Taking into account the motion of the nucleus of a hydrogen atom , find the expressions for the electron's binding energy in the ground state and for the Rydberg constant. How much (in percent) do the binding energy and the Rydberg constant , obtained without taking into account corresponding values of these of these quantities?

Explanation

If mass of nucleus is considered (not infinite), then the reduced mass of nucleus-electron system can be taken as
`mu = (m M)/(m + M) e V`
Here `m` is the mass electron and `M` is that of the nucleus. The binding energy atom cannow be given as
`E = + (2 pi^(2) K^(2) e^(4) mu)/(h^(2)) = 13.6 xx (mu)/(m) e V`

We have hydrogen atom Rydberg constant given as
`R = (2 pi^(2) K^(2) e^(4) m)/(ch^(2))`
If effect of mass of nucleus is considered , the new value of Rydberg constant can be given as
` R_(mu) = (2 pi^(2) K^(2) e^(4) mu)/(ch^(2)) or R_(mu) = (RM)/(m + M)`
Percentage different in the value of `R` and `R_(mu)`given as ` (Delta R)/(R) xx 100 = ((R_(mu - R) xx 100)/(R_(mu))) = (m)/(M) xx 100 = 0.055%`