University Physics III · topic UPIII-11.08
Fusion, the Coulomb barrier, and tunnelling
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.
Fusion, the Coulomb barrier, and tunnelling: predict a finite change
Fusion is the merging of two light nuclei into a more tightly bound one; at thermal energies it happens only when a pair from the high-energy tail of the Maxwell distribution tunnels through the Coulomb barrier that keeps it out of range of the strong force. Use the supplied topic model Y_out = 3*(eg/(4*kt))^(1/3), where Y_out is PEAK EXPONENT τ = 3E₀/kT (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. kt is Thermal energy kT, measured in keV. Keep eg (Gamow energy E_G) = 493 keV fixed. For this exercise use only 3.01875 keV ≤ kt ≤ 17.9812 keV 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 kt = 6.3105 keV and 8.40525 keV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δkt across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
PEAK EXPONENT τ = 3E₀/kT against kt
Predictions from the supplied Fusion, the Coulomb barrier, and tunnelling 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): PEAK EXPONENT τ = 3E₀/kT
Horizontal axis (x): kt: Thermal energy kT / keV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| kt: Thermal energy kT / keV | PEAK EXPONENT τ = 3E₀/kT |
|---|---|
| 3.01875 | 10.330179024286256 |
| 6.7593749999999995 | 7.896137766330991 |
| 10.5 | 6.817949775652791 |
| 14.240624999999998 | 6.159430330694692 |
| 17.98125 | 5.698714996433874 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of kt separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At kt = 6.3105 keV, substitute into Y_out = 3*(eg/(4*kt))^(1/3): Y₀ = 8.07909.
- At kt = 8.40525 keV, the same fixed-parameter model gives Y₁ = 7.34288.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.736203; the input change is Δkt = 2.09475 keV.
- Average sensitivity is ΔY_out/Δkt = -0.351452 1 per keV. 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 = 8.07909, Y₁ = 7.34288; ΔY_out = -0.736203; average sensitivity = -0.351452 1 per keV.
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.
Fusion, the Coulomb barrier, and tunnelling: infer an input with uncertainty
Fusion is the merging of two light nuclei into a more tightly bound one; at thermal energies it happens only when a pair from the high-energy tail of the Maxwell distribution tunnels through the Coulomb barrier that keeps it out of range of the strong force. Use the supplied topic model Y_out = 3*(eg/(4*kt))^(1/3), where Y_out is PEAK EXPONENT τ = 3E₀/kT (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. kt is Thermal energy kT, measured in keV. Keep eg (Gamow energy E_G) = 493 keV fixed. For this exercise use only 3.01875 keV ≤ kt ≤ 17.9812 keV 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 = 6.4949223, with a bounded reading uncertainty ±0.11943. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown kt by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00149625 keV, 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.
PEAK EXPONENT τ = 3E₀/kT against kt
Predictions from the supplied Fusion, the Coulomb barrier, and tunnelling 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): PEAK EXPONENT τ = 3E₀/kT
Horizontal axis (x): kt: Thermal energy kT / keV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| kt: Thermal energy kT / keV | PEAK EXPONENT τ = 3E₀/kT |
|---|---|
| 3.01875 | 10.330179024286256 |
| 6.7593749999999995 | 7.896137766330991 |
| 10.5 | 6.817949775652791 |
| 14.240624999999998 | 6.159430330694692 |
| 17.98125 | 5.698714996433874 |
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(kt) = 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(3.01875) = 10.3302 and Y_out(17.9812) = 5.69871. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives kt_line = 3.01875 + (6.49492 − 10.3302)(17.9812 − 3.01875)/(5.69871 − 10.3302) = 15.409 keV. 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 kt = 12.1459 keV; substituting back gives Y_out = 6.4949223.
- The interpolation error in the input is |kt_line − kt| = 3.26313 keV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 6.3754923 and 6.6143523. Sort the roots even if the response decreases: 11.4998 keV ≤ kt ≤ 12.8413 keV.
- The lower and upper input deviations are 0.646117 keV and 0.69544 keV. 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 kt ≈ 12.1459 keV; compatible interval [11.4998, 12.8413] keV. Endpoint interpolation gives 15.409 keV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.