University Physics IV · topic UPIV-01.06
The photoelectric effect and Einstein's light quantum
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 light quantum: predict a finite change
The photoelectric effect is the emission of electrons from an illuminated surface in which one light quantum of energy hν is absorbed by one electron, which then pays the work function φ to escape: K_max = hν − φ. Use the supplied topic model Y_out = 0.4136*n, where Y_out is PHOTON ENERGY hν (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. n is Photon frequency ν, measured in ×10¹⁴ Hz. Keep pc (Cathode work function φ_C) = 2.1 eV; pa (Anode work function φ_A) = 2.6 eV fixed. For this exercise use only 5.0625 ×10¹⁴ Hz ≤ n ≤ 12.9375 ×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 n = 6.795 ×10¹⁴ Hz and 7.8975 ×10¹⁴ Hz. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δn across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
PHOTON ENERGY hν against n
Predictions from the supplied The photoelectric effect and the light quantum 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): PHOTON ENERGY hν / eV
Horizontal axis (x): n: Photon frequency ν / ×10¹⁴ Hz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| n: Photon frequency ν / ×10¹⁴ Hz | PHOTON ENERGY hν / eV |
|---|---|
| 5.0625 | 2.09385 |
| 7.03125 | 2.908125 |
| 9 | 3.7224000000000004 |
| 10.96875 | 4.536675000000001 |
| 12.9375 | 5.35095 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of n separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At n = 6.795 ×10¹⁴ Hz, substitute into Y_out = 0.4136*n: Y₀ = 2.81041 eV.
- At n = 7.8975 ×10¹⁴ Hz, the same fixed-parameter model gives Y₁ = 3.26641 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.455994 eV; the input change is Δn = 1.1025 ×10¹⁴ Hz.
- Average sensitivity is ΔY_out/Δn = 0.4136 eV per ×10¹⁴ Hz. Its positive sign means that this response increases 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 = 2.81041 eV, Y₁ = 3.26641 eV; ΔY_out = 0.455994 eV; average sensitivity = 0.4136 eV 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 light quantum: infer an input with uncertainty
The photoelectric effect is the emission of electrons from an illuminated surface in which one light quantum of energy hν is absorbed by one electron, which then pays the work function φ to escape: K_max = hν − φ. Use the supplied topic model Y_out = 0.4136*n, where Y_out is PHOTON ENERGY hν (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. n is Photon frequency ν, measured in ×10¹⁴ Hz. Keep pc (Cathode work function φ_C) = 2.1 eV; pa (Anode work function φ_A) = 2.6 eV fixed. For this exercise use only 5.0625 ×10¹⁴ Hz ≤ n ≤ 12.9375 ×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 = 4.080681 eV, with a bounded reading uncertainty ±0.19054 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown n by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0007875 ×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.
PHOTON ENERGY hν against n
Predictions from the supplied The photoelectric effect and the light quantum 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): PHOTON ENERGY hν / eV
Horizontal axis (x): n: Photon frequency ν / ×10¹⁴ Hz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| n: Photon frequency ν / ×10¹⁴ Hz | PHOTON ENERGY hν / eV |
|---|---|
| 5.0625 | 2.09385 |
| 7.03125 | 2.908125 |
| 9 | 3.7224000000000004 |
| 10.96875 | 4.536675000000001 |
| 12.9375 | 5.35095 |
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(n) = 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(5.0625) = 2.09385 eV and Y_out(12.9375) = 5.35095 eV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives n_line = 5.0625 + (4.08068 − 2.09385)(12.9375 − 5.0625)/(5.35095 − 2.09385) = 9.86625 ×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 n = 9.86625 ×10¹⁴ Hz; substituting back gives Y_out = 4.080681 eV.
- The interpolation error in the input is |n_line − n| = 3.55271e-15 ×10¹⁴ Hz. This is within the requested numerical tolerance.
- Invert both limits Y_out = 3.890141 eV and 4.271221 eV. Sort the roots even if the response decreases: 9.40556 ×10¹⁴ Hz ≤ n ≤ 10.3269 ×10¹⁴ Hz.
- The lower and upper input deviations are 0.460687 ×10¹⁴ Hz and 0.460687 ×10¹⁴ Hz. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer n ≈ 9.86625 ×10¹⁴ Hz; compatible interval [9.40556, 10.3269] ×10¹⁴ Hz. Endpoint interpolation gives 9.86625 ×10¹⁴ Hz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.