a single photon turns into an electron-positron pair, and a pair turns into photons, conserving energy and momentum like particles.
Linking the two natures
The particle properties of a photon are fixed by its wave properties:
The two natures complement each other: light travels (spreads, interferes, diffracts) as a wave, but it is emitted and absorbed as particles, one photon at a time. Interference and diffraction experiments show the wave behaviour; photoelectric and Compton experiments show the particle behaviour.
| Wave nature | Particle nature |
|---|---|
| Interference | Photoelectric effect |
| Diffraction | Compton effect |
| Polarization | Pair production and annihilation |
| Described by \lambda and f | Described by E=hf and p=h/\lambda |
Extension to matter
In 1924 de Broglie suggested that the duality works both ways: particles such as electrons have a wavelength \lambda=h/mv. Electron diffraction confirmed this, so wave-particle duality is a general property of both radiation and matter.
Photon idea
In 1900 Planck showed that atoms emit and absorb radiation only in discrete packets (quanta) of energy hf. In 1905 Einstein proposed that radiation itself travels as these packets, called photons. A photon is a particle of electromagnetic radiation with:
- energy E=hf=hc/\lambda,
- momentum p=E/c=h/\lambda,
- zero rest mass and speed c.
A beam of light is a stream of photons; its intensity is set by the number of photons per second, and the energy of each photon by its frequency.
Evidence 1: photoelectric effect
Electrons are emitted only if f\ge f_0, emission is instant, and K.E_{\max} depends on frequency but not on intensity. These results follow from hf=\Phi+K.E_{\max}, where one photon is absorbed by one electron. A continuous wave cannot explain them.
Evidence 2: Compton effect
X-ray photons scattered by electrons have a longer wavelength, \Delta\lambda=\dfrac{h}{m_0c}(1-\cos\theta). This is explained only by treating each photon as a particle with momentum h/\lambda colliding with an electron, conserving energy and momentum.
Evidence 3: pair production and annihilation
A single gamma photon of energy at least 1.02\,\mathrm{MeV} turns into an electron and a positron, and an electron-positron pair turns into two photons of 0.511\,\mathrm{MeV} each. Radiation behaves as discrete particles of definite energy and momentum.
Evidence 4: radiation pressure and line spectra
Light exerts a force F=P/c on a surface because each photon carries momentum. Atoms emit and absorb light only at definite wavelengths because each transition creates or destroys one photon of energy hf=E_n-E_p.
Conclusion
All these phenomena show that electromagnetic radiation is emitted, absorbed and scattered in discrete units (photons). Interference and diffraction show its wave nature, so radiation has a dual nature, but its interaction with matter is particulate.
Principle
An electron microscope uses a beam of fast electrons instead of light. It works because moving electrons behave as waves (de Broglie), and their wavelength is extremely short.
Resolution depends on wavelength
Resolution is the ability to show two close points as separate. Because of diffraction, no microscope can resolve details much smaller than the wavelength of the radiation it uses. Visible light has \lambda=400 to 700\,\mathrm{nm}, so an optical microscope cannot resolve details below about 200\,\mathrm{nm}.
Wavelength of the electrons
Derivation Wavelength of an accelerated electron
Aim. Find the de Broglie wavelength of electrons accelerated through a p.d. V.
- Kinetic energy gained from the field.\tfrac12mv^{2}=eV\ \Rightarrow\ p=mv=\sqrt{2meV}
- de Broglie relation.\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2meV}}
- For V=10\,000\,\mathrm{V}:\lambda=\dfrac{6.63\times10^{-34}}{\sqrt{2(9.11\times10^{-31})(1.6\times10^{-19})(10^{4})}}=1.2\times10^{-11}\,\mathrm{m}=0.12\,\text{\AA}
Higher accelerating voltage gives shorter wavelength. At 10\,\mathrm{kV} the electron wavelength is tens of thousands of times shorter than that of visible light.
In practice imperfect magnetic lenses limit the resolution to about 0.1 to 0.2\,\mathrm{nm}, still about a thousand times better than an optical microscope. Very small objects such as viruses, and even rows of atoms in crystals, can be seen.
Construction and working
- A heated tungsten filament emits electrons; a high positive potential on the anode accelerates them into a narrow beam.
