 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2016/s3/3/4/solution

The relative orbital speed is

Z e^{2} / (4 π ϵ_{0} ℏ n)

. Divide the circumference by this speed to find

T_{n} = \frac{4 ϵ _{0}^{2} h ^{3} n ^{3}}{Z ^{2} μ e ^{4}} .

n

here labels principal levels with

n = 1

the ground state; if “nth excited state” is read literally as excitation count, replace

n

n + 1

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