4.10.5Electrons, Waves and Photons
Maximum Kinetic Energy and Intensity — OCR A Level Physics A H556 Revision Notes & Topical Questions
5 min read 6 parts Lesson 17 of 62 Unit quiz
Part 1 of 6Intensity in the Photon Model
Photon rate
- The intensity of radiation depends on the number of photons arriving per unit area per second, each of energy hf.
One-to-one
- Each photon frees at most one electron, so the number of photoelectrons per second is proportional to the photon rate.
Photocurrent
- Above the threshold frequency, doubling the intensity doubles the number of electrons per second and so the photocurrent.
Energy unchanged
- Increasing intensity does not change the energy of any single photon, so KEmax stays the same.
Part 2 of 6Effect of Changing Frequency
Effect of Changing Frequency
Higher frequency
- Each photon carries more energy, so each photoelectron can leave with a larger maximum kinetic energy.
Linear link
- KEmax = hf − Φ means KEmax rises in proportion to the excess of f over the threshold frequency.
Number of electrons
- At constant power, a higher frequency means fewer photons per second, so fewer electrons are emitted.
Below threshold
- At any intensity, no electrons are emitted if the frequency is below f₀.
| Change (above f₀) | KEmax | Electrons per second |
|---|---|---|
| Intensity increased | Unchanged | Increases |
| Frequency increased, same photon rate | Increases | Unchanged |
| Frequency below f₀ | No emission | Zero |
Part 3 of 6Applying It to a Question
1
State whether the frequency is above or below the threshold frequency.
2
If it is above, say that each photon gives its energy to one electron.
3
Explain intensity as the number of photons per second, which controls the number of electrons emitted per second.
4
Explain that KEmax depends only on frequency through hf − Φ.
5
Conclude with the effect on the photocurrent and on the energy of the electrons.
Part 4 of 6Common Graph Shapes
Photocurrent against intensity
- Above threshold this is a straight line through the origin, since the current is proportional to the photon rate.
KEmax against intensity
- This is a horizontal line, because the energy of the electrons is independent of intensity.
KEmax against frequency
- This is a straight line with gradient h that starts at the threshold frequency.
Different metals
- Different work functions give parallel lines shifted along the frequency axis.
Part 5 of 6Worked Example
Question
- Radiation of wavelength 500 nm has a power of 2.0 mW.
- Calculate the number of photons arriving per second.
- (h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m/s)
- Energy of one photon E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (500 × 10⁻⁹) = 3.98 × 10⁻¹⁹ J.
- Power = 2.0 × 10⁻³ W = 2.0 × 10⁻³ J per second.
- Photons per second = power / energy per photon = 2.0 × 10⁻³ / 3.98 × 10⁻¹⁹ = 5.0 × 10¹⁵.
Answer About 5.0 × 10¹⁵ photons per second.
Part 6 of 6Exam Checklist & Tips
4.10.5 – Key Points for the Exam
- Intensity Power per unit area, which depends on the number of photons arriving per second per unit area.
- Photocurrent The current due to photoelectrons leaving the surface.
- Maximum kinetic energy Greatest kinetic energy of emitted photoelectrons, equal to hf − Φ.
- Photon rate The number of photons arriving each second.
- Threshold frequency Minimum frequency that allows photoelectric emission.
Common Mistakes
- Saying more intense light gives faster electrons; it gives more electrons at the same maximum energy.
- Forgetting that a photon below threshold energy cannot be helped by raising the intensity.
- Writing that photocurrent depends on frequency alone; it depends on the photon rate as well.
- Using the word power instead of photons per second when explaining intensity in the photon model.
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