Senior 5 practice book · Unit 7 of 11

Wave and Particle Nature of Light

50 questions that complete the Senior 5 quiz for this unit: 27 core and 23 challenge. Easy and hard questions are mixed. Try each question, then tap “Show answer”.

Common misconceptions

Brighter light makes photoelectrons come out faster.Brightness (intensity) only changes how many photons arrive each second; the energy of each electron depends on the frequency of the light.
A black object is a blackbody, and a blackbody does not glow.A blackbody is any perfect absorber; when hot it is also the best possible emitter, which is why the Sun and a small hole in a hot oven are good blackbodies.
Only light behaves like a wave; electrons are just tiny balls.Electrons have a de Broglie wavelength λ = h/mv and give diffraction patterns, so matter also shows wave behaviour.
A hotter body gives out less long-wavelength (red and infrared) radiation because its peak moves to shorter wavelengths.A hotter body emits MORE at every wavelength; only the position of the peak moves to shorter wavelengths (Wien).
An electron can collect energy from several weak photons until it has enough to escape.An electron absorbs one whole photon at a time; if hf is below the work function, no electron escapes however long the light shines.

What this unit covers

Topics marked new are not tested much in the quiz, so this book gives them extra questions.

Blackbody radiation: perfect absorber/emitter and the intensity–wavelength curves
Stefan–Boltzmann law P = σAT⁴, including net radiation and finding temperature or size
Wien's displacement law λ_max T = 2.9 × 10⁻³ m K and the colour of hot bodies
Planck's quantum hypothesis, photon energy E = hf = hc/λ and the electronvolt
Photon flux: number of photons per second from a source of given power (P = nhf)
Photoelectric effect: threshold frequency, work function, Einstein's equation, stopping potential
Photocell experiments: current–voltage graphs, saturation current, effect of intensity and frequency
Applications of the photoelectric effect: photocells, light sensors, alarms, solar cells
Photon momentum p = h/λ and the Compton effect
de Broglie wavelength λ = h/p = h/mv and electron diffraction
Wave–particle duality and the principle of complementarity
Electron microscope: why short matter waves give fine detail
Effects of single photons on matter: sunburn, film, photosynthesis, radio photons

Go to the questions

Questions (1–50)

Easier and harder questions are mixed, just like in a real exam. The 23 harder ones are marked ★ Challenge.

1True or false
A street light controlled by a photocell can switch on at dusk because the photocurrent falls as the daylight intensity falls.
Show answer
Answer: True
Fewer photons per second free fewer electrons, so the current falls; a circuit then switches the lamp on.
Common mistakeSome think the light colour at dusk falls below threshold; the main change is the intensity, which controls the size of the current.
2True or false · ★ Challenge
In Compton scattering the largest wavelength shift happens when the photon bounces straight back (scattering angle 180°).
Show answer
Answer: True
Δλ = (h/m_e c)(1 − cos θ) is largest when cos θ = −1, giving 2h/m_e c.
Common mistakeSome think the shift is largest at 90°; 1 − cos θ keeps increasing up to 180°, where the electron receives the most momentum.
3True or false
In a double-slit experiment with very dim light, photons arrive at the screen one at a time as separate dots, yet after many photons the dots build up an interference pattern.
Show answer
Answer: True
Each photon is detected like a particle at one point, but where it is likely to land is set by the wave pattern.
Common mistakeLearners often think interference needs many photons interacting with each other; the pattern forms even when photons pass one at a time.
4True or false · ★ Challenge
Two lamps each give out 1.0 W of light, one red (700 nm) and one violet (400 nm). The red lamp emits more photons per second.
Show answer
Answer: True
Red photons carry less energy (E = hc/λ), so more of them are needed for the same power: 700/400 = 1.75 times as many.
Common mistakeMany learners think more energetic light must mean more photons; for equal power, higher-energy photons means FEWER of them.
5True or false
In a photocell, a small current can still flow when the anode voltage is zero, because some electrons leave the cathode with enough kinetic energy to reach the anode by themselves.
Show answer
Answer: True
Electrons leave with KE up to KE_max, so some cross the gap without help; only a reverse voltage of V_s stops them all.
Common mistakeMany learners think a current needs a voltage; here the photons give the electrons the energy to move.
