The ionization energy for the hydrogen atom is `13.6 eV` then calculate the required energy in `eV` to excite it from the ground state to `1^