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22.1Photons and the photoelectric effect

Photons

Light is said to be an electromagnetic wave a transverse wave moving at the speed of light in vacuum and consisting of oscillations of an electric and a magnetic field at right angles to each other consisting of oscillating electric and magnetic fields. This was Maxwell’s great discovery in the nineteenth century. The wave has a wavelength λ and a frequency f and, as with all waves, the wave speed c is given by:

c=f λ

where in this case the wave speed is the speed of light.

Through Maxwell’s theory, complex phenomena such as diffraction, interference, polarisation and others could be understood. The successful application of Maxwell’s theory meant that light was definitely and without any doubt a wave. It therefore came as a shock that in a phenomenon known as the photoelectric effect, light did not behave as a wave should.

As we will see, Einstein suggested that light should be thought of as a collection of quanta, or bundles of energy. Each quantum a small quantity of something, for example, energy or bundle of light has energy E given by E = hf, where f is the frequency of the light and h is Planck’s constant. A beam of light of frequency f is now to be thought of as a very large number of these quanta moving at the speed of light. The total energy of the beam is then the product of hf (the energy of one quantum) times N, the number of quanta in the beam. The energy of the beam is therefore an integral multiple of the basic unit hf. These quanta have energy and momentum and are localised in space; this means they behave as particles. But the theory of relativity states that if a particle moves at the speed of light it has to have zero mass. So this quantum of light, which came to be known as the photon the particle of light (a quantum of energy), a massless particle moving at the speed of light, is a particle with zero mass and zero electric charge.

Check yourself 1

Which has more energy, a photon of blue light or a photon of red light?

In Chapter 21 we saw that a photon will be emitted when an atom makes a transition from a high to a lower energy state. Its energy is the energy difference of the two levels. A photon can also be absorbed by an atom. An atom in a low energy state can absorb a photon of just the right energy and make a transition to a higher energy level. When we look at the light from a light bulb we see a continuous emission of light. But if we could slow down the process by a few billion times, the continuity in the emission of light would stop. We would see different spots on the filament emit tiny flashes of light (photons) at random intervals of time; the spots on the filament would be on (emitting) and off (not emitting) randomly. The discreteness of energy we talked about in Chapter 21 would surface again.

In Einstein’s theory of special relativity the total energy E, the momentum p and the mass m of a particle are related according to:

E 2=p 2 c 2+m 2 c 4

The mass of the photon is zero, so E = pc. The photon therefore has momentum p=E c. (The conventional Newtonian formula for momentum, p = mv, does not apply to particles with zero mass.) So the momentum of the photon is:

p=E c=h f c=h λ

Check yourself 2

The wavelength of a beam of light is decreased, but the intensity of the beam stays the same. What happens to the number of photons in the beam?

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Worked Example 22.1

Estimate how many photons of wavelength 5.0 × 10−7 m are emitted per second by a 60 W lamp, assuming that 10% of the energy of the lamp goes into photons of this wavelength.

Answer

Let there be N photons per second emitted. Then the energy emitted is N h c λ in one second.

This has to be 10% of 60 J, that is 6.0 J.

So:

N h c λ=6.0

N=6.0 λ h c

N=6.0×5.0×10−7 6.63×10−34×3.0×10 8

⇒N=1.5×10 19 photons per second

Worked Example 22.2

All the photons from worked example 1 are incident normally on a mirror of area 0.5 m2 and are reflected by it. Estimate the pressure these photons exert on the mirror.

Answer

Each photon has momentum h λ. The momentum change upon reflection is 2 h λ (momentum is a vector!).

Since there are N such reflections per second, the force F on the mirror is:

F=2 N h λ F=2×1.5×10 19×6.63×10−34 5.0×10−7⇒F=4.0×10−8 N

The pressure is thus:

F A=8.0×10−8 Nm−2

(Note that if the photons were absorbed rather than reflected, the pressure would be half that obtained here.)

The photoelectric effect

The photoelectric effect the emission of electrons from a metallic surface when light is incident on the surface is the phenomenon in which light (or other forms of electromagnetic radiation) incident on a metallic surface causes electrons to be emitted from the surface.

To investigate the facts about the photoelectric effect, apparatus like the one in Figure 22.1 may be used.

