University Physics II · topic UPII-15.04
Photons and the photoelectric effect
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.
Photons and the photoelectric effect: predict a finite change
A photon is the quantum of an electromagnetic wave of frequency f: a massless packet carrying energy E = hf and momentum p = h/λ, absorbed or scattered whole rather than drained gradually. Use the supplied topic model Y_out = 1239.8/phi, where Y_out is THRESHOLD λ₀ = hc/φ (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. phi is Work function φ, measured in eV. Keep f (Photon frequency f) = 7.4 ×10¹⁴ Hz fixed. For this exercise use only 1.76563 eV ≤ phi ≤ 3.73438 eV 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 phi = 2.19875 eV and 2.47438 eV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δphi across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
THRESHOLD λ₀ = hc/φ against phi
Predictions from the supplied Photons and the photoelectric effect 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): THRESHOLD λ₀ = hc/φ / nm
Horizontal axis (x): phi: Work function φ / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| phi: Work function φ / eV | THRESHOLD λ₀ = hc/φ / nm |
|---|---|
| 1.765625 | 702.187610619469 |
| 2.2578125 | 549.1155709342561 |
| 2.75 | 450.8363636363636 |
| 3.2421875 | 382.3961445783132 |
| 3.734375 | 331.99665271966524 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of phi separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At phi = 2.19875 eV, substitute into Y_out = 1239.8/phi: Y₀ = 563.866 nm.
- At phi = 2.47438 eV, the same fixed-parameter model gives Y₁ = 501.056 nm.
- Signed change: ΔY_out = Y₁ − Y₀ = -62.81 nm; the input change is Δphi = 0.275625 eV.
- Average sensitivity is ΔY_out/Δphi = -227.882 nm per eV. 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 = 563.866 nm, Y₁ = 501.056 nm; ΔY_out = -62.81 nm; average sensitivity = -227.882 nm per eV.
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.
Photons and the photoelectric effect: infer an input with uncertainty
A photon is the quantum of an electromagnetic wave of frequency f: a massless packet carrying energy E = hf and momentum p = h/λ, absorbed or scattered whole rather than drained gradually. Use the supplied topic model Y_out = 1239.8/phi, where Y_out is THRESHOLD λ₀ = hc/φ (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. phi is Work function φ, measured in eV. Keep f (Photon frequency f) = 7.4 ×10¹⁴ Hz fixed. For this exercise use only 1.76563 eV ≤ phi ≤ 3.73438 eV 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 = 417.92479 nm, with a bounded reading uncertainty ±12.889 nm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown phi by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000196875 eV, 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.
THRESHOLD λ₀ = hc/φ against phi
Predictions from the supplied Photons and the photoelectric effect 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): THRESHOLD λ₀ = hc/φ / nm
Horizontal axis (x): phi: Work function φ / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| phi: Work function φ / eV | THRESHOLD λ₀ = hc/φ / nm |
|---|---|
| 1.765625 | 702.187610619469 |
| 2.2578125 | 549.1155709342561 |
| 2.75 | 450.8363636363636 |
| 3.2421875 | 382.3961445783132 |
| 3.734375 | 331.99665271966524 |
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(phi) = 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(1.76563) = 702.188 nm and Y_out(3.73438) = 331.997 nm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives phi_line = 1.76563 + (417.925 − 702.188)(3.73438 − 1.76563)/(331.997 − 702.188) = 3.27739 eV. 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 phi = 2.96656 eV; substituting back gives Y_out = 417.92479 nm.
- The interpolation error in the input is |phi_line − phi| = 0.310829 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 405.03579 nm and 430.81379 nm. Sort the roots even if the response decreases: 2.87781 eV ≤ phi ≤ 3.06096 eV.
- The lower and upper input deviations are 0.088753 eV and 0.0944016 eV. 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 phi ≈ 2.96656 eV; compatible interval [2.87781, 3.06096] eV. Endpoint interpolation gives 3.27739 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.