- A magnetic condenser lens (current-carrying coil) focuses the beam onto a very thin specimen.
- The objective lens forms a magnified image; the intermediate and projector lenses magnify it further.
- The image is formed on a fluorescent screen or detector and can be photographed (electron micrograph).
- The whole column is kept in a vacuum so the electrons are not scattered by air.
Comparison
| Optical microscope | Electron microscope | |
|---|---|---|
| Radiation | Light, \lambda\approx500\,\mathrm{nm} | Electrons, \lambda\approx0.01\,\mathrm{nm} |
| Lenses | Glass | Magnetic coils |
| Resolution | about 200\,\mathrm{nm} | about 0.1\,\mathrm{nm} |
| Medium | Air | Vacuum |
Thus the very short de Broglie wavelength of fast electrons is what gives the electron microscope its very high resolution.
Discrete energy levels
The electrons of an isolated atom (for example in a gas at low pressure) can occupy only certain allowed orbits. Each orbit has a definite energy E_1, E_2, E_3, ... These are the energy levels; E_1 is the ground state. An electron can change level only by gaining or losing exactly the energy difference between two levels.
E_n: higher level, E_p: lower level.
Because the levels are discrete, only certain energy differences exist, so only photons of certain definite frequencies can be emitted or absorbed. The spectra prove this.
Formation of the emission line spectrum
Atoms of a gas in a discharge tube are excited (by electric current or heating) to higher levels. When the electrons fall back to lower levels, each transition emits a photon of energy hf=E_n-E_p. The light is passed through a slit and a prism or diffraction grating.
Appearance: a set of separate bright lines of definite wavelengths on a dark background. There is no continuous band, only lines, because only certain energy jumps are possible.
Formation of the absorption line spectrum
White light (containing all wavelengths) from a tungsten lamp is passed through the cool vapour of the element. Atoms in the ground state absorb only those photons whose energy exactly equals E_n-E_1 for some level, and jump up. The rest of the light passes through. The excited atoms re-emit the absorbed energy in all directions, so little of it continues in the original direction.
Appearance: a continuous coloured spectrum crossed by dark lines at definite wavelengths.
Conclusion
- Both spectra contain only discrete wavelengths, never a continuous range, so the energies of the electrons in the atom must be discrete.
- The dark absorption lines occur at exactly the same wavelengths as the bright emission lines of the same element, because both come from jumps between the same fixed energy levels.
- Each element has its own set of levels, so its line spectrum is unique and can be used to identify it.
Numerical problemsp. 140
Constants used, as in the book: h=6.626\times10^{-34}\,\mathrm{J\,s} (or 6.63\times10^{-34}), c=3\times10^{8}\,\mathrm{m\,s^{-1}}, e=1.6\times10^{-19}\,\mathrm{C}, m_0=9.11\times10^{-31}\,\mathrm{kg}, m_0c^{2}=0.511\,\mathrm{MeV}.
- \Phi = 2.1\,\mathrm{eV} work function
- V_0 = 0.43\,\mathrm{V} stopping potential
- h = 6.626\times10^{-34}\,\mathrm{J\,s}
- c = 3\times10^{8}\,\mathrm{m\,s^{-1}}
- wavelength \lambda
- The stopping potential gives the maximum kinetic energy: K.E_{\max}=eV_0. With V_0 in volts the energy in eV has the same number.K.E_{\max}=eV_0=0.43\,\mathrm{eV}
- Einstein's equation: photon energy = work function + maximum kinetic energy.E=\Phi+K.E_{\max}=2.1+0.43=2.53\,\mathrm{eV}
- Change into joules.E=2.53\times1.6\times10^{-19}=4.05\times10^{-19}\,\mathrm{J}
- Use E=hc/\lambda and make \lambda the subject.\lambda=\dfrac{hc}{E}=\dfrac{6.626\times10^{-34}\times3\times10^{8}}{4.05\times10^{-19}}=4.91\times10^{-7}\,\mathrm{m}
Book answer. 5.91\times10^{-7}\,\mathrm{m}
Note. The book's answer 5.91\times10^{-7}\,\mathrm{m} is wrong: it is hc/\Phi, the threshold wavelength of potassium, which ignores the stopping potential. Including K.E_{\max}=0.43\,\mathrm{eV} gives 4.91\times10^{-7}\,\mathrm{m}.
