A Level Physics (A2) chapter 11: Quantum physics MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
100 exam-style questions on Quantum physics, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.
Energy levels in atoms and line spectra24 questions
Ten Quantum physics questions with answers
Try each one, then open the answer. These ten stay the same; the test draws new ones every time.
1. The energy levels of an atom are −9.0 eV (ground state), −4.2 eV, −2.6 eV and −1.2 eV. A beam of photons passes through a gas of these atoms, all in their ground state. Which photon energy can be absorbed?
A
5.0 eV
B
3.0 eV
C
1.6 eV
D
6.4 eV
Show answer
Answer: D. From the ground state an atom can absorb only 4.8, 6.4, 7.8 eV (to each higher level). 3.0 eV and 1.6 eV are differences between two excited levels, so they cannot be absorbed by atoms in the ground state; 5.0 eV matches no transition.
2. Light of wavelength 600 nm and intensity 120 W m−2 falls normally on a photocell of area 5.0 cm2. How many photons fall on the photocell each second?
A
1.8 × 1015
B
1.8 × 1017
C
1.8 × 1019
D
3.6 × 1020
Show answer
Answer: B. Power = 120 × 5.0 × 10−4 = 0.060 W; photon energy = hc/λ = 3.3 × 10−19 J; number per second = 0.060 ÷ 3.3 × 10−19 = 1.8 × 1017. Converting 5.0 cm2 as 5.0 × 10−2 m2 gives 1.8 × 1019; ignoring the area gives 3.6 × 1020.
3. A neutron of mass 1.67 × 10−27 kg moves at 2.0 × 103 m s−1. What is its de Broglie wavelength?
A
9.9 × 10−14 m
B
9.9 × 10−11 m
C
2.0 × 10−10 m
D
3.6 × 10−7 m
Show answer
Answer: C. λ = h/mv = 6.63 × 10−34 ÷ (1.67 × 10−27 × 2.0 × 103) = 2.0 × 10−10 m. Using the electron mass gives 3.6 × 10−7 m; including a stray factor of 2 gives 9.9 × 10−11 m.
4. A monochromatic source emits electromagnetic radiation of constant power. The frequency of the radiation is doubled but the power is kept the same. What happens to the number of photons emitted per second?
A
It is halved.
B
It stays the same.
C
It doubles.
D
It is four times as large.
Show answer
Answer: A. Each photon now carries twice the energy (E = hf). For the same power, the number of photons per second = P/hf is halved.
5. Which statement explains photoelectric emission?
A
an electron gathers energy from several photons over a period of time until it has enough to escape from the metal
B
each photon shares out its energy among many electrons near the surface of the metal
C
the wave energy spreads over the surface, and the electrons escape after a short delay
D
one photon gives all its energy to one electron, which can escape if this energy is at least the work function
Show answer
Answer: D. The interaction is one photon with one electron. Some of the photon energy hf is used to overcome the work function; the rest becomes kinetic energy of the electron.
6. An electron and a proton are each accelerated from rest through the same potential difference. What is the ratio (de Broglie wavelength of the electron) ÷ (de Broglie wavelength of the proton)?
A
1/43
B
1
C
43
D
1800
Show answer
Answer: C. Both gain the same kinetic energy eV, and p = √(2mE), so λ = h/p is proportional to 1/√m. The ratio is √(mp/me) = √(1.67 × 10−27 ÷ 9.11 × 10−31) = 43. 1800 forgets the square root.
7. A photon has an energy of 2.0 eV. What is its frequency?
A
2.1 × 10−15 Hz
B
1.2 × 1014 Hz
C
4.8 × 1014 Hz
D
3.0 × 1033 Hz
Show answer
Answer: C. E = 2.0 × 1.60 × 10−19 = 3.2 × 10−19 J, so f = E/h = 3.2 × 10−19 ÷ 6.63 × 10−34 = 4.8 × 1014 Hz. Using 2.0 without converting eV to J gives 3.0 × 1033 Hz.
8. A photon has energy 3.3 × 10−15 J. In which region of the electromagnetic spectrum is it?
A
infrared
B
visible
C
ultraviolet
D
X-ray
Show answer
Answer: D. λ = hc/E = (6.63 × 10−34 × 3.00 × 108) ÷ (3.3 × 10−15) = 6.0 × 10−11 m, which is in the X-ray region (visible light is about 4 × 10−7 to 7 × 10−7 m).
9. The speed of an electron is doubled, and it remains much less than the speed of light. What happens to its de Broglie wavelength?
A
it doubles
B
it falls to a quarter
C
it does not change
D
it halves
Show answer
Answer: D. λ = h/(mv): doubling v doubles the momentum and halves the wavelength. It is the kinetic energy, not the wavelength, that changes by a factor of four.
10. The work function of a metal is 2.3 eV. What is the threshold wavelength for photoelectric emission from this metal?
A
5.4 × 10−7 m
B
8.6 × 10−26 m
C
1.9 × 106 m
D
1.8 × 10−15 m
Show answer
Answer: A. At the threshold the photon energy equals the work function: hc/λ0 = Φ. λ0 = 6.63 × 10−34 × 3.00 × 108 ÷ (2.3 × 1.60 × 10−19) = 5.4 × 10−7 m. Forgetting to convert eV to joules gives 8.6 × 10−26 m.
100, each with the answer and a short explanation, across 4 topics: Energy and momentum of a photon, Photoelectric effect, Wave-particle duality, Energy levels in atoms and line spectra. Every chapter test draws ten of them, so you can take it many times.
Is the A Level Physics (A2) chapter 11 test free?
Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.
How is the chapter test marked?
The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.
