Question 1Multiple choice · 1 mark
Light of wavelength 200 nm falls on a metal with work function 4.3 eV.
What is the stopping potential?
- A 1.9 V
- B 4.1 V
- C 4.3 V
- D 6.2 V
Show answer
Answer: A
Free E.2 Quantum physics questions for IB Physics (SL & HL), HL only, part of E Nuclear & quantum physics. Each question has the answer and the marking points an examiner would look for.
Generate a full E.2 Quantum Physics worksheet →All E Nuclear & quantum physics questionsLight of wavelength 200 nm falls on a metal with work function 4.3 eV.
What is the stopping potential?
Answer: A
In a photoelectric experiment the intensity of the light is doubled while its frequency is kept constant. What happens to the maximum kinetic energy of the photoelectrons and the photocurrent?
| maximum kinetic energy | photocurrent | |
|---|---|---|
| A | unchanged | unchanged |
| B | doubles | doubles |
| C | unchanged | doubles |
| D | doubles | unchanged |
Answer: C
Light of wavelength 400 nm falls on a metal surface of work function 2.3 eV.
Calculate the photon energy in eV.
Calculate the maximum kinetic energy of the emitted electrons and the stopping potential.
Calculate the threshold frequency.
The intensity of the light is doubled. State the effect on the stopping potential and on the photocurrent.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 3.1 eV hc/λ = 4.97 × 10−19 J = 3.1 eV ✓ | 1 |
| (b) | 0.81 eV; 0.81 V Ek = 3.11 − 2.3 ✓ = 0.81 eV (1.3 × 10−19 J) ✓ stopping potential 0.81 V ✓ | 3 |
| (c) | 5.6 × 1014 Hz f0 = Φ/h ✓ = 5.6 × 1014 Hz ✓ | 2 |
| (d) | unchanged; doubles stopping potential unchanged ✓ photocurrent doubles ✓ | 2 |
In a photoelectric experiment, light of wavelength 400 nm gives a stopping potential of 1.05 V and light of wavelength 250 nm gives 2.91 V. e = 1.60 × 10−19 C, c = 3.00 × 108 m s−1, ε₀ = 8.85 × 10−12 F m−1.
Use the two results, without using a known value of h, to determine the Planck constant and the work function of the metal.
A 2.0 mW beam of 250 nm light produces a photocurrent of 0.12 μA. Determine the fraction of photons that release an electron.
The same beam falls on an isolated sphere of this metal, of radius 5.0 cm, which initially carries a charge of −2.0 × 10−11 C. Determine the final potential of the sphere and the minimum time taken to reach it.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | h ≈ 6.6 × 10⁻³⁴ J s; φ ≈ 2.05 eV eV_s = hc/λ − φ for each ✓ subtract: e(2.91 − 1.05) = hc(1/250 nm − 1/400 nm) ✓ h = 6.6 × 10−34 J s ✓ φ = hc/λ − eV_s ✓ φ ≈ 3.3 × 10−19 J ≈ 2.05 eV ✓ | 5 |
| (b) | ≈ 3 × 10⁻⁴ photons per second = Pλ/hc = 2.5 × 1015 ✓ electrons per second = I/e = 7.5 × 1011 ✓ ratio ≈ 3.0 × 10−4 ✓ | 3 |
| (c) | +2.91 V; ≈ 3.0 × 10⁻⁴ s emission stops when the sphere’s potential equals the stopping potential, +2.91 V ✓ while the sphere is negative every emitted electron escapes, so the current is 0.12 μA for 2.0 × 10−11/0.12 × 10−6 = 1.7 × 10−4 s ✓ then a further Q = 4πε₀rV = 1.6 × 10−11 C must be lost ✓ current now at most 0.12 μA (it falls as V rises), so at least 1.3 × 10−4 s more ✓ minimum total ≈ 3.0 × 10−4 s ✓ | 5 |
In a photoelectric experiment, light of wavelength 200 nm gives a stopping potential of 4.02 V and light of wavelength 280 nm gives a stopping potential of 2.24 V.
Use these data to determine a value for the Planck constant.
Determine the work function of the metal in eV.
Explain why the stopping potential does not depend on the intensity of the light.
Calculate the de Broglie wavelength of the fastest electrons emitted by the 200 nm light.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 6.6 × 10−34 J s eΔV = hcΔ(1/λ) ✓ Δ(1/λ) = 1.43 × 106 m−1 ✓ h ≈ 6.6 × 10−34 J s ✓ | 3 |
| (b) | 2.2 eV Φ = hf − eVs ✓ = 2.2 eV ✓ | 2 |
| (c) | photon energy fixed each electron absorbs one photon whose energy depends only on frequency ✓ intensity changes only the number of photons per second ✓ | 2 |
| (d) | 6.1 × 10−10 m p = √(2meEk) ✓ λ = h/p = 6.1 × 10−10 m ✓ | 2 |
An electron is accelerated from rest through a potential difference of 150 V.
Calculate its momentum.
Calculate its de Broglie wavelength and suggest why such electrons can be diffracted by crystals.
A proton has the same de Broglie wavelength. Calculate its speed.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 6.6 × 10−24 kg m s−1 p = √(2meV) ✓ = 6.6 × 10−24 kg m s−1 ✓ | 2 |
| (b) | 1.0 × 10−10 m λ = h/p = 1.0 × 10−10 m ✓ similar to the spacing of atoms ✓ | 2 |
| (c) | 4.0 × 103 m s−1 same momentum ✓ v = 6.6 × 10−24/1.67 × 10−27 = 4.0 × 103 m s−1 ✓ | 2 |
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