Figure 22.1: Apparatus for investigating the photoelectric effect. The collecting plate is negatively charged and so decelerates the emitted electrons. At the right voltage the electrons can be stopped, resulting in zero current in the circuit.

It consists of an evacuated tube, inside which is the photo-surface technical term for the metallic surface that releases electrons when light is incident on it (the metallic surface that light is incident on). Light passes through the tube and falls on the photo-surface, which emits electrons. The emitted electrons are called photoelectrons the electrons emitted in the photoelectric effect. Some of the emitted electrons arrive at the collecting plate. The photo-surface and the collecting plate are part of a circuit as shown. Those electrons that make it to the collecting plate complete the circuit and so we have an electric current that is recorded by the sensitive galvanometer.

Notice that in Figure 22.1 the terminal of the variable power supply connected to the collecting plate is the negative terminal. This means that the collecting plate actually repels the emitted electrons. Only the very energetic electrons will make it to the plate. As the magnitude of the voltage is increased (i.e. made more negative) fewer and fewer electrons make it to the plate; eventually no electron will arrive there and at that point the current becomes zero. The voltage at which the current becomes zero is called the stopping voltage the voltage in the photoelectric effect that makes the current zero, V s. Its significance is that the maximum kinetic energy of the emitted electrons is e V s. We see this as follows: the maximum kinetic energy of an electron is E max as it leaves the photo-surface; the electrical potential energy is (−e)V s, and so the total energy at the photo-surface is E max − e V s. At the collecting plate the total energy is zero (zero kinetic and zero potential) and so e V s = E max.

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We now connect the positive terminal of the power supply to the collecting plate. The electrons are now attracted to the collecting plate and the current increases. As the voltage is increased even more the current saturates, i.e. it approaches a constant value. This is because the collecting plate is so positive that it attracts every single emitted electron (even those that were not directed at the collected plate). So we have a current–voltage graph like the one in Figure 22.2.

Figure 22.2: When the collecting plate is connected to the negative terminal of the power supply, there is a voltage at which the current becomes zero (Vs). This is 0.4 V in this example.

Exam tip

It is important to understand that the stopping voltage is related to the maximum kinetic energy of the emitted electrons.

Worked Example 22.3

Using the graph of Figure 22.2 determine:

athe stopping voltage

bthe maximum energy of the emitted electrons

cthe maximum speed of the emitted electrons.

Answer

aThe current becomes zero when the voltage is −0.40 V, so the stopping voltage is 0.40 V.

bThe maximum kinetic energy of the emitted electrons is 0.40 eV = 6.4 × 10−20 J.

cFrom E max=1 2 m v 2, we find v=2 E max m, giving:

v=2×6.4×10−20 9.1×10−31=3.8×10 5 m s−1

The results of this experiment reveal two immediate surprises:

The first is that changing the intensity of the light (without changing the wavelength) does not affect the stopping voltage. Light from a candle and light from an airport searchlight release electrons of the same kinetic energy, and so the stopping voltage is the same, Figure 22.3. Higher intensity means more electrons are emitted with no change in kinetic energy.

Figure 22.3: The stopping voltage for weak light (thin line) and intense light (thick line) of the same frequency. The stopping voltage is independent of the intensity of the light source.

The second is that the kinetic energy of the emitted electrons increases with increasing light frequency. This means the stopping voltage also increases with frequency. This is shown in Figure 22.4: the violet curve corresponds to violet light and the green curve to green light of lower frequency. The stopping voltages are 0.40 V for green and 0.90 V for violet.

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Figure 22.4: The stopping voltages for green and violet light.

Check yourself 3

Which of the graphs, X or Y, in Figure 22.5 corresponds to light of higher frequency?

Figure 22.5: For check yourself question 3.

If we plot the kinetic energy of the electrons (which equals e V s) versus frequency, we find a straight line as shown in Figure 22.6a.

The puzzling feature of this graph is that there exists a frequency, called the critical (or threshold) frequency f c, such that no electrons at all are emitted if the frequency of the light source is less than f c. This is true even if very intense light is allowed to fall on the photo-surface. When the experiment is repeated with a different photo-surface and the kinetic energy of the electrons is plotted versus frequency, a line parallel to the first is obtained, as shown in Figure 22.6b.