6Fill in the blank · ★ Challenge
The filament of a torch bulb radiates most strongly at 1.16 μm. Taking Wien's constant as 2.9 × 10⁻³ m K, the filament temperature is about ______ K.
Show answer
Answer: 2500
T = 2.9 × 10⁻³ ÷ (1.16 × 10⁻⁶) = 2500 K.
Common mistakeForgetting that 1 μm = 10⁻⁶ m gives an absurd temperature of a few millikelvin.
7True or false
A small hole in the wall of a closed hollow box behaves almost like a perfect blackbody, because radiation entering the hole is reflected many times inside and is almost all absorbed.
Show answer
Answer: True
Each reflection absorbs some radiation, so almost nothing that enters comes back out: the hole is a near-perfect absorber, and so also a near-perfect emitter when the box is hot.
Common mistakeLearners often think a blackbody must be painted black; it is the absorbing behaviour, not the colour of paint, that matters.
8True or false
A photon of blue light carries more energy than a photon of red light, yet a bright red lamp can give out more energy per second than a dim blue lamp.
Show answer
Answer: True
Energy per photon depends on frequency (E = hf), but power also depends on how many photons are emitted each second.
Common mistakeLearners often mix up energy per photon with total power; a bright source of low-energy photons can still have a large power.
9Multiple choice · ★ Challenge
A student writes: 'Light of 300 nm (4.1 eV) falls on sodium (φ = 2.3 eV), so KE = 4.1 − 2.3 = 1.8 eV and every emitted electron has 1.8 eV.' What is wrong with the statement?
1. AThe KE should be 2.3 − 4.1 = −1.8 eV, so none escape
2. B1.8 eV is only the largest KE; most electrons have less
3. CThe KE should be 4.1 + 2.3 = 6.4 eV for each electron
4. DAt 300 nm each photon has only 0.24 eV, so none escape
Show answer
Answer: B. 1.8 eV is only the largest KE; most electrons have less
hf − φ gives KE_max; electrons from below the surface lose extra energy on the way out, so they have any KE from 0 up to 1.8 eV.
Common mistakeChoosing '4.1 + 2.3' adds the work function; the work function is energy the electron must spend to escape, so it is subtracted.
10Multiple choice
Photographic darkrooms use a dim red safelight because
1. Ared light travels most slowly, so it reaches the film too late
2. Bred light is never intense enough to reach the film at all
3. Cred light is absorbed by the air before it reaches the film
4. Dred photons have too little energy to change the film chemicals
Show answer
Answer: D. red photons have too little energy to change the film chemicals
The film reaction needs photons above a minimum energy; red photons are below it, so even a red lamp does not fog the film.
Common mistakeChoosing 'not intense enough' misses the point: even bright red light would not expose the film, because each red photon is too weak.
11Short answer · ★ Challenge
Suggest two reasons why stopping potentials measured with a school photocell may be inaccurate, and one improvement for each.
Show answer
Model answer: Any two, with improvements: (1) stray room light also reaches the cathode – work in a darkened room or shield the cell; (2) the coloured filters let through a band of wavelengths, not one – use a line source such as a mercury lamp with narrow filters; (3) the current near V_s is tiny and hard to read – use a sensitive microammeter or electrometer and approach V_s slowly; (4) light falling on the anode causes a small reverse current – shade the anode.
Common mistakeLearners often say "use brighter light"; brighter light does not change V_s, it only changes the current.
12Short answer · ★ Challenge
"Light travels like a wave but is absorbed like a particle." Explain this statement, using one experiment for each half of it.
Show answer
Model answer: Travelling: interference and diffraction (e.g. Young's double slit) can only be explained if light spreads out as a wave and the parts overlap. Absorbed: in the photoelectric effect each electron takes the energy of one photon hf at a single point, with no delay and with a threshold frequency, which only the particle model explains. Both models are needed (complementarity).
Common mistakeA common error is to say light "changes" from wave to particle; it has both properties all the time and an experiment shows only one of them.
13Fill in the blank
If the speed of a (non-relativistic) electron is doubled, its de Broglie wavelength is multiplied by ______.
Show answer
Answer: ½ (0.5)
λ = h/mv, so doubling v halves λ.
