University Quantum Mechanics I · topic UQMI-01.03
The photoelectric effect and the photon hypothesis
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
The photoelectric effect and the photon hypothesis: predict a finite change
The photoelectric effect is the emission of electrons from a surface under light, in which each electron absorbs the whole energy hν of one quantum, so the fastest leave with hν − φ however bright the light. Use the supplied topic model Y_out = 2998/nu, where Y_out is WAVELENGTH λ (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. nu is Frequency ν, measured in 10¹⁴ Hz. Keep phi (Work function φ) = 2.3 eV; pw (Beam power P) = 1 μW fixed. For this exercise use only 4.85 10¹⁴ Hz ≤ nu ≤ 11.15 10¹⁴ Hz on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at nu = 6.236 10¹⁴ Hz and 7.118 10¹⁴ Hz. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δnu across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
WAVELENGTH λ against nu
Predictions from the supplied The photoelectric effect and the photon hypothesis model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): WAVELENGTH λ / nm
Horizontal axis (x): nu: Frequency ν / 10¹⁴ Hz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| nu: Frequency ν / 10¹⁴ Hz | WAVELENGTH λ / nm |
|---|---|
| 4.85 | 618.1443298969073 |
| 6.424999999999999 | 466.614785992218 |
| 7.999999999999999 | 374.75000000000006 |
| 9.575 | 313.1070496083551 |
| 11.149999999999999 | 268.8789237668162 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of nu separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At nu = 6.236 10¹⁴ Hz, substitute into Y_out = 2998/nu: Y₀ = 480.757 nm.
- At nu = 7.118 10¹⁴ Hz, the same fixed-parameter model gives Y₁ = 421.186 nm.
- Signed change: ΔY_out = Y₁ − Y₀ = -59.5712 nm; the input change is Δnu = 0.882 10¹⁴ Hz.
- Average sensitivity is ΔY_out/Δnu = -67.541 nm per 10¹⁴ Hz. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 480.757 nm, Y₁ = 421.186 nm; ΔY_out = -59.5712 nm; average sensitivity = -67.541 nm per 10¹⁴ Hz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
The photoelectric effect and the photon hypothesis: infer an input with uncertainty
The photoelectric effect is the emission of electrons from a surface under light, in which each electron absorbs the whole energy hν of one quantum, so the fastest leave with hν − φ however bright the light. Use the supplied topic model Y_out = 2998/nu, where Y_out is WAVELENGTH λ (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. nu is Frequency ν, measured in 10¹⁴ Hz. Keep phi (Work function φ) = 2.3 eV; pw (Beam power P) = 1 μW fixed. For this exercise use only 4.85 10¹⁴ Hz ≤ nu ≤ 11.15 10¹⁴ Hz on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 344.87519 nm, with a bounded reading uncertainty ±11.399 nm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown nu by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00063 10¹⁴ Hz, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
WAVELENGTH λ against nu
Predictions from the supplied The photoelectric effect and the photon hypothesis model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): WAVELENGTH λ / nm
Horizontal axis (x): nu: Frequency ν / 10¹⁴ Hz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| nu: Frequency ν / 10¹⁴ Hz | WAVELENGTH λ / nm |
|---|---|
| 4.85 | 618.1443298969073 |
| 6.424999999999999 | 466.614785992218 |
| 7.999999999999999 | 374.75000000000006 |
| 9.575 | 313.1070496083551 |
| 11.149999999999999 | 268.8789237668162 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(nu) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(4.85) = 618.144 nm and Y_out(11.15) = 268.879 nm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives nu_line = 4.85 + (344.875 − 618.144)(11.15 − 4.85)/(268.879 − 618.144) = 9.77919 10¹⁴ Hz. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives nu = 8.693 10¹⁴ Hz; substituting back gives Y_out = 344.87519 nm.
- The interpolation error in the input is |nu_line − nu| = 1.08619 10¹⁴ Hz. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 333.47619 nm and 356.27419 nm. Sort the roots even if the response decreases: 8.41487 10¹⁴ Hz ≤ nu ≤ 8.99015 10¹⁴ Hz.
- The lower and upper input deviations are 0.278133 10¹⁴ Hz and 0.297147 10¹⁴ Hz. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer nu ≈ 8.693 10¹⁴ Hz; compatible interval [8.41487, 8.99015] 10¹⁴ Hz. Endpoint interpolation gives 9.77919 10¹⁴ Hz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.