- \lambda = 5000\,\text{\AA}=5000\times10^{-10}\,\mathrm{m}=5\times10^{-7}\,\mathrm{m}
- \Phi = 1.9\,\mathrm{eV}
- h = 6.626\times10^{-34}\,\mathrm{J\,s}
- E in eV
- Photon energy in joules.E=\dfrac{hc}{\lambda}=\dfrac{6.626\times10^{-34}\times3\times10^{8}}{5\times10^{-7}}=3.98\times10^{-19}\,\mathrm{J}
- Divide by 1.6\times10^{-19}\,\mathrm{J} per eV.E=\dfrac{3.976\times10^{-19}}{1.6\times10^{-19}}=2.48\,\mathrm{eV}
- Einstein's equation, working in eV.K.E_{\max}=E-\Phi=2.48-1.9=0.58\,\mathrm{eV}
- In joules, if required.K.E_{\max}=0.58\times1.6\times10^{-19}=9.3\times10^{-20}\,\mathrm{J}
- K.E_{\max}=eV_0, so V_0=K.E_{\max}/e.V_0=\dfrac{0.58\times1.6\times10^{-19}\,\mathrm{J}}{1.6\times10^{-19}\,\mathrm{C}}=0.58\,\mathrm{V}
Book answer. 2.48\,\mathrm{eV}, 0.58\,\mathrm{eV}, 0.58\,\mathrm{V}
Note. With h=6.63\times10^{-34}\,\mathrm{J\,s} the photon energy comes out as 2.49\,\mathrm{eV} and K.E_{\max}=0.59\,\mathrm{eV}; the book's values use h=6.626\times10^{-34}\,\mathrm{J\,s}.
For each part use E=hc/\lambda with hc=6.626\times10^{-34}\times3\times10^{8}=1.988\times10^{-25}\,\mathrm{J\,m}, then divide by 1.6\times10^{-19} to get eV.
- Energy in joules.E=\dfrac{1.988\times10^{-25}}{413\times10^{-9}}=4.81\times10^{-19}\,\mathrm{J}
- Energy in eV.E=\dfrac{4.81\times10^{-19}}{1.6\times10^{-19}}=3.0\,\mathrm{eV}
- Energy in joules, with \lambda=0.1\times10^{-9}=10^{-10}\,\mathrm{m}.E=\dfrac{1.988\times10^{-25}}{1\times10^{-10}}=1.988\times10^{-15}\,\mathrm{J}
- Energy in eV.E=\dfrac{1.988\times10^{-15}}{1.6\times10^{-19}}=1.24\times10^{4}\,\mathrm{eV}\approx12\,424\,\mathrm{eV}
- Energy in joules.E=\dfrac{1.988\times10^{-25}}{10}=1.99\times10^{-26}\,\mathrm{J}
- Energy in eV.E=\dfrac{1.988\times10^{-26}}{1.6\times10^{-19}}=1.24\times10^{-7}\,\mathrm{eV}
Book answer. 3\,\mathrm{eV}; 12424\,\mathrm{eV}; 1.24\times10^{-7}\,\mathrm{eV}
The photon gives energy to a free electron by Compton scattering. We need the smallest photon energy E for which the scattered photon can be left with only half its energy, E'=E/2.
- E' = \tfrac12E energy of scattered photon
- m_0c^{2} = 0.511\,\mathrm{MeV} rest energy of the electron
- minimum photon energy E
- Half the energy means the scattered photon has half the frequency, so its wavelength doubles: \lambda'=2\lambda.E'=\dfrac{E}{2}\ \Rightarrow\ \lambda'=2\lambda\ \Rightarrow\ \Delta\lambda=\lambda'-\lambda=\lambda
- Compton equation: the shift needed equals the incident wavelength.\lambda=\dfrac{h}{m_0c}(1-\cos\theta)
- The largest possible shift is at \theta=180^\circ, where (1-\cos\theta)=2. So the incident wavelength can be at most 2h/m_0c.\lambda\le\dfrac{2h}{m_0c}
- Energy is inversely proportional to wavelength, so the minimum energy corresponds to this maximum wavelength (photon scattered straight back).E_{\min}=\dfrac{hc}{\lambda_{\max}}=\dfrac{hc}{2h/m_0c}=\dfrac{m_0c^{2}}{2}
- Substitute the rest energy of the electron.E_{\min}=\dfrac{0.511}{2}=0.256\,\mathrm{MeV}=4.1\times10^{-14}\,\mathrm{J}
Book answer. 0.25\,\mathrm{MeV}
Note. The exact value is m_0c^{2}/2=0.2555\,\mathrm{MeV}; the book rounds it down to 0.25\,\mathrm{MeV}.