Common mistakeAnswering 4 or ¼ confuses speed with kinetic energy; λ is inversely proportional to v itself.
14Multiple choice
A shop burglar alarm uses a beam of light falling on a photocell across the doorway. What sets off the alarm?
1. AThe beam gets brighter, so the photocell overheats
2. BThe intruder's body heat raises the threshold frequency
3. CThe beam is blocked, so the photocurrent stops
4. DThe intruder reflects extra electrons into the photocell
Show answer
Answer: C. The beam is blocked, so the photocurrent stops
While light falls on the cell a photocurrent flows; when someone breaks the beam the current drops to zero and a relay switches on the alarm.
Common mistakeChoosing the body-heat option is wrong: the threshold frequency is a property of the metal and does not change with an intruder's warmth.
15Fill in the blank · ★ Challenge
The work function of tungsten is 4.5 eV. Its threshold frequency is ______ × 10¹⁵ Hz. (h = 6.63 × 10⁻³⁴ J s, 1 eV = 1.6 × 10⁻¹⁹ J)
Show answer
Answer: 1.1
φ = 4.5 × 1.6 × 10⁻¹⁹ = 7.2 × 10⁻¹⁹ J; f₀ = φ/h = 7.2 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ ≈ 1.1 × 10¹⁵ Hz.
Common mistakeDividing 4.5 by h without first converting eV to joules gives a meaningless 6.8 × 10³³ Hz.
16Multiple choice
In Compton scattering, an X-ray photon bounces off a free electron. Which statement is correct?
1. AThe photon gains energy and its wavelength becomes shorter
2. BThe photon's speed falls below c after the collision
3. CThe photon is completely absorbed, as in the photoelectric effect
4. DThe scattered photon has less energy and a longer wavelength
Show answer
Answer: D. The scattered photon has less energy and a longer wavelength
The electron recoils and takes some energy and momentum, so the photon leaves with lower energy E = hc/λ, i.e. a longer wavelength; it still travels at c.
Common mistakeChoosing 'speed falls below c' treats the photon like a ball; a photon always moves at c and loses energy by a drop in frequency instead.
17Short answer · ★ Challenge
A blacksmith in Nyabugogo heats an iron bar in a charcoal fire. Describe how the colour of the glowing bar changes as it gets hotter and explain this using the blackbody curve.
Show answer
Model answer: It first glows dull red, then orange, yellow and finally almost white. As T rises the whole curve rises and its peak moves to shorter wavelengths, so at first only the long-wavelength (red) tail reaches the visible range, and at higher T more of the shorter visible wavelengths are emitted until all colours mix to look white.
Common mistakeMany learners say the bar turns blue first; the shortest visible colours only become strong at very high temperatures, so red appears first.
18Multiple choice
A star's radiation is strongest at 290 nm, in the ultraviolet. What colour does the star appear to the eye?
1. AInvisible, because its peak is outside the visible range
2. BViolet only, because only its shortest waves are visible
3. CBlue-white, as it gives out more blue than red light
4. DRed, because only its long-wavelength tail is visible
Show answer
Answer: C. Blue-white, as it gives out more blue than red light
The curve of a very hot star falls steadily across the visible range from violet to red, so it gives out all colours with more blue than red: it looks blue-white.
Common mistakeChoosing 'invisible' forgets that a blackbody curve is broad: a star peaking in the UV still emits strongly at all visible wavelengths.
19Multiple choice · ★ Challenge
A metal with work function 2.0 eV gives a stopping potential of 1.1 V. What wavelength of light was shining on it? (hc = 1240 eV nm)
1. A1130 nm
2. B1380 nm
3. C400 nm
4. D620 nm
Show answer
Answer: C. 400 nm
Photon energy = φ + eV_s = 2.0 + 1.1 = 3.1 eV; λ = 1240 ÷ 3.1 = 400 nm.
Common mistakeChoosing 1380 nm subtracts (2.0 − 1.1 = 0.9 eV); the photon must supply the work function AND the kinetic energy, so they add.
