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

The total energy of the `H-atom` in an arbitrary frame is
`E=(1)/(2)mvec(V)_(2)^(2)-(e^(2))/((4pi epsilon_(0))|r_(1)vec(r)_(2)|)`
Here `vec(V)_(1)=vec(r )_(1),vec(V)_(2)=vec(r )_(2), vec(r )& vec(r )_(2)` are the coordinates of the electron and protons.
we define
`vec(R )=(mvec(r )_(1)+Mvec(r )_(2))/(M+m)`
`vec(r )=vec(r )_(1)-vec(r )_(2)`
Then `vec( V)=(mvec(V)_(1)+Mvec(V)_(2))/(m+M)`
`vec(v)=vec(V)_(1)-vec(V)_(2)`
or `vec(V)_(1)=vec(V)+(M)/(m+M)vec(v)`
and we get `E=(1)/(2) (m+M)vec(V)_(2)+(1)/(2)(mM)/(m+(M))v^(2)-(e^(2))/(4 pi epsilon_(r ))`
In the frame `vec(V)=0`, this reduces to the energy of a particel of mass
`mu=(mM)/(m+M)`
`mu` is called the reduced mass.
Then
`E_(b)=(mue^(4))/(2 ħ^(2))` and `R=(mue^(4))/(2 ħh^(3))` Since `mu=(m)/(1+(m)/(M))~~m(1-(m)/(M))`
these values differ by `(m)/(M) (~~0.54%)` from the values obtained without considering nuclear motion. `(M= 1837)`