Photoelectric effect & spectra — Physical Sciences

4 exam-style question sets on Photoelectric effect & spectra (Paper 1), each with a hint and a fully worked answer. The app holds 20 questions on this section in total, including variants of every set below, and lets you mark yourself part by part.

Multi-colour photoemission & line spectraPhotoelectric effectMetals compared at one fixed frequencyEinstein's photoelectric equation & graphs

Practise this section in the app →

Question 1

Multi-colour photoemission & line spectraComprehension15 marksPaper 1

Monochromatic red, green and violet light shine simultaneously on a sodium plate inside a vacuum tube. Measurements show that the ejected electrons have only TWO possible maximum kinetic energies, namely 3,12 × 10⁻²⁰ J and 1,31 × 10⁻¹⁹ J. Each ejected electron has one of these two maximum kinetic energies.

The frequency of the green light is 5,9 × 10¹⁴ Hz.

aDefine the term photoelectric effect. (2)
bWhich colour of light ejects the electrons that have a maximum kinetic energy of 3,12 × 10⁻²⁰ J? (1)
cExplain the answer to QUESTION b. Refer to BOTH the number of possible kinetic energies and the frequencies of the colours. (2)
dCalculate the frequency of the light that ejects the electrons with a maximum kinetic energy of 1,31 × 10⁻¹⁹ J. (5)
eThe intensity of the RED light is now doubled, while the other intensities remain the same. What effect does this have on the number of electrons ejected per second? Choose from INCREASES, DECREASES or REMAINS THE SAME. Explain the answer. (2)
fThe light emitted by a glowing gas-discharge tube in a darkened room is viewed through a spectroscope. Bright coloured lines on a dark background are observed. Name this type of spectrum. (1)
gExplain why only lines of certain specific colours appear in the spectrum in QUESTION f. (2)

Hint

Only two kinetic energies from three colours means one colour is below the threshold; find the work function from the green data first.

Worked answer

a) The photoelectric effect is the process whereby electrons are ejected from a metal surface when light of suitable frequency shines on that surface. ✓✓

b) Green light. ✓

c) Only TWO kinetic energies occur, so one colour (red, the lowest frequency) is below the threshold frequency and ejects no electrons ✓. Of the two colours that do eject electrons, green has the lower frequency, so its photons carry less energy and give the smaller maximum kinetic energy ( $E_{k (ma x)} = h f - W_{0}$ ) ✓.

d) For the green light:
$W_{0} = h f - E_{k (ma x)}$ ✓ $= (6, 63 \times 1 0^{- 34}) (5, 9 \times 1 0^{14}) - 3, 12 \times 1 0^{- 20} = 3, 60 \times 1 0^{- 19}$ J ✓
For the violet light:
$h f = W_{0} + E_{k (ma x)}$ ✓ $= 3, 60 \times 1 0^{- 19} + 1, 31 \times 1 0^{- 19}$ ✓
$f = \frac{4 , 91 \times 1 0 ^{- 19}}{6 , 63 \times 1 0 ^{- 34}} = 7, 41 \times 1 0^{14}$ Hz ✓

e) REMAINS THE SAME ✓ — red light is below the threshold frequency, so its photons cannot eject electrons no matter how many photons arrive per second (intensity does not change the energy of each photon) ✓.

f) (Atomic) line emission spectrum. ✓

g) Excited electrons in the gas atoms drop from higher to lower energy levels ✓. Because the energy levels are discrete, only photons with specific energies (and therefore specific frequencies/colours) are emitted ✓.

Modelled on Nov 2024 Paper 1, Question 10 — same skills, new scenario.

Question 2

Photoelectric effectComprehension2 marksPaper 1

Monochromatic light with a frequency ABOVE the threshold frequency shines on a metal surface and photoelectrons are ejected. Consider the statements:

(i) Increasing the intensity increases the number of photoelectrons ejected per second.
(ii) Increasing the intensity increases the maximum kinetic energy of the photoelectrons.
(iii) Increasing the frequency increases the maximum kinetic energy of the photoelectrons.

Which statement(s) is/are TRUE?

Hint

Intensity controls how many photons arrive each second; frequency controls the energy of each photon.

Worked answer

C. Greater intensity means more photons per second, so more electrons are ejected per second — (i) is true. Each photon's energy is $E = h f$ , so only a higher frequency raises the maximum kinetic energy ( $E_{k (ma x)} = h f - W_{0}$ ) — (iii) is true and (ii) is false. A misses the frequency effect. B and D both accept (ii), the classic misconception that brighter light gives faster electrons.