Figure 22.6: a The graph of kinetic energy versus frequency is a straight line. The horizontal intercept is the critical frequency, f c. b When another photo-surface is used, a line parallel to the first is obtained.

The final puzzling observation in these experiments is that the electrons are emitted immediately after the light is incident on the photo-surface, with no apparent time delay.

KEY POINT

We now have four surprising observations:

●The intensity of the incident light does not affect the energy of the emitted electrons.

●The electron kinetic energy depends on the frequency of the incident light.

●There is a certain minimum frequency of light below which no electrons are emitted.

●Electrons are emitted with no time delay.

Check yourself 4

The intensity of light incident on a metal is increased without any change to the wavelength. State the effect this change has on:

athe kinetic energy of the emitted electrons

bthe rate at which electrons are emitted.

These four observations cannot be understood in terms of light as a wave for several reasons:

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●If light is a wave, then an intense beam of light carries a lot of energy, and so it should cause the emission of more energetic electrons.

●The formula for the energy of a light wave does not include the frequency, and so frequency should play no role in the energy of the emitted electrons. For the same reason, there can be no explanation for why we have a critical frequency.

●Finally, a very low intensity beam of light carries little energy. An electron might have to wait for a considerable length of time before it accumulated enough energy to escape from the metal. This would cause a delay in its emission.

Exam tip

A simple analogy to see the difference between light as a wave and light as a particle is the following: imagine winning one million euro in a lottery. If this were to be given to you in the wave model of light you would have to wait 10 years to get all the money if the money were paid to you at a rate of 100 000 a year. In the photon model, all the money would be given to you at once.

Einstein’s explanation

The explanation of all these strange observations was provided by Albert Einstein in 1905 (his ‘annus mirabilis’).

Einstein’s mechanism for the photoelectric effect assumes that light is made up of photons, i.e. particles. A single photon of frequency f is absorbed by a single electron in the photo-surface, and so the electron’s energy increases by hf. The electron will have to spend a certain amount of energy, ϕ, to free itself from the pull of the nuclei of the atoms of the photo-surface. The electron will be emitted (become free) if hf is bigger than ϕ. The difference E K = hf – ϕ will be the kinetic energy E K of the (now) free electron, Figure 22.7.

Figure 22.7: a A single photon of light may release a single electron from a metal. b A more tightly bound electron needs more energy to release it from the metal.

The value of ϕ (called the work function ϕ, the amount of energy required to free an electron from a photo-surfrace) is read off the graph in Figure 22.6, from the intercept of the straight line with the vertical axis. Note that the work function and the critical frequency are related by h f c = ϕ, since E K = 0 when f = f c.

In the photoelectric apparatus, the maximum kinetic energy of the electrons is measured to be e V s = E max.

So, from E max = h f − ϕ, it follows that:

e V s=h f−ϕ

V s=h e f−ϕ e.

That is, in a graph of stopping voltage against frequency, the graph is a straight line with slope h e.

Worked Example 22.4

A photo-surface has a work function of 1.50 eV.

aDetermine the critical frequency.

bLight of frequency 6.10 × 1014 Hz falls on this surface. Calculate the energy and speed of the emitted electrons.

Answer

aThe critical frequency f c is given in terms of the work function by h f c = ϕ and thus:

f c=ϕ h=1.50×1.6×10−19 6.63×10−34=3.62×10 14 Hz

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bThe maximum kinetic energy of the electron is E max = h f − ϕ, so:

E max=h f−h f c=h(f−f c)=6.63×10−34×((6.10−3.62)×10 14)=1.64×10−19 J

From E=1 2 m v 2 we find:

v=2 E max m=2×1.64×10−19 9.1×10−31=6.0×10 5 m s−1

Exam tip

Remember to use energy in joules for calculations.

Worked Example 22.5

Monochromatic light of intensity I and wavelength 4.0 × 10−7 m falling on a photo-surface whose critical frequency is 6.0 × 1014 Hz releases 2.0 × 1010 electrons per second.

aDetermine the current leaving the surface.

bThe intensity of the light is increased to 2I. Predict the value of the new current.

cLight of intensity 2I and wavelength 6.0 × 10−7 m falls on this photo-surface. Determine the current in this case.

