IB · E · extended response
A photoelectric threshold measured with stopping voltage
Nuclear and quantum physics · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Photoelectric stopping potential
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Stopping potential / V
Horizontal axis (x): Frequency / 10¹⁴ Hz
Read the exact data as a table
| Frequency / 10¹⁴ Hz | Stopping potential / V |
|---|---|
| 5 | 0 |
| 6 | 0.414 |
| 7 | 0.828 |
| 8 | 1.242 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
A metal's stopping-potential measurements against incident-light frequency are supplied. Use electron charge e = 1.60×10⁻¹⁹ C and Kmax = eVs. The stopping-potential fit extrapolates to Vs = 0 at 5.0×10¹⁴ Hz. Assume one-photon photoemission and no contact-potential correction. Supplied data (Frequency / 10¹⁴ Hz; Stopping potential / V): (5, 0), (6, 0.414), (7, 0.828), (8, 1.242).
- (a)
Estimate Planck's constant and the work function from the graph; find maximum electron kinetic energy at 8.0×10¹⁴ Hz.
3 marks - (b)
Predict the effect of doubling intensity at fixed frequency above threshold, and of using intense light below threshold.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Estimate Planck's constant and the work function from the graph; find maximum electron kinetic energy at 8.0×10¹⁴ Hz.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Einstein's relation gives Vs = (h/e)f−φ/e. Graph slope is 0.414/(1.0×10¹⁴) = 4.14×10⁻¹⁵ V s.
- 2
Thus h = e×slope = 6.624×10⁻³⁴ J s. At threshold, φ = hf0 = 3.312×10⁻¹⁹ J, or 2.07 eV.
- 3
At 8.0×10¹⁴ Hz, Kmax = eVs = (1.60×10⁻¹⁹)(1.242)≈1.987×10⁻¹⁹ J. The graph intercept is an energy-per-charge condition, not a delay.
(b)
Predict the effect of doubling intensity at fixed frequency above threshold, and of using intense light below threshold.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
At fixed frequency, each photon carries the same hf. Doubling intensity supplies more photons per second, which can increase photocurrent if collection and material conditions remain unchanged.
- 2
Stopping potential depends on the maximum energy per emitted electron, so it does not double with intensity.
- 3
Below threshold each photon lacks the work-function energy. Increasing photon arrival rate does not fix that within the stated single-photon model; multiphoton regimes are outside this approximation.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- Einstein's relation gives Vs = (h/e)f−φ/e. Graph slope is 0.414/(1.0×10¹⁴) = 4.14×10⁻¹⁵ V s.
- Thus h = e×slope = 6.624×10⁻³⁴ J s. At threshold, φ = hf0 = 3.312×10⁻¹⁹ J, or 2.07 eV.
- At 8.0×10¹⁴ Hz, Kmax = eVs = (1.60×10⁻¹⁹)(1.242)≈1.987×10⁻¹⁹ J. The graph intercept is an energy-per-charge condition, not a delay.
- At fixed frequency, each photon carries the same hf. Doubling intensity supplies more photons per second, which can increase photocurrent if collection and material conditions remain unchanged.
- Stopping potential depends on the maximum energy per emitted electron, so it does not double with intensity.
- Below threshold each photon lacks the work-function energy. Increasing photon arrival rate does not fix that within the stated single-photon model; multiphoton regimes are outside this approximation.
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