In Hydrogen and Hydrogen like atoms, the ratio of difference of energies `E_(2n) -E_n and E_(2n) E_n` varies with its atomic number Z and n as

Options

A. `Z^2/n^n`
B. `Z^4/n^4`
C. `Z/n`
D. `(n^2/Z^2)`

Answer

D

Explanation

`E_n = (-13.6)/n^2 (Z^2), E_(2n)=-(13.6)/((2n^2))Z^2 = (-13.6)/4 (Z^2/n^2)`
`E_(2n) -E_n=[(-(13.6)/4)Z^2/n^2]-[13.6(Z^2/n^2)],E_(2n)-E_n=Z^2/n^2(13.6 -(13.6)/4)=(10.2) (Z^2/n^2)`
`(E_(2n))(E_n)=((13.6)^2Z^4)/4 Z^4/n^4, (E_(2n)-E_n)/((E_(2n))-(E_n))=((10.2)(Z^2/n^2))/(((13.6)^2)/4(Z^4/n))implies((E_(2n)-E_n))/((E_(2n))(E_n))alpha(n^2/Z^2)`