20Multiple choice
In Planck's hypothesis, an oscillating atom in the wall of a hot body, with frequency f, can only have energies equal to
1. Aany value between zero and hf, changing smoothly
2. Bthe fractions hf/2, hf/4, hf/8, … of one quantum
3. Cthe single value hf, whatever the temperature is
4. Dwhole-number multiples of hf: 0, hf, 2hf, 3hf, …
Show answer
Answer: D. whole-number multiples of hf: 0, hf, 2hf, 3hf, …
Planck assumed the energy is quantised: E = nhf with n = 0, 1, 2, 3, …
Common mistakeChoosing 'any value' is the classical idea that led to the ultraviolet catastrophe; Planck's step was to forbid values between the multiples of hf.
21Short answer · ★ Challenge
Using energy conservation, explain why the electrons emitted in the photoelectric effect have a range of kinetic energies from zero up to KE_max, even when the light has a single frequency.
Show answer
Model answer: Each electron receives the same energy hf from one photon. The work function is the least energy needed to free an electron, which applies to electrons at the surface; these leave with KE_max = hf − φ. Electrons from deeper inside lose extra energy in collisions on the way out, so they leave with less, down to zero.
Common mistakeA common error is to think different photons in single-frequency light carry different energies; they all carry hf, and the spread comes from the electrons.
22Multiple choice
A photocell must detect a red laser beam (wavelength 650 nm, photon energy 1.9 eV). Which cathode metal could work?
1. AA metal with a work function of 2.3 eV
2. BA metal with a work function of 4.3 eV
3. CA metal with a work function of 1.8 eV
4. DAny metal, if the laser is bright enough
Show answer
Answer: C. A metal with a work function of 1.8 eV
Electrons are emitted only if hf > φ: 1.9 eV > 1.8 eV, but 1.9 eV < 2.3 eV and 4.3 eV.
Common mistakeChoosing 'any metal if bright enough' is the classical wave idea; brightness gives more photons, not more energy per photon.
23Fill in the blank · ★ Challenge
The Compton shift is Δλ = (h/m_e c)(1 − cos θ). For X-rays scattered through 90° by electrons, Δλ = ______ pm. (h = 6.63 × 10⁻³⁴ J s, m_e = 9.11 × 10⁻³¹ kg, c = 3.0 × 10⁸ m/s)
Show answer
Answer: 2.4
cos 90° = 0, so Δλ = h/(m_e c) = 6.63 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 3.0 × 10⁸) = 2.4 × 10⁻¹² m = 2.4 pm.
Common mistakeUsing cos 90° = 1 gives zero shift; the cosine of 90° is 0, so the bracket equals 1.
24Multiple choice
Why can an electron microscope show much finer detail than a light microscope?
1. AElectrons travel faster than light, so images form faster
2. BElectrons are smaller than photons, so they fit between atoms
3. CIts electrons have a much shorter wavelength than light
4. DIts glass lenses bend electrons more strongly than light
Show answer
Answer: C. Its electrons have a much shorter wavelength than light
The smallest detail that can be seen is about one wavelength; fast electrons have λ around 10⁻¹¹ m, thousands of times shorter than visible light.
Common mistakeChoosing 'faster than light' is impossible; resolution depends on wavelength, not speed of image formation. Electron microscopes use magnetic lenses, not glass.
25Fill in the blank
A piece of glowing charcoal cools from 900 K to 300 K. If its surface area stays the same, it now radiates ______ times less power.
Show answer
Answer: 81
P ∝ T⁴, and (900/300)⁴ = 3⁴ = 81.
Common mistakeAnswering 3 forgets the fourth power; answering 9 squares the ratio instead of raising it to the power 4.
26Multiple choice · ★ Challenge
In a photocell experiment the light intensity is doubled while its frequency stays the same. How does the current–voltage graph change?
1. ASaturation current doubles; stopping potential also doubles
2. BSaturation current stays the same; stopping potential doubles
3. CBoth stay the same, because only the frequency matters
4. DSaturation current doubles; stopping potential stays the same
Show answer
Answer: D. Saturation current doubles; stopping potential stays the same
Twice as many photons per second free twice as many electrons, but each photon still has the same hf, so KE_max and V_s are unchanged.
Common mistakeChoosing 'stopping potential doubles' links intensity to electron energy; V_s depends only on hf − φ.
27Multiple choice
Why does ultraviolet light from the Sun cause sunburn, while much more intense infrared from a charcoal stove does not?