- \Delta x = 1\,\text{\AA}=1\times10^{-10}\,\mathrm{m} size of the box
- m = 9.1\times10^{-31}\,\mathrm{kg}
- h = 6.63\times10^{-34}\,\mathrm{J\,s}
- speed v
- The electron is somewhere inside the box, so the uncertainty in its position is the size of the box.\Delta x\approx1\times10^{-10}\,\mathrm{m}
- By the uncertainty principle (rough form \Delta x\,\Delta p\approx h), its momentum cannot be known better than h/\Delta x. A particle whose momentum is uncertain by this much must have a momentum of at least this size.p\approx\Delta p\approx\dfrac{h}{\Delta x}=\dfrac{6.63\times10^{-34}}{1\times10^{-10}}=6.63\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}
- Speed from p=mv.v=\dfrac{p}{m}=\dfrac{6.63\times10^{-24}}{9.1\times10^{-31}}=7.29\times10^{6}\,\mathrm{m\,s^{-1}}
Book answer. 7.29\times10^{6}\,\mathrm{m\,s^{-1}}
- E = 50\,\mathrm{keV}=50\times10^{3}\times1.6\times10^{-19}=8.0\times10^{-15}\,\mathrm{J}
- \theta = 90^\circ
- \dfrac{h}{m_0c} = 2.43\times10^{-12}\,\mathrm{m} Compton wavelength
- energy E' of the scattered X-ray
- Wavelength of the incident X-ray from E=hc/\lambda.\lambda=\dfrac{hc}{E}=\dfrac{6.626\times10^{-34}\times3\times10^{8}}{8.0\times10^{-15}}=2.485\times10^{-11}\,\mathrm{m}
- Compton shift at 90^\circ, where \cos90^\circ=0.\Delta\lambda=\dfrac{h}{m_0c}(1-\cos90^\circ)=2.43\times10^{-12}\,\mathrm{m}
- Wavelength of the scattered X-ray.\lambda'=\lambda+\Delta\lambda=2.485\times10^{-11}+0.243\times10^{-11}=2.728\times10^{-11}\,\mathrm{m}
- Energy is inversely proportional to wavelength, so scale the incident energy.E'=E\,\dfrac{\lambda}{\lambda'}=50\times\dfrac{2.485}{2.728}=45.5\,\mathrm{keV}
Book answer. 45.5\,\mathrm{keV}
FBISE 2026 questions
Questions from the FBISE HSSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.
- AKinetic energy of electron
- BStopping potential
- CNumber of emitted electronsAnswer
- DWork function
Brighter light means more photons per second, and each photon frees one electron, so more electrons are emitted per second (larger current). The kinetic energy and stopping potential depend on frequency, not intensity, and the work function is a property of the metal.
- A\lambda=\dfrac{3h}{mv}
- B\lambda=\dfrac{h}{mv}Answer
- C\lambda=\dfrac{2h}{mv}
- D\lambda=\dfrac{h}{2mv}
de Broglie's relation is \lambda=h/p with momentum p=mv, so \lambda=h/mv. The other options put wrong numerical factors in front.
- ARedAnswer
- BGreen
- CBlue
- DViolet
Red light has the longest wavelength in the visible band (about 750\,\mathrm{nm}). Since f=c/\lambda, it has the lowest frequency. Violet has the highest.
Pair production (1 mark)
Pair production is the creation of a particle and its antiparticle from energy. When a high-energy gamma-ray photon passes near a heavy nucleus, it is completely absorbed and an electron-positron pair is created. Charge, energy and momentum are conserved; energy is converted into mass.
