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Past exam of the advanced quantum physics course of the University of Cambridge 2017 4 Solution by
Codex 0 2026-10-18
For elastic scattering by the Yukawa potential, take and use the first Born approximation in the plane-wave normalizationChoose the polar axis along the momentum transfer . The angular integral isThe radial volume element cancels the remaining powers of , leavingFor real , the differential scattering cross-section is thereforeThis also has a finite forward limit when . The Born amplitude for a Yukawa potential provides a screened route to the Coulomb potential by taking at fixed .
For an alpha particle of charge scattering from a heavy nucleus of charge , set . Neglect recoil, so that the reduced mass is approximately the projectile mass, , and elastic geometry givesThe Coulomb-limit scattering amplitude becomeswhich gives the Rutherford differential cross-sectionThe unscreened Coulomb potential has long range, so taking this limit does not justify the Born approximation for its full scattering state, particularly for a strongly charged target. The exact Rutherford scattering amplitude differs by a phase and has the same modulus. Thus the resulting differential scattering cross-section remains applicable to the ideal nonrelativistic Coulomb problem beyond the weak-potential regime.
To relate the detector counts to the atomic number, first compute the number of target atoms per illuminated area. With molar mass , the number density integrated through the foil isThe detector's transverse area is already its area normal to the scattered ray. Its small solid angle isThere is no extra factor: the foil is traversed normally and the detector area is transverse to its line of sight.
For a thin foil with negligible multiple scattering and energy loss, the detected fraction is . Thus the measurements implyUsing in the Rutherford differential cross-section gives
The stated measurements do not reproduce gold's atomic number 79 in this model. The values and units above agree with the original PDF. Substituting instead would predict with the other inputs unchanged, rather than . No stated correction accounts for that discrepancy; forcing the answer to 79 would require changing the data or adding an experimental correction not specified in the problem.
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