1. AUV light from the Sun is more intense than stove infrared
2. BInfrared photons are absorbed only by clothes, not by skin
3. CUV light travels faster than infrared through the air
4. DEach UV photon has enough energy to damage skin molecules
Show answer
Answer: D. Each UV photon has enough energy to damage skin molecules
Damage to a molecule needs a single photon with enough energy; UV photons (> 3 eV) can break bonds, infrared photons (< 1.6 eV) cannot, however many arrive.
Common mistakeChoosing 'more intense' confuses intensity with photon energy; the stove's infrared is more intense, but its photons are individually too weak.
28Multiple choice · ★ Challenge
A learner's skin (area 1.6 m², treated as a blackbody) is at 34 °C in a classroom whose walls are at 24 °C. Using P_net = σA(T⁴ − T₀⁴) with σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, what is the net rate of heat loss by radiation?
1. A806 W
2. B0.091 W
3. C9.1 × 10⁻⁴ W
4. D100 W
Show answer
Answer: D. 100 W
T = 307 K, T₀ = 297 K: P_net = 5.67 × 10⁻⁸ × 1.6 × (307⁴ − 297⁴) = 5.67 × 10⁻⁸ × 1.6 × 1.10 × 10⁹ ≈ 100 W.
Common mistakeChoosing 0.091 W comes from putting 34 and 24 (°C) into T⁴; the law needs kelvin. 806 W forgets that the body also absorbs radiation from the walls.
29Fill in the blank
One electronvolt is the energy gained by an electron accelerated through a potential difference of 1 V, so 1 eV = ______ J.
Show answer
Answer: 1.6 × 10⁻¹⁹
W = QV = 1.6 × 10⁻¹⁹ C × 1 V = 1.6 × 10⁻¹⁹ J.
Common mistakeSome learners think the electronvolt is a unit of voltage; it is a unit of energy equal to e × 1 V.
30Multiple choice · ★ Challenge
In an electron microscope, electrons are accelerated from rest through 12 kV. What is their de Broglie wavelength? (h = 6.63 × 10⁻³⁴ J s, m = 9.11 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C)
1. A3.5 × 10⁻¹⁰ m
2. B1.1 × 10⁻¹¹ m
3. C1.0 × 10⁻¹⁰ m
4. D1.6 × 10⁻¹¹ m
Show answer
Answer: B. 1.1 × 10⁻¹¹ m
p = √(2meV) = √(2 × 9.11 × 10⁻³¹ × 1.6 × 10⁻¹⁹ × 12 000) = 5.9 × 10⁻²³ kg m/s; λ = h/p = 1.1 × 10⁻¹¹ m.
Common mistakeChoosing 3.5 × 10⁻¹⁰ m uses 12 V instead of 12 000 V; 1.0 × 10⁻¹⁰ m uses the photon formula hc/E, which does not apply to electrons.
31Fill in the blank
A source emits 2.0 × 10¹⁸ photons per second, each with an energy of 4.0 × 10⁻¹⁹ J. Its light output power is ______ W.
Show answer
Answer: 0.80
P = n × E = 2.0 × 10¹⁸ × 4.0 × 10⁻¹⁹ = 0.80 W.
Common mistakeDividing instead of multiplying gives 5 × 10³⁶ W; power is (photons per second) × (energy per photon).
32Multiple choice · ★ Challenge
Through what potential difference must electrons be accelerated from rest to give them a de Broglie wavelength of 0.050 nm? (h = 6.63 × 10⁻³⁴ J s, m = 9.11 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C)
1. A1200 V
2. B600 V
3. C2.5 × 10⁴ V
4. D300 V
Show answer
Answer: B. 600 V
p = h/λ = 1.33 × 10⁻²³ kg m/s; eV = p²/2m, so V = p²/(2me) = (1.33 × 10⁻²³)² ÷ (2 × 9.11 × 10⁻³¹ × 1.6 × 10⁻¹⁹) ≈ 600 V.
Common mistakeChoosing 2.5 × 10⁴ V treats the electron like a photon (E = hc/λ); for a particle with mass, KE = p²/2m.