Answer

aThe definition of electric current is Δ q Δ t.

In a time of 1 second, the number of electrons emitted is 2.0 × 1010 and so the magnitude of the charge they carry is e × 2.0 × 1010.

The current is thus e × 2.0 × 1010, i.e.

3.2×10−9 A.

bIf the intensity doubles, the number of photons will double, and so the number of electrons emitted will double. Thus, so will the current, giving 6.4 × 10−9 A.

cThe critical frequency f c is 6.0 × 1014 Hz. The incident frequency is:

f=c λ=3×10 8 6.0×10−7=5.0×10 14 Hz

This is less than the critical frequency and so no electrons will be emitted at all.

Worked Example 22.6

The green light in Figure 22.4 has a wavelength of 496 nm.

aDetermine the work function of the photo-surface.

bEstimate the wavelength of the violet light in that experiment.

Answer

aThe stopping voltage is 0.40 V and so using e V s = h f − ϕ we deduce that:

ϕ=h c λ−e V s=1.24×10−6 4.96×10−7−0.40=2.10 eV

bWe again use e V s = h f − ϕ to get that h c λ=e V s+ϕ.

The stopping voltage is 0.90 V and so:

h c λ=0.90+2.1=3.0 eV

Hence:

λ=h c 3.0=1.24×10−6 3.0=4.1×10−7 m

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Exam tip

Notice the use of hc = 1.24 × 10−6 eV m, which makes calculations much faster.

Compton scattering

We saw how Einstein’s explanation of the photoelectric effect requires that light must be thought of as a stream of particles called photons. Even more convincing evidence for the particle nature of light is provided by Compton scattering the scattering of light off an electron. This is the scattering of light off an electron. The scattered light bounces off the electron with a longer wavelength. We can understand this in terms of photons easily: the light gives some energy to the electron which recoils with kinetic energy. The energy of the light has therefore been reduced. Thinking of light in terms of photons means that the energy of the photon is given by E=h c λ and so a reduced photon energy means an increased wavelength, which is what was observed in experiments carried out by Arthur Holly Compton in 1923.

Figure 22.8 shows a photon of wavelength λ that is incident on an electron at rest. The photon scatters off the electron in a direction making an angle θ with the original direction of the photon. The wavelength of the scattered photon is λ′. The electron gains kinetic energy and moves off in some direction.

Figure 22.8: A photon incident on an electron scatters off the electron; the electron recoils.

This diagram should be compared to Figure 4.22 in Chapter 4. It is an identical diagram to the diagram we drew for a two-dimensional collision of two balls. Here the one ‘ball’ is replaced by a photon. The methods of Chapter 4 can be applied here as well, namely applying the laws of conservation of momentum and energy to derive the formula

λ′−λ=h m e c(1−cos θ)

which relates the change in the wavelength to the scattering angle θ and the electron mass m e. It is in this sense that Compton scattering is solid evidence that light behaves as particles. We treated the photon as if it were an ordinary particle with momentum in mechanics.

This formula is an interesting one. The presence of Planck’s constant implies this is a quantum effect; this is not surprising since we are treating light as particles, photons, each of energy hf as Einstein said. But the speed of light c also enters the formula. Right away, this tells us that this is a relativistic effect which means we need the theory of relativity to derive this formula. (An end-of-chapter problem guiding to this derivation is provided for those interested.)

The largest wavelength shift occurs when θ = π. In this case:

λ′−λ=h m e c(1−cos π)=h m e c(1−(−1))=2 h m e c,

and the energy transferred to the electron is the largest possible. The energy that is transferred to the electron is h c λ−h c λ′. There is no shift when θ = 0.

The quantity h m e c is known as the Compton wavelength. Its numerical value for the electron is

h m e c=6.63×10−34 9.11×10−31×3.0×10 8≈2.4×10−12 m.

The smallness of this number means that observable effects are possible only with high-energy photons whose wavelength is comparable to this number (X-rays and gamma rays).

Check yourself 5

Photons of energy 40 keV and 60 keV scatter off an electron with the same scattering angle. How do the wavelength shifts compare?

