The ionisation potential of Hydrogen is `2.17 xx 10^(-11)` erg/atom. The energy of the electron in the second orbit of the hydrogen atom is

Options

A. `-(2.17 xx 10^(-11))/(2) ` erg/atom
B. `-(2.17 xx 10^(-11))/(2^2) ` erg/atom
C. `-(2.17 xx 10^(-17))/(2^2) ` erg/atom
D. `-(2.17 xx 10^(11))/(2^2) ` erg/atom

Answer

B

Explanation

`E = - IE =(2.18 xx 10^(-11))/((2)^2) ` erg/atom