Minimum energy (2 marks)
- Energy conservation: the photon supplies the rest mass energy of both particles and their kinetic energies.hf=2m_0c^{2}+K.E_{e^-}+K.E_{e^+}
- The energy is minimum when the particles are created with zero kinetic energy.hf_{\min}=2m_0c^{2}
- Substitute m_0=9.11\times10^{-31}\,\mathrm{kg} and c=3\times10^{8}\,\mathrm{m\,s^{-1}}.hf_{\min}=2(9.11\times10^{-31})(3\times10^{8})^{2}=1.64\times10^{-13}\,\mathrm{J}
- Change into MeV (1\,\mathrm{MeV}=1.6\times10^{-13}\,\mathrm{J}).hf_{\min}=\dfrac{1.64\times10^{-13}}{1.6\times10^{-13}}\,\mathrm{MeV}=1.02\,\mathrm{MeV}
The resolution of a microscope (its ability to show two close points separately) is limited by diffraction: it cannot show details much smaller than the wavelength of the radiation used. So the shorter the wavelength, the higher the resolution.
An electron microscope uses fast electrons, which behave as waves with de Broglie wavelength
where V is the accelerating potential difference. For V=10\,\mathrm{kV}, \lambda\approx1.2\times10^{-11}\,\mathrm{m} (0.12\,\text{\AA}), whereas visible light has \lambda\approx4000 to 7000\,\text{\AA}. The electron wavelength is thousands of times shorter, so the electron microscope resolves details down to about 0.1 to 0.2\,\mathrm{nm}, compared with about 200\,\mathrm{nm} for an optical microscope. Increasing V makes \lambda even shorter and the resolution even higher.
Photoelectric effect (1 mark)
The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency falls on it. The emitted electrons are called photoelectrons.
No emission below threshold frequency (2 marks)
According to Einstein, light consists of photons of energy hf, and one photon is absorbed by one electron. The electron needs at least the work function \Phi to escape:
If f<f_0, the energy of each photon is less than the work function (hf<\Phi), so no single photon can free an electron. An electron cannot store the energy of one photon and add the energy of another, so increasing the intensity (sending more photons) still gives no emission. Hence no electrons are emitted below the threshold frequency, however bright the light.
Effect of intensity on photoelectric current (3 marks)
Keeping the frequency (above f_0) and the potential difference fixed, the photoelectric current increases linearly with the intensity. Intensity is the number of photons arriving per second; each photon releases one electron, so the number of photoelectrons per second, and hence the current, is proportional to intensity.
When the current is plotted against the collector potential for different intensities, brighter light gives a larger saturation current, but the current falls to zero at the same stopping potential V_0 for all intensities. So intensity increases the number of electrons but not their maximum kinetic energy, K.E_{\max}=eV_0.
- AGamma raysAnswer
- BUltraviolet
- CMicrowaves
- DInfrared
Gamma rays have the shortest wavelength (about 10^{-12}\,\mathrm{m}) in the electromagnetic spectrum, so from f=c/\lambda they have the highest frequency (and photon energy). The order of increasing frequency is microwaves, infrared, ultraviolet, gamma rays.
- A1.3\times10^{-34}\,\mathrm{mm}
- B1.3\times10^{-24}\,\mathrm{mm}
- C1.3\times10^{-24}\,\mathrm{m}
- D1.3\times10^{-34}\,\mathrm{m}Answer
This is the book's own example in section 21.6. Watch the unit: the answer is in metres, not millimetres.
- AEnergy of incident light
- BWavelength of incident light
- CIntensity of Incident lightAnswer
- DFrequency of incident light
Each photon releases one electron, so the number of electrons emitted per second is proportional to the number of photons per second, that is, to the intensity. Frequency, wavelength and photon energy decide the kinetic energy of the electrons (and whether emission happens at all), not their number.
Pair annihilation (1 mark)
Pair annihilation is the process in which an electron and a positron (its antiparticle) combine and disappear, producing two gamma-ray photons. Mass is converted into energy according to E=mc^{2}. It is the reverse of pair production.
Conservation of energy and momentum (2 marks)
Energy: the total energy of the pair (rest mass energy plus kinetic energy) appears as the energy of the photons:
For a pair at rest each photon has hf=m_0c^{2}=0.511\,\mathrm{MeV}, so the total energy before and after is 1.02\,\mathrm{MeV}.
