University Physics IV · topic UPIV-11.09
Inner-shell vacancies, X-ray spectra, and Moseley's law
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 X-rays and Moseley's law: predict a finite change
Moseley's law states that the square root of a characteristic X-ray frequency rises linearly with atomic number, √ν = A(Z − σ), where σ is one screening constant standing in for every other electron in the atom. Use the supplied topic model Y_out = 1.2398/max(v,0.01), where Y_out is CUT-OFF WAVELENGTH (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. v is Tube accelerating voltage, measured in kV. Keep z (Anode atomic number Z) = 42 fixed. For this exercise use only 19.5625 kV ≤ v ≤ 90.4375 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 = 35.155 kV and 45.0775 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.
CUT-OFF WAVELENGTH against v
Predictions from the supplied Inner-shell X-rays and Moseley's law 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): CUT-OFF WAVELENGTH / nm
Horizontal axis (x): v: Tube accelerating voltage / kV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Tube accelerating voltage / kV | CUT-OFF WAVELENGTH / nm |
|---|---|
| 19.5625 | 0.06337635782747604 |
| 37.28125 | 0.03325532271584242 |
| 55 | 0.022541818181818184 |
| 72.71875 | 0.017049247958745167 |
| 90.4375 | 0.013708914996544575 |
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 = 35.155 kV, substitute into Y_out = 1.2398/max(v,0.01): Y₀ = 0.0352667 nm.
- At v = 45.0775 kV, the same fixed-parameter model gives Y₁ = 0.0275037 nm.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.00776293 nm; the input change is Δv = 9.9225 kV.
- Average sensitivity is ΔY_out/Δv = -0.000782357 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 = 0.0352667 nm, Y₁ = 0.0275037 nm; ΔY_out = -0.00776293 nm; average sensitivity = -0.000782357 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 X-rays and Moseley's law: infer an input with uncertainty
Moseley's law states that the square root of a characteristic X-ray frequency rises linearly with atomic number, √ν = A(Z − σ), where σ is one screening constant standing in for every other electron in the atom. Use the supplied topic model Y_out = 1.2398/max(v,0.01), where Y_out is CUT-OFF WAVELENGTH (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. v is Tube accelerating voltage, measured in kV. Keep z (Anode atomic number Z) = 42 fixed. For this exercise use only 19.5625 kV ≤ v ≤ 90.4375 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 = 0.019743217 nm, with a bounded reading uncertainty ±0.00090515 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 0.0070875 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.
CUT-OFF WAVELENGTH against v
Predictions from the supplied Inner-shell X-rays and Moseley's law 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): CUT-OFF WAVELENGTH / nm
Horizontal axis (x): v: Tube accelerating voltage / kV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Tube accelerating voltage / kV | CUT-OFF WAVELENGTH / nm |
|---|---|
| 19.5625 | 0.06337635782747604 |
| 37.28125 | 0.03325532271584242 |
| 55 | 0.022541818181818184 |
| 72.71875 | 0.017049247958745167 |
| 90.4375 | 0.013708914996544575 |
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(19.5625) = 0.0633764 nm and Y_out(90.4375) = 0.0137089 nm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives v_line = 19.5625 + (0.0197432 − 0.0633764)(90.4375 − 19.5625)/(0.0137089 − 0.0633764) = 81.8266 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 = 62.7963 kV; substituting back gives Y_out = 0.019743217 nm.
- The interpolation error in the input is |v_line − v| = 19.0304 kV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.018838067 nm and 0.020648367 nm. Sort the roots even if the response decreases: 60.0435 kV ≤ v ≤ 65.8135 kV.
- The lower and upper input deviations are 2.75276 kV and 3.0173 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 ≈ 62.7963 kV; compatible interval [60.0435, 65.8135] kV. Endpoint interpolation gives 81.8266 kV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.