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Worked Example 22.7

A photon of energy 54 keV is scattered off an electron at rest with a scattering angle of 82°.

aWhat is the shift in the photon wavelength?

bWhat energy is transferred to the electron?

Answer

aThe shift is λ′−λ=h m c(1−cos θ)=2.4×10−12×(1−cos 82°)=2.1×10−12 m.

bThe incident photon wavelength is found from h c λ=54×10 3⇒λ=h c 54×10 3=1.24×10−6 54×10 3 eVm eV=2.3×10−11 m. The scattered wavelength is λ′=2.3×10−11+2.1×10−12=2.5×10−11 m. The difference in photon energies is therefore h c λ−h c λ′=54×10 3−1.24×10−6 2.5×10−11=4.4 keV. This is the energy transferred to the electron; it is about 8% of the original photon energy.

international mindedness

This list contains a few of the names associated with the development of quantum theory in the twentieth century: A. Einstein, N. Bohr, M. Planck, L. de Broglie, M. Born, E. Schrödinger, W. Heisenberg, W. Pauli, P. Dirac, R. Feynman. How many nationalities are represented here? In the twenty-first century, at the Large Hadron Collider at CERN alone, more than 10 000 scientists and engineers from dozens of laboratories and universities from more than 100 countries contribute to its development and operation. What does this say about how international mindedness in physics has evolved?

TEST YOUR UNDERSTANDING

1A light source has power 60 W and emits photons of wavelength 550 nm. How many photons per second are emitted?

2A beam of light of wavelength 620 nm is incident normally on a surface. The intensity of the beam at the surface is 25 W m−2. How many photons are incident on the surface per unit time per unit area?

3A beam of photons of intensity 660 W m−2 and wavelength 480 nm is incident normally on a surface. The photons are absorbed by the surface. What is the pressure exerted on the surface?

4In an experiment, the stopping voltage was measured to be 0.32 V. What is the maximum kinetic energy of the emitted electrons?

5In an experiment, electrons are emitted from a photo-surface with kinetic energy 1.8 eV. The photo-surface is at zero potential, and the collecting plate is at potential +2.2 V. What is the kinetic energy of the electrons as they arrive at the collecting plate?

6In an experiment, electrons are emitted from a photo-surface with kinetic energy 2.1 eV. The photo-surface is at potential + 1.2 V and the collecting plate is at zero potential. What is the kinetic energy of the electrons as they arrive at the collecting plate?

7In photoelectric experiments with different photo-surfaces, the graphs of maximum electron kinetic energy versus light frequency are parallel straight lines. Why is that?

8Light is incident on a photo-surface.

State and explain the change, if any, to the stopping voltage when the following two independent changes are made:

athe wavelength of the light stays the same but the intensity of the light increases

bthe wavelength is decreased and the intensity of the light increases.

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9A photo-surface is moving away at high speed from a source of monochromatic light that emits in all directions. State and explain what happens to:

athe energy of the emitted electrons

bthe rate at which electrons are being emitted.

10A photo-surface has a work function of 3.00 eV. Determine the critical frequency.

11 aWhat evidence is there for the existence of photons?

bA photo-surface has a critical frequency of 2.25 × 1014 Hz. Radiation of frequency 3.87 × 1014 Hz falls on this surface. Deduce the stopping voltage.

12Light of wavelength 5.4 × 10−7 m falls on a photo-surface and causes the emission of electrons of maximum kinetic energy 2.1 eV at a rate of 1015 per second. The light is emitted by a 60 W light bulb.

aExplain how light causes the emission of electrons.

bCalculate the electric current that leaves the photo-surface.

cDetermine the work function of the surface.

dEstimate the maximum kinetic energy of the electrons when the power of the light becomes 120 W.

eEstimate the current from the photo-surface when the power is 120 W.

13 aState three aspects of the photoelectric effect that cannot be explained by the wave theory of light. For each, outline how the photon theory provides an explanation.

bLight of wavelength 2.08 × 10−7 m falls on a photo-surface. The stopping voltage is 1.40 V.

iOutline what is meant by stopping voltage.

iiCalculate the largest wavelength of light that will result in emission of electrons from this photo-surface.

