University Quantum Mechanics II · topic UQMII-13.04
The 2p lifetime of hydrogen computed
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 2p lifetime of hydrogen computed: predict a finite change
The 2p lifetime is τ = 1/A, the inverse of the spontaneous-emission rate to 1s, which is fixed by the cube of the Lyman-α frequency and by the squared dipole matrix element e²|⟨100|r|21m⟩|², the same for all three m. Use the supplied topic model Y_out = 1.29027*(1-exp(-1.5*z*r)*(1+z*r*(1.5+z*r*(1.125+z*r*(0.5625+0.2109375*z*r)))))/z, where Y_out is RADIAL INTEGRAL FROM 0 TO R (a₀). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. r is Upper limit R of the radial integral, measured in a₀. Keep z (Nuclear charge Z (hydrogen-like ion)) = 1 fixed. For this exercise use only 1.72188 a₀ ≤ r ≤ 10.7781 a₀ 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 = 3.71425 a₀ and 4.98212 a₀. 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.
RADIAL INTEGRAL FROM 0 TO R against r
Predictions from the supplied The 2p lifetime of hydrogen computed 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): RADIAL INTEGRAL FROM 0 TO R / a₀
Horizontal axis (x): r: Upper limit R of the radial integral / a₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Upper limit R of the radial integral / a₀ | RADIAL INTEGRAL FROM 0 TO R / a₀ |
|---|---|
| 1.721875 | 0.15503511029679815 |
| 3.9859374999999995 | 0.9188143244691869 |
| 6.249999999999999 | 1.2340717826720429 |
| 8.5140625 | 1.2845834649366372 |
| 10.778125 | 1.2898156243003407 |
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 = 3.71425 a₀, substitute into Y_out = 1.29027*(1-exp(-1.5*z*r)*(1+z*r*(1.5+z*r*(1.125+z*r*(0.5625+0.2109375*z*r)))))/z: Y₀ = 0.843185 a₀.
- At r = 4.98212 a₀, the same fixed-parameter model gives Y₁ = 1.11734 a₀.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.274151 a₀; the input change is Δr = 1.26787 a₀.
- Average sensitivity is ΔY_out/Δr = 0.216229 a₀ per a₀. 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 = 0.843185 a₀, Y₁ = 1.11734 a₀; ΔY_out = 0.274151 a₀; average sensitivity = 0.216229 a₀ per a₀.
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 2p lifetime of hydrogen computed: infer an input with uncertainty
The 2p lifetime is τ = 1/A, the inverse of the spontaneous-emission rate to 1s, which is fixed by the cube of the Lyman-α frequency and by the squared dipole matrix element e²|⟨100|r|21m⟩|², the same for all three m. Use the supplied topic model Y_out = 1.29027*(1-exp(-1.5*z*r)*(1+z*r*(1.5+z*r*(1.125+z*r*(0.5625+0.2109375*z*r)))))/z, where Y_out is RADIAL INTEGRAL FROM 0 TO R (a₀). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. r is Upper limit R of the radial integral, measured in a₀. Keep z (Nuclear charge Z (hydrogen-like ion)) = 1 fixed. For this exercise use only 1.72188 a₀ ≤ r ≤ 10.7781 a₀ 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 = 1.2689892 a₀, with a bounded reading uncertainty ±0.003124 a₀. 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 0.000905625 a₀, 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.
RADIAL INTEGRAL FROM 0 TO R against r
Predictions from the supplied The 2p lifetime of hydrogen computed 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): RADIAL INTEGRAL FROM 0 TO R / a₀
Horizontal axis (x): r: Upper limit R of the radial integral / a₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Upper limit R of the radial integral / a₀ | RADIAL INTEGRAL FROM 0 TO R / a₀ |
|---|---|
| 1.721875 | 0.15503511029679815 |
| 3.9859374999999995 | 0.9188143244691869 |
| 6.249999999999999 | 1.2340717826720429 |
| 8.5140625 | 1.2845834649366372 |
| 10.778125 | 1.2898156243003407 |
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(1.72188) = 0.155035 a₀ and Y_out(10.7781) = 1.28982 a₀. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives r_line = 1.72188 + (1.26899 − 0.155035)(10.7781 − 1.72188)/(1.28982 − 0.155035) = 10.6119 a₀. 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 = 7.24619 a₀; substituting back gives Y_out = 1.2689892 a₀.
- The interpolation error in the input is |r_line − r| = 3.36573 a₀. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.2658652 a₀ and 1.2721132 a₀. Sort the roots even if the response decreases: 7.10957 a₀ ≤ r ≤ 7.40317 a₀.
- The lower and upper input deviations are 0.136615 a₀ and 0.156986 a₀. 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 r ≈ 7.24619 a₀; compatible interval [7.10957, 7.40317] a₀. Endpoint interpolation gives 10.6119 a₀.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.