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

Answer

B

Explanation

Energy `E = K [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` (`K =` constant)
`n_(1) = 2` and `n_(2) = 3`, so `E = K [(1)/(2^(2)) - (1)/(3^(2))] = K [(5)/(36)]`
For removing an electro `n_(1) = 1` to `n_(2) = oo`
Energy `E_(1) = K [1] (36)/(5) E = 7.2 E`
`:.` Ionization energy `= 7.2 E`