Modelled on Nov 2024 Paper 1, Question 1.10 — same skill, new scenario.

Question 3

Metals compared at one fixed frequencyApplication15 marksPaper 1

Learners shine ultraviolet light of a SINGLE frequency, 1,25 × 10¹⁵ Hz, onto the clean surfaces of four different metals, one at a time, and measure the maximum kinetic energy of the emitted electrons.

The results are shown in the table below.

METAL	ELECTRONS EMITTED	MAXIMUM KINETIC ENERGY (J)
P	Yes	0
Q	Yes	2,29 × 10⁻¹⁹
R	Yes	4,79 × 10⁻¹⁹
T	No	–

aDefine the term work function. (2)
bWrite down the threshold frequency of metal P. (1)
cHow does the work function of metal Q compare with that of metal R? Choose from SMALLER THAN, GREATER THAN or EQUAL TO. Explain the answer. (3)
dLight of the SAME frequency but with a HIGHER intensity is now incident on metal T. Will electrons now be emitted? Choose from YES or NO. Give a reason for the answer. (2)
eCalculate the work function of metal Q. (4)
fLight of frequency 8,0 × 10¹⁴ Hz is now incident on metal R. Show, by means of a calculation, whether electrons will be emitted from metal R. (3)

Hint

At one fixed frequency every photon carries the same energy hf, so a larger measured E_k(max) means a smaller work function.

Worked answer

a) The work function is the minimum energy that a photon must have to eject an electron from the surface of a metal. ✓✓

b) $1, 25 \times 1 0^{15}$ Hz ✓ (electrons are emitted with zero kinetic energy, so this frequency is exactly the threshold).

c) GREATER THAN. ✓ The photon energy $h f$ is the same for both metals; since $W_{0} = h f - E_{k (ma x)}$ ✓ and Q's electrons have the SMALLER maximum kinetic energy, Q's work function must be larger than R's. ✓

d) NO. ✓ Increasing the intensity only increases the NUMBER of photons per second; each photon still has energy $h f$ , which is less than the work function of T. ✓

e) $E = W_{0} + E_{k (ma x)}$ ✓
$W_{0} = h f - E_{k (ma x)} = (6, 63 \times 1 0^{- 34}) (1, 25 \times 1 0^{15}) - 2, 29 \times 1 0^{- 19}$ ✓✓
$W_{0} = 8, 2875 \times 1 0^{- 19} - 2, 29 \times 1 0^{- 19} = 6, 00 \times 1 0^{- 19}$ J ✓

f) $W_{0} (R) = 8, 2875 \times 1 0^{- 19} - 4, 79 \times 1 0^{- 19} = 3, 50 \times 1 0^{- 19}$ J ✓
Photon energy: $E = h f = (6, 63 \times 1 0^{- 34}) (8, 0 \times 1 0^{14}) = 5, 30 \times 1 0^{- 19}$ J ✓
$5, 30 \times 1 0^{- 19}$ J $> 3, 50 \times 1 0^{- 19}$ J, so electrons WILL be emitted. ✓

Modelled on Nov 2025 Paper 1, Question 10 — same skills, new scenario.

Question 4

Einstein's photoelectric equation & graphsComprehension2 marksPaper 1

Monochromatic light of steadily increasing FREQUENCY shines on the same metal surface, and the maximum kinetic energy Eₖ(max) of the emitted photoelectrons is recorded.

For frequencies above the threshold frequency f₀, the graph of Eₖ(max) against frequency f is ...

Hint

Rearrange $E_{k (ma x)} = h f - W_{0}$ into the form $y = m x + c$ .

Worked answer

B. $E_{k (ma x)} = h f - W_{0}$ is linear in $f$ with gradient $h$ ; $E_{k (ma x)} = 0$ when $h f = W_{0}$ , i.e. at $f = f_{0}$ , so the line cuts the frequency axis at the threshold frequency. A would require a zero work function. C is wrong — the relationship is linear, not saturating. D would mean the photoelectron energy ignores the light's frequency, contradicting the photon model. Modelled on Nov 2025 Paper 1, Question 1.10 — same skills, new scenario.

Open Photoelectric effect & spectra in the app →

Grade 12 Physical Sciences: Photoelectric effect & spectra

Question 1

Question 2

Question 3

Question 4

About this material