Energy associated with nth orbit of H-atom is given by the expression, `E_(n)=-(13.6)/(n^(2))eV`. Show that `E_((n+l))-E_(n)=(13.6xx2)/(n^(3))eV~, when 'n' in very large.

Explanation

`E_((n+1))-E_(n)=[-(13.6)/((n+1)^(2))-(-(13.6)/(n^(2)))]`
`=[(13.6)/(n^(2))-(13.6)/((n+1)^(2))]eV=(13.6(2n+1))/(n^(2)(n+1)^(2))`
If the value of n is very large, then (2n+1)=2n and `(n+1)=n" "therefore E_((n+1))-E_(n)=(13.6xx2n)/(n^(2)xxn^(2))=(13.6xx2)/(n^(3))eV`