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Photoelectric Effect Practice Paper 2: photon energy, threshold and stopping voltage
Photoelectric Effect Practice Paper 2: photon energy, threshold and stopping voltage
This paper walks the equation hv = W + Kmax from left to right: photon energy from frequency or wavelength, then the threshold from the work function, then the work function back from a stopping voltage. Write the conservation equation first, substitute second; intensity changes the number of electrons, not their maximum energy. Every value is checked by computer.
Quick answer From 6.0e14 Hz the photon carries 2.49 eV; 500 nm gives the same 2.49 eV; threshold 5.31e14 Hz; threshold wavelength 622 nm; 400 nm against 2.0 eV leaves 1.11 eV; 0.80 V corresponds to 1.28e-19 J; current doubles while energy stays; work function 2.94 eV; work function 1.66 eV; the misconception item is answered as no change in energy.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | 25 minutes |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: Photoelectric Effect |
| Level | S6 |
02 Questions (10 items)
- FoundationWhat is the energy in eV of a photon of frequency 6.0e14 Hz?
- FoundationWhat is the energy in eV of a photon of wavelength 500 nm?
- FoundationA metal has a work function of 2.2 eV. What is its threshold frequency?
- FoundationWhat is the threshold wavelength of a metal with work function 2.0 eV?
- IntermediateLight of 400 nm falls on a metal of work function 2.0 eV. What is the maximum kinetic energy of the photoelectrons?
- IntermediateA stopping voltage of 0.80 V is measured. What is the maximum kinetic energy in joules?
- IntermediateThe frequency is unchanged and the intensity doubles. What happens to the saturation current and the maximum kinetic energy?
- ChallengeLight of 300 nm gives a stopping voltage of 1.20 V. Find the work function of the metal.
- ChallengeA metal has a threshold frequency of 4.0e14 Hz. Find its work function.
- Error redoA student answers that increasing the intensity raises the maximum kinetic energy of the photoelectrons. Where is the error and what is the correct conclusion?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 3.98e-19 J, that is 2.49 eV. | Use E = hf = 6.63e-34 times 6.0e14, then divide by 1.60e-19 to convert into eV. |
| 2 | 3.98e-19 J, that is 2.49 eV. | Use E = hc over lambda = 1.989e-25 divided by 5.00e-7, which matches the 6.0e14 Hz case. |
| 3 | 5.31e14 Hz. | At the threshold hf0 = W, so f0 is 3.52e-19 divided by 6.63e-34. |
| 4 | About 622 nm. | Use lambda0 = hc over W = 1.989e-25 divided by 3.20e-19. |
| 5 | 1.77e-19 J, that is 1.11 eV. | The photon carries 4.97e-19 J (3.11 eV); subtracting 2.0 eV leaves 1.11 eV. |
| 6 | 1.28e-19 J, that is 0.80 eV. | Kmax = e times Vs = 1.60e-19 times 0.80, so the value in eV equals the stopping voltage. |
| 7 | The saturation current doubles and the maximum kinetic energy stays the same. | Doubling the intensity doubles the photon count and hence the electron count, while each electron energy depends only on frequency. |
| 8 | About 2.94 eV, that is 4.71e-19 J. | The photon carries 6.63e-19 J (4.14 eV); subtracting 1.20 eV leaves 2.94 eV. |
| 9 | About 1.66 eV, that is 2.65e-19 J. | At the threshold the emitted electron has zero energy, so W = hf0 = 6.63e-34 times 4.0e14. |
| 10 | Frequency fixes the energy of one photon, so the maximum kinetic energy depends only on frequency; doubling the intensity leaves it unchanged and only increases the number of electrons. | Treating intensity as photon energy is the commonest error in this chapter; always ask first whether the frequency changed. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.
