Calculate the difference between the ionisation potentials of atomic hydrogen and atomic deuterium.

Explanation

Wave number `barv` of lines in the spectrum of an atom is given by
`barv =(mue^4 Z^2)/(8epsilon_0^2 ch^3)[1/n_1^2 -1/n_2^2]`
where `mu=(Mm)/(M+m)`
The ionisation energy is the energy required to liberate an electron from a given orbit to an infinite distance from nucleus. It corresponds to energy transition from `n_1 =1 " to " n_2 = 0`
`barvc=v=(mue^4Z^2)/(8epsilon_0^2h^3)`
For hydrogen atom (Z=1),
`v_H=(e^4)/(8epsilon_0^2h^3)[(M_Hm)/(M_H+m)] [1/1^2-1/oo^2]`
For deuterium atom (Z = 1)
`v_D=e^4/(8epsilon_0^2h^3)[(M_Dm)/(M_D+m)][1/1^2-1/oo^2]`
Hence difference between the ionisation energies of the two atoms,
`h(v_D-v_H) =(e^4)/(8epsilon_0^2h^2)[(M_Dm)/(M_D+m)-(M_Hm)/(M_H+m)]`
`=(e^4)/(8epsilon_0^2h^2)[(3680mxxm)/(3680m+m)-(1840mxxm)/(1840m+m)]`
`=(e^4)/(8epsilon_0^2h^2)[3680/3681-1840/1841]`
`=Rhc[3680/3681-1840/1841]`
`= 5.885 xx10^(-22)J = (5.885 xx10^(-22))/(1.6 xx10^(-19))eV`
`= 3.68 xx10^(-3) eV`