14 aThe intensity of the light incident on a photo-surface is doubled while the wavelength of light stays the same. For the emitted electrons, discuss the effect of this, if any, on:

ithe energy of the electrons emitted

iithe number of electrons emitted.

bTo determine the work function of a given photo-surface, light of wavelength 2.3 × 10−7 m is directed at the surface and the stopping voltage, V s, recorded. When light of wavelength 1.8 × 10−7 m is used, the stopping voltage is twice as large as the previous one. Determine the work function.

15Light falling on a metallic surface of work function 3.0 eV gives energy to the surface at a rate of 5.0 × 10−4 W per square metre of the metal’s surface. Assume that an electron on the metal surface can absorb energy from an area of about 1.0 × 10−18 m2.

aEstimate how long will it will take the electron to absorb an amount of energy equal to the work function.

bOutline the implication of this.

cDescribe how the photon theory of light explains the fact that electrons are emitted almost instantaneously with the incoming photons.

16 aFrom the graph (Figure 22.9) of electron kinetic energy E K versus frequency of incoming radiation, deduce:

ithe critical frequency of the photo-surface

iithe work function.

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bWhat is the kinetic energy of an electron ejected when light of frequency f = 8.0 × 1014 Hz falls on the surface?

cAnother photo-surface has a critical frequency of 6.0 × 1014 Hz. On a copy of Figure 22.9, sketch the variation with frequency of the emitted electrons’ kinetic energy.

Figure 22.9: For question 16.

17This question will look at the intensity of radiation in a bit more detail. The intensity of light, I, incident normally on an area A is defined to be I=P A, where P is the power carried by the light.

aShow that I = Φh f, where Φ is the photon flux density, i.e. the number of photons incident on the surface per second per unit area, and f is the frequency of the light.

bCalculate the intensity of light of wavelength λ = 5.0 × 10–7 m incident on a surface when the photon flux density is Φ = 3.8 × 1018 m–2 s–1.

cThe wavelength of the light is decreased to λ = 4.0 × 10–7 m. Calculate the new photon flux density so that the intensity of light incident on the surface is the same as that found in b.

dHence explain why light of wavelength λ = 4.0 × 10–7 m and of the same intensity as that of light of wavelength λ = 5.0 × 10–7 m will result in fewer electrons being emitted from the surface per second.

eState one assumption made in reaching this conclusion.

18If a photon scatters off an entire atom instead of a single electron, how would you expect the wavelength shifts to compare for the same scattering angle?

19What is the wavelength shift when a photon of wavelength 4.0 × 10–12 m scatters off an electron at rest with a scattering angle of 120°?

20What is the wavelength shift when a photon of wavelength 4.0 × 10–12 m scatters off a proton at rest with a scattering angle of 120°?

21In a Compton scattering experiment the incident photon has wavelength λ and the scattered photon has wavelength λ′. Show that the fraction of the incident photon energy that is transferred to the electron is given by λ′−λ λ′.

22How much energy is transferred to the electron for a photon scattering angle of 0°?

23A photon of wavelength 3.0 × 10–12 m scatters off an electron at rest with a scattering angle of 90°. How much energy is transferred to the electron?

24 aUsing your graphics calculator or otherwise, find the angle at which the energy transferred to an electron during Compton scattering is a maximum.

bIf the maximum energy that can be transferred to an electron is 30 keV, what is the wavelength of the incident photon?

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25Derivation of the Compton formula—not for exam purposes.

Refer to Figure 22.8.

aShow by applying conservation of momentum that the momentum components of the electron after the photon scatters off it are p x=h λ−h λ′cos θ and p y=h h′sin θ.

bHence deduce that p 2=p x 2+p y 2=(h λ′)2+(h λ)2−2 h 2 cos θ λ λ′.

cNow apply energy conservation. You will need the result from relativity that the total energy of an electron of momentum p is given by E=p 2 c 2+(m c 2)2. What is the total energy of an electron at rest? Therefore deduce that energy conservation leads to h c λ+m c 2=h c λ′+p 2 c 2+(m c 2)2

dIsolate the term with the square root, square both sides and then use the result in b. After a few lines of algebra you should get the Compton formula.