33Multiple choice
What is the momentum of a photon of orange light of wavelength 600 nm? (h = 6.63 × 10⁻³⁴ J s)
1. A1.1 × 10⁻²⁷ kg m/s
2. B4.0 × 10⁻⁴⁰ kg m/s
3. C1.1 × 10⁻³⁶ kg m/s
4. D3.3 × 10⁻¹⁹ kg m/s
Show answer
Answer: A. 1.1 × 10⁻²⁷ kg m/s
p = h/λ = 6.63 × 10⁻³⁴ ÷ 600 × 10⁻⁹ = 1.1 × 10⁻²⁷ kg m/s.
Common mistakeChoosing 1.1 × 10⁻³⁶ kg m/s divides by 600 without converting nm to m; 3.3 × 10⁻¹⁹ is the photon energy in joules, not its momentum.
34Multiple choice · ★ Challenge
In a photocell experiment the frequency of the light is increased, while the number of photons arriving per second is kept the same. How does the current–voltage graph change?
1. ALarger stopping potential; saturation current about the same
2. BSmaller stopping potential; saturation current becomes larger
3. CSame stopping potential; saturation current larger
4. DBoth the same, because the photon rate is unchanged
Show answer
Answer: A. Larger stopping potential; saturation current about the same
Higher hf gives a larger KE_max, so a larger reverse voltage is needed to stop the electrons; the same number of photons per second gives about the same saturation current.
Common mistakeChoosing 'both the same' forgets that each photon now carries more energy, which raises KE_max and so V_s.
35Multiple choice
The light from an electric welding arc is strongest at a wavelength of 290 nm. Treating the arc as a blackbody (Wien constant 2.9 × 10⁻³ m K), what is its temperature?
1. A1.0 × 10⁴ K
2. B1.0 × 10⁷ K
3. C1.0 × 10⁻⁴ K
4. D1.0 × 10⁻⁵ K
Show answer
Answer: A. 1.0 × 10⁴ K
T = 2.9 × 10⁻³ ÷ (290 × 10⁻⁹) = 1.0 × 10⁴ K.
Common mistakeChoosing 1.0 × 10⁻⁵ K comes from putting 290 into the formula without converting nm to m; 1.0 × 10⁻⁴ K divides λ by the constant instead of the constant by λ.
36Short answer · ★ Challenge
The largest possible Compton shift is about 4.9 pm. Explain why the Compton effect is easy to observe with X-rays of wavelength 0.071 nm but not with visible light of 500 nm.
Show answer
Model answer: For X-rays, 4.9 pm ÷ 71 pm ≈ 7%, a change that can be measured. For visible light, 4.9 pm ÷ 500 000 pm ≈ 0.001%, far too small to detect. The shift is the same size for all wavelengths, so it only shows up when the wavelength itself is very short.
Common mistakeLearners often think visible photons do not collide with electrons at all; they do, but the fractional change in wavelength is too small to notice.
37Multiple choice
A photon of infrared light has an energy of 2.4 × 10⁻¹⁹ J. What is this energy in electronvolts? (1 eV = 1.6 × 10⁻¹⁹ J)
1. A3.8 × 10⁻³⁸ eV
2. B1.5 eV
3. C0.67 eV
4. D2.4 eV
Show answer
Answer: B. 1.5 eV
E = 2.4 × 10⁻¹⁹ ÷ 1.6 × 10⁻¹⁹ = 1.5 eV.
Common mistakeChoosing 3.8 × 10⁻³⁸ eV multiplies by 1.6 × 10⁻¹⁹ instead of dividing; to go from J to eV you divide.
38Short answer
Explain how a solar panel on a health centre uses photons to produce an electric current, and why it produces no current at night even though the panel is still warm.
Show answer
Model answer: Each photon with enough energy frees an electron inside the semiconductor, and the built-in electric field of the cell drives these electrons round the circuit as a current. At night there are no photons of visible light with enough energy; the warm panel gives out and receives only infrared photons whose energy is too small to free electrons.
Common mistakeA common error is to think the panel works on heat; it works on individual photons above a minimum energy, not on temperature.
39Multiple choice · ★ Challenge
The Sun gives out a total power of 3.9 × 10²⁶ W and has a radius of 7.0 × 10⁸ m. Treating it as a blackbody (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴), estimate its surface temperature.