Momentum: the electron and positron are nearly at rest, so their total momentum is about zero. A single photon always carries momentum hf/c, so one photon cannot be produced. Two photons of equal energy moving in opposite directions have equal and opposite momenta (+hf/c and -hf/c), giving zero total momentum. Hence momentum is conserved. Charge is also conserved: (+e)+(-e)=0.
Reason (1.5 marks)
The resolution of any microscope is limited by the wavelength of the radiation it uses: details smaller than about one wavelength cannot be seen. An electron microscope uses fast electrons whose de Broglie wavelength \lambda=h/\sqrt{2meV} is extremely short (about 0.01\,\mathrm{nm} at 10\,\mathrm{kV}), thousands of times shorter than visible light (400 to 700\,\mathrm{nm}). So it can resolve much finer detail.
Comparison (1.5 marks)
| Optical microscope | Electron microscope | |
|---|---|---|
| Illumination | Visible light | Beam of fast electrons |
| Wavelength | 400 to 700\,\mathrm{nm} | about 0.01\,\mathrm{nm} |
| Lenses | Glass lenses | Magnetic lenses (coils) |
| Resolution | about 200\,\mathrm{nm} | about 0.1 to 0.2\,\mathrm{nm} |
| Medium and specimen | Air; living specimens possible | Vacuum; thin, non-living specimen |
Photoelectric effect (1 mark)
The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency falls on it. The emitted electrons are called photoelectrons.
Work function and Einstein's equation (3 marks)
The work function \Phi of a metal is the minimum energy needed to remove an electron from its surface. It is different for different metals (for example about 2.3\,\mathrm{eV} for sodium and 4.3\,\mathrm{eV} for zinc). It is related to the threshold frequency f_0 by \Phi=hf_0.
Derivation Einstein's photoelectric equation
Aim. Relate photon energy, work function and maximum kinetic energy.
- Light is a stream of photons; one photon of frequency f gives its whole energy hf to one electron.E=hf
- Part of the energy is used as the work function to free the electron; the rest is its kinetic energy (maximum for surface electrons).hf=\Phi+K.E_{\max}
- At threshold, K.E_{\max}=0 and f=f_0, so \Phi=hf_0. Substituting:K.E_{\max}=\tfrac12mv_{\max}^{2}=hf-hf_0
Emission occurs only when hf\ge\Phi; the extra energy appears as kinetic energy.
Effect of frequency on K.E of photoelectrons (2 marks)
From K.E_{\max}=hf-\Phi: below the threshold frequency f_0 no electrons are emitted. Above f_0, the maximum kinetic energy increases linearly with the frequency; the graph is a straight line of slope h, cutting the frequency axis at f_0. Increasing the frequency (for example from red to violet to ultraviolet light) gives faster photoelectrons, whatever the intensity.
Definition (1 mark)
One electronvolt is the energy gained (or lost) by an electron when it moves through a potential difference of one volt.
Relation with the joule (2 marks)
- Work done (energy gained) when a charge q moves through a p.d. V.W=qV
- For an electron, q=e=1.6\times10^{-19}\,\mathrm{C}, and V=1\,\mathrm{V}.1\,\mathrm{eV}=(1.6\times10^{-19}\,\mathrm{C})(1\,\mathrm{V})
- Since 1\,\mathrm{C}\times1\,\mathrm{V}=1\,\mathrm{J}:1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}
Also 1\,\mathrm{keV}=1.6\times10^{-16}\,\mathrm{J} and 1\,\mathrm{MeV}=1.6\times10^{-13}\,\mathrm{J}. The electronvolt is a convenient small unit for atomic and nuclear energies.
The maximum kinetic energy of photoelectrons is found by measuring the stopping potential V_0. In the photoelectric set-up the collector is made negative with respect to the emitter (battery reversed), and the retarding potential is increased until the photoelectric current just becomes zero. At this potential even the fastest electrons are stopped, so their kinetic energy has all been used in doing work eV_0 against the field:
For example, if V_0=2\,\mathrm{V} then K.E_{\max}=2\,\mathrm{eV}=3.2\times10^{-19}\,\mathrm{J}. If the frequency and the work function are known instead, the same energy follows from Einstein's equation K.E_{\max}=hf-\Phi.
Photoelectric effect (2 marks)
The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency falls on it. The emitted electrons are called photoelectrons and the current they form is the photoelectric current.
