University Quantum Mechanics II · topic UQMII-09.08
The 21 cm line: F = 1 and F = 0
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 21 cm line: F = 1 and F = 0: predict a finite change
The 21 cm line is the photon hydrogen emits when its ground state drops from F = 1, where the electron and proton spins combine into a triplet, to the F = 0 singlet 5.87 μeV below it. Use the supplied topic model Y_out = 5.877444*r*(1-0.001632073*rm)*(1+0.001159652*ge), where Y_out is ΔE = E(F=1) − E(F=0) (μeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r is Nuclear moment μ/μₚ, dimensionless. Keep rm (Reduced mass (0 off, 1 on)) = 0; ge (Electron g: 0 sets 2, 1 sets 2.00232) = 0 fixed. For this exercise use only -0.76625 ≤ r ≤ 0.96625 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 r = -0.3851 and -0.14255. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δr across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
ΔE = E(F=1) − E(F=0) against r
Predictions from the supplied The 21 cm line: F = 1 and F = 0 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): ΔE = E(F=1) − E(F=0) / μeV
Horizontal axis (x): r: Nuclear moment μ/μₚ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Nuclear moment μ/μₚ | ΔE = E(F=1) − E(F=0) / μeV |
|---|---|
| -0.76625 | -4.5035914649999995 |
| -0.333125 | -1.9579235325 |
| 0.09999999999999998 | 0.5877443999999998 |
| 0.533125 | 3.1334123324999994 |
| 0.9662499999999999 | 5.679080265 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of r separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At r = -0.3851, substitute into Y_out = 5.877444*r*(1-0.001632073*rm)*(1+0.001159652*ge): Y₀ = -2.2634 μeV.
- At r = -0.14255, the same fixed-parameter model gives Y₁ = -0.83783 μeV.
- Signed change: ΔY_out = Y₁ − Y₀ = 1.42557 μeV; the input change is Δr = 0.24255.
- Average sensitivity is ΔY_out/Δr = 5.87744 μeV per unit input. 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 = -2.2634 μeV, Y₁ = -0.83783 μeV; ΔY_out = 1.42557 μeV; average sensitivity = 5.87744 μeV per unit input.
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 21 cm line: F = 1 and F = 0: infer an input with uncertainty
The 21 cm line is the photon hydrogen emits when its ground state drops from F = 1, where the electron and proton spins combine into a triplet, to the F = 0 singlet 5.87 μeV below it. Use the supplied topic model Y_out = 5.877444*r*(1-0.001632073*rm)*(1+0.001159652*ge), where Y_out is ΔE = E(F=1) − E(F=0) (μeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r is Nuclear moment μ/μₚ, dimensionless. Keep rm (Reduced mass (0 off, 1 on)) = 0; ge (Electron g: 0 sets 2, 1 sets 2.00232) = 0 fixed. For this exercise use only -0.76625 ≤ r ≤ 0.96625 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 = 1.7078383 μeV, with a bounded reading uncertainty ±0.59569 μeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00017325, 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.
ΔE = E(F=1) − E(F=0) against r
Predictions from the supplied The 21 cm line: F = 1 and F = 0 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): ΔE = E(F=1) − E(F=0) / μeV
Horizontal axis (x): r: Nuclear moment μ/μₚ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Nuclear moment μ/μₚ | ΔE = E(F=1) − E(F=0) / μeV |
|---|---|
| -0.76625 | -4.5035914649999995 |
| -0.333125 | -1.9579235325 |
| 0.09999999999999998 | 0.5877443999999998 |
| 0.533125 | 3.1334123324999994 |
| 0.9662499999999999 | 5.679080265 |
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(r) = 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(-0.76625) = -4.50359 μeV and Y_out(0.96625) = 5.67908 μeV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives r_line = -0.76625 + (1.70784 − -4.50359)(0.96625 − -0.76625)/(5.67908 − -4.50359) = 0.290575. 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 r = 0.290575; substituting back gives Y_out = 1.7078383 μeV.
- The interpolation error in the input is |r_line − r| = 0. This is within the requested numerical tolerance.
- Invert both limits Y_out = 1.1121483 μeV and 2.3035283 μeV. Sort the roots even if the response decreases: 0.189223 ≤ r ≤ 0.391927.
- The lower and upper input deviations are 0.101352 and 0.101352. 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 r ≈ 0.290575; compatible interval [0.189223, 0.391927]. Endpoint interpolation gives 0.290575.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.