In hydrogen atom, if the difference in the energy of the electron in `n = 2` and `n = 3` orbits is `E`, the ionization energy of hydrogen atom is

Options

A. 3.2 E
B. 5.6 E
C. 7.2 E
D. 13.2 E

Answer

C

Explanation

`E_n = - (13.6)/(n^2) therefore E_3 = - (13.6)/(9) = -K/9`
and `E_2 = -(13.6)/(4) =- K/4`
`therefore E = E_3 - E_2 = K [1/4 - 1/9] = (5K)/(36)`
`therefore K=(36E)/(5) " " ..(1)`
and for ionisation of the atom, the electron must be removed from n = 1 to ` n= oo`
`therefore E_1 = K [ (1)/(1^2) - 1/oo] = K (1) = K`
`therefore E_1 = 36/5 E = 7.2 E`