 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/2017/s3/1/a/solution

Substitute the quantized radius in

m v r = n ℏ

to get

v_{n} = e^{2} / (4 π ϵ_{0} ℏ n)

- e^{2} / (4 π ϵ_{0} r_{n})

and the stipulated virial relation makes the total one half of this value. Hence

v_{n} = \frac{α c}{n}, E_{n} = - \frac{m e ^{4}}{2 ( 4 π ϵ _{0} ) ^{2} ℏ ^{2} n ^{2}} = - \frac{Ry}{n ^{2}} .

The negative sign means a bound state relative to a separated stationary proton and electron.

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