Failure of classical physics (2 marks)
Classical physics treats light as a continuous wave whose energy depends on intensity. It fails to explain:
- Threshold frequency: no emission below f_0 however intense the light (wave theory predicts emission at any frequency if the light is bright enough).
- Kinetic energy independent of intensity: K.E_{\max} does not increase with intensity.
- Kinetic energy depends on frequency: K.E_{\max} increases linearly with f.
- No time lag: emission is instantaneous even in very dim light.
Einstein's photoelectric equation (3 marks)
Derivation Einstein's photoelectric equation
Aim. Derive the relation between photon energy, work function and maximum kinetic energy.
- Light consists of photons, each of energy hf. One photon is absorbed completely by one electron.E=hf
- The electron uses energy \Phi (work function) to escape from the surface; by conservation of energy the remainder is its maximum kinetic energy.hf=\Phi+K.E_{\max}
- At the threshold frequency the kinetic energy is zero, so the work function is hf_0.\Phi=hf_0
- Substitute and write the kinetic energy in terms of speed and stopping potential.\tfrac12mv_{\max}^{2}=hf-hf_0=eV_0
This equation explains every observation: no emission if hf<\Phi; K.E_{\max} rises linearly with f (slope h) and is independent of intensity; the number of electrons depends on the number of photons.
Exam focus
- Photoelectric effect: definition, observations with graphs, failure of classical physics and derivation of Einstein's equation hf=\Phi+K.E_{\max}. (long)The most examined topic of the unit: a Section C question in all three 2026 papers (Set B Q.5 OR, Set D Q.5 OR, old syllabus Q.4).Go to answer
- Draw and explain the graphs: current against collector potential for different intensities, current against intensity, and K.E_{\max} against frequency. (long)Graphs earn marks in Section C (Set B Q.5 OR asked for the effect of intensity with a graph; Set D asked for the effect of frequency).Go to answer
- Stopping potential and K.E_{\max}=eV_0; how to find \Phi and h from a V_0 against f graph. (short)Old-syllabus 2026 Q.2 (i) OR asked how kinetic energy of photoelectrons is calculated; book short question 5 asks about the graph.Go to answer
- Photoelectric numericals: photon energy in eV, K.E_{\max}, stopping potential and wavelength. (numerical)Direct use of hf=\Phi+eV_0; book numericals 1 and 2 (note the misprinted answer of numerical 1).Go to answer
- Pair production: definition, conservation laws and minimum energy 2m_0c^{2}=1.02\,\mathrm{MeV}. (short)Asked in 2026 Set B Q.2 (vii) (1 + 2 marks); also a comprehensive question in the book.Go to answer
- Pair annihilation: why two photons, conservation of energy and momentum, 0.511\,\mathrm{MeV} per photon. (short)Asked in 2026 Set D Q.2 (vii) (1 + 2 marks).Go to answer
- Electron microscope: why electron wavelength gives very high resolution; comparison with an optical microscope. (short)Asked in both 2026 new-syllabus sets (Set B and Set D Q.2 (viii) OR).Go to answer
- de Broglie wavelength \lambda=h/mv and its numericals (convert km/h and mg to SI units). (mcq)Two 2026 MCQs (Set B MCQ 12 and Set D MCQ 13) and book examples 21.4 and assignment 21.4.Go to answer
- Compton effect: explanation with photon collisions, Compton shift \Delta\lambda=\dfrac{h}{m_0c}(1-\cos\theta) and its significance. (long)Book comprehensive question 2 and several short questions (8, 9, 10) and numericals 4 and 6 come from this section.Go to answer
- Heisenberg's uncertainty principle: statement, \Delta x\,\Delta p\ge\hbar, \Delta E\,\Delta t\ge\hbar and simple numericals. (long)Book comprehensive question 3, MCQ 8 and numerical 5; state which form of the principle you use.Go to answer
- Electronvolt: definition and derivation of 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}; converting MeV to joules. (short)Old-syllabus 2026 Q.2 (i) (1 + 2 marks) and book short question 2.Go to answer
- Discrete energy levels and the formation of emission and absorption line spectra, with hf=E_n-E_p. (long)Book comprehensive question 9 and two learning outcomes; energy level diagrams earn marks.Go to answer