University Physics III · topic UPIII-10.09
Inner-shell transitions and X-ray spectra
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.
Inner-shell transitions and X-ray spectra: predict a finite change
An inner-shell transition is the refilling of a K- or L-shell vacancy by an electron from a higher shell; the X-ray photon it emits carries the difference between the two binding energies. Use the supplied topic model Y_out = 1.23984/v, where Y_out is CUTOFF λ_min = hc/eV (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. v is Tube voltage V, measured in kV. Keep z (Anode atomic number Z) = 42 fixed. For this exercise use only 14.25 kV ≤ v ≤ 45.75 kV 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 v = 21.18 kV and 25.59 kV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δv across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
CUTOFF λ_min = hc/eV against v
Predictions from the supplied Inner-shell transitions and X-ray spectra 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): CUTOFF λ_min = hc/eV / nm
Horizontal axis (x): v: Tube voltage V / kV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Tube voltage V / kV | CUTOFF λ_min = hc/eV / nm |
|---|---|
| 14.25 | 0.08700631578947368 |
| 22.125 | 0.05603796610169492 |
| 30 | 0.041328000000000004 |
| 37.875 | 0.03273504950495049 |
| 45.75 | 0.02710032786885246 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of v separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At v = 21.18 kV, substitute into Y_out = 1.23984/v: Y₀ = 0.0585382 nm.
- At v = 25.59 kV, the same fixed-parameter model gives Y₁ = 0.0484502 nm.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0100881 nm; the input change is Δv = 4.41 kV.
- Average sensitivity is ΔY_out/Δv = -0.00228754 nm per kV. 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 = 0.0585382 nm, Y₁ = 0.0484502 nm; ΔY_out = -0.0100881 nm; average sensitivity = -0.00228754 nm per kV.
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.
Inner-shell transitions and X-ray spectra: infer an input with uncertainty
An inner-shell transition is the refilling of a K- or L-shell vacancy by an electron from a higher shell; the X-ray photon it emits carries the difference between the two binding energies. Use the supplied topic model Y_out = 1.23984/v, where Y_out is CUTOFF λ_min = hc/eV (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. v is Tube voltage V, measured in kV. Keep z (Anode atomic number Z) = 42 fixed. For this exercise use only 14.25 kV ≤ v ≤ 45.75 kV 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 = 0.037048857 nm, with a bounded reading uncertainty ±0.0014923 nm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00315 kV, 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.
CUTOFF λ_min = hc/eV against v
Predictions from the supplied Inner-shell transitions and X-ray spectra 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): CUTOFF λ_min = hc/eV / nm
Horizontal axis (x): v: Tube voltage V / kV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Tube voltage V / kV | CUTOFF λ_min = hc/eV / nm |
|---|---|
| 14.25 | 0.08700631578947368 |
| 22.125 | 0.05603796610169492 |
| 30 | 0.041328000000000004 |
| 37.875 | 0.03273504950495049 |
| 45.75 | 0.02710032786885246 |
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(v) = 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(14.25) = 0.0870063 nm and Y_out(45.75) = 0.0271003 nm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives v_line = 14.25 + (0.0370489 − 0.0870063)(45.75 − 14.25)/(0.0271003 − 0.0870063) = 40.5188 kV. 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 v = 33.465 kV; substituting back gives Y_out = 0.037048857 nm.
- The interpolation error in the input is |v_line − v| = 7.05383 kV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.035556557 nm and 0.038541157 nm. Sort the roots even if the response decreases: 32.1692 kV ≤ v ≤ 34.8695 kV.
- The lower and upper input deviations are 1.29575 kV and 1.40452 kV. 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 v ≈ 33.465 kV; compatible interval [32.1692, 34.8695] kV. Endpoint interpolation gives 40.5188 kV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.