1. A8200 K
2. B5800 K
3. C3.3 × 10⁷ K
4. D1.0 × 10⁵ K
Show answer
Answer: B. 5800 K
A = 4πR² = 6.16 × 10¹⁸ m²; T⁴ = P/(σA) = 3.9 × 10²⁶ ÷ (5.67 × 10⁻⁸ × 6.16 × 10¹⁸) = 1.12 × 10¹⁵ K⁴, so T ≈ 5800 K.
Common mistakeChoosing 8200 K comes from using the area of a flat disc πR² instead of the surface of a sphere 4πR²; 3.3 × 10⁷ K takes a square root instead of a fourth root.
40Short answer
Explain why the Stefan–Boltzmann law must be used with temperatures in kelvin, not degrees Celsius. Use a block of ice at 0 °C as an example.
Show answer
Model answer: P = σAT⁴ uses absolute temperature. With T in °C, ice at 0 °C would be predicted to radiate nothing, but at 273 K it radiates σT⁴ = 5.67 × 10⁻⁸ × 273⁴ ≈ 315 W per m². Every body above 0 K radiates.
Common mistakeUsing °C gives zero or even negative "temperatures to the fourth", which have no meaning; only kelvin starts at true zero of thermal energy.
41Multiple choice · ★ Challenge
A 60 W sodium street lamp turns 30% of its electrical power into yellow light of wavelength 589 nm. How many photons of yellow light does it give out per second? (h = 6.63 × 10⁻³⁴ J s, c = 3.0 × 10⁸ m/s)
1. A1.8 × 10²⁰
2. B5.3 × 10²⁸
3. C1.9 × 10⁻²⁰
4. D5.3 × 10¹⁹
Show answer
Answer: D. 5.3 × 10¹⁹
Light power = 0.30 × 60 = 18 W; E = hc/λ = 3.38 × 10⁻¹⁹ J; n = 18 ÷ 3.38 × 10⁻¹⁹ ≈ 5.3 × 10¹⁹ per second.
Common mistakeChoosing 1.8 × 10²⁰ uses the full 60 W, but only 30% of the power leaves as light; 5.3 × 10²⁸ forgets to change nm into m.
42Multiple choice
On a graph of photocurrent I against anode voltage V, the current levels off at a "saturation current" for large positive voltages. What does the saturation current show?
1. AThe frequency of the light has reached the threshold value
2. BAll the electrons emitted each second are reaching the anode
3. CThe emitted electrons have reached the speed of light
4. DThe cathode has run out of electrons that can be freed
Show answer
Answer: B. All the electrons emitted each second are reaching the anode
Once the anode collects every emitted electron, a larger voltage cannot increase the current; it is limited by the number of photons arriving per second.
Common mistakeChoosing 'run out of electrons' is wrong: the cathode is connected to the circuit and is continually resupplied; the limit is the photon arrival rate.
43Short answer · ★ Challenge
A 60 kg learner runs at 5.0 m/s through a classroom door 0.90 m wide. Calculate her de Broglie wavelength and explain why she shows no diffraction. (h = 6.63 × 10⁻³⁴ J s)
Show answer
Model answer: λ = h/mv = 6.63 × 10⁻³⁴ ÷ (60 × 5.0) = 2.2 × 10⁻³⁶ m. Diffraction is only noticeable when the gap is comparable to λ; the door is about 10³⁵ times wider, so the spreading is far too small to observe.
Common mistakeSome learners say she is "not a wave"; all matter has a de Broglie wavelength, but for everyday objects it is immeasurably small.
44Multiple choice
Electrons passing through a thin graphite film produce rings on a screen. The accelerating voltage is increased. What happens to the rings?
1. AThey get smaller, because the electron wavelength gets shorter
2. BThey get larger, because faster electrons spread out more widely
3. CThey stay the same size but become much brighter
4. DThey disappear, because fast electrons act only as particles
Show answer
Answer: A. They get smaller, because the electron wavelength gets shorter
Higher voltage → larger momentum → shorter λ = h/p → smaller diffraction angles, so the rings shrink.
Common mistakeChoosing 'larger' links faster with more spreading; in diffraction, shorter wavelength always means LESS spreading.
45Short answer · ★ Challenge
At night a thermal camera shows the engine of a parked bus (90 °C) much brighter than the road (25 °C). Find the wavelength at which the engine radiates most strongly, and the ratio of the power per square metre from the engine to that from the road. (Wien constant 2.9 × 10⁻³ m K)
Show answer
Model answer: T = 363 K: λ_max = 2.9 × 10⁻³ ÷ 363 ≈ 8.0 × 10⁻⁶ m (8.0 μm, infrared). Ratio = (363/298)⁴ ≈ 2.2, so the engine gives out about twice as much power per m² and looks brighter.
Common mistakeUsing 90 and 25 in °C gives a ratio of about 170 and a peak wavelength of 32 μm; both T values must be in kelvin.
46Short answer
Give two disadvantages of an electron microscope compared with a light microscope for a hospital laboratory.
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Model answer: Any two: it is very expensive to buy and maintain; the beam must travel in a vacuum, so living cells cannot be viewed; samples need long special preparation (thin slices, metal coating); it needs high voltages and trained staff; the image is black and white.
Common mistakeSaying "it has lower magnification" is wrong; its advantage is higher resolution, and its drawbacks are cost, vacuum and sample preparation.
47Multiple choice
Two metals P (φ = 2.0 eV) and Q (φ = 3.0 eV) are tested with light of many frequencies. On one graph of KE_max against frequency, their two straight lines are
1. Aparallel, with the line for Q further to the right
2. Bparallel, with the line for Q further to the left
3. Cof different slopes, both passing through the origin
4. Done single line, because the same light is used
Show answer
Answer: A. parallel, with the line for Q further to the right
Both lines have gradient h; Q has the larger work function, so its threshold frequency f₀ = φ/h is larger and its line is shifted to the right.
Common mistakeChoosing different slopes forgets that the gradient is Planck's constant, which is the same for every metal.
48Short answer · ★ Challenge
Sunlight delivers 800 W to each square metre of a solar panel on a house in Nyagatare. Taking an average wavelength of 550 nm, estimate how many photons hit each square metre per second, and explain why we never notice light "arriving in lumps". (h = 6.63 × 10⁻³⁴ J s, c = 3.0 × 10⁸ m/s)
Show answer
Model answer: E = hc/λ = 6.63 × 10⁻³⁴ × 3.0 × 10⁸ ÷ 550 × 10⁻⁹ = 3.6 × 10⁻¹⁹ J. n = 800 ÷ 3.6 × 10⁻¹⁹ ≈ 2.2 × 10²¹ photons per second. So many photons arrive that the flow looks perfectly smooth and continuous.
Common mistakeDividing the photon energy by the power (instead of power by photon energy) gives a tiny number with no meaning.
49Short answer · ★ Challenge
Compare a photon from an FM radio station (100 MHz) with a photon of green light (5.6 × 10¹⁴ Hz). Find the ratio of their energies and the energy of the radio photon in eV, and explain why radio waves cannot break chemical bonds. (h = 6.63 × 10⁻³⁴ J s, 1 eV = 1.6 × 10⁻¹⁹ J)
Show answer
Model answer: E ∝ f, so the ratio is 5.6 × 10¹⁴ ÷ 1.0 × 10⁸ = 5.6 × 10⁶. E(radio) = hf = 6.63 × 10⁻²⁶ J ≈ 4.1 × 10⁻⁷ eV. Breaking a bond needs a few eV from one photon; radio photons are millions of times too weak.
Common mistakeSome argue that a powerful transmitter can break bonds; more power means more photons, but each still carries far too little energy.
50Multiple choice
Blackbody curves for 3000 K, 4000 K and 5000 K are drawn on the same axes of intensity against wavelength. How does the 5000 K curve compare with the other two?
1. AIt is higher at every wavelength and peaks at a shorter wavelength
2. BIt peaks at a longer wavelength and has the largest area under its curve
3. CIt has the same peak height, only moved to shorter wavelengths
4. DIt is higher only near its peak and lower at long wavelengths
Show answer
Answer: A. It is higher at every wavelength and peaks at a shorter wavelength
A hotter blackbody emits more at every wavelength (total power ∝ T⁴) and its peak moves to shorter wavelength (λ_max ∝ 1/T).
Common mistakeChoosing 'same peak height' ignores that the total power rises as T⁴, so the hotter curve lies above the cooler ones everywhere.