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

When motion of nucleus is considered, replace mass of electron by reduced mass of electron and nucleus.
Reduced mass of electron and nucleus :
`(1)/(mu)=(1)/(m)+(1)/(M) implies mu =(mM)/(m+M)`
where
m : mass of electron
M : mass of nucleus.
B.E.E `prop` m (nucleus has infinite mass)
`E' prop mu` (mass of nucleus is considered )
`(E')/(E)=(mu)/(m)=(M)/(m+M)`
`(E)/(E')-1=(m)/(M)=(1)/(1837)`
`((E)/(E')-1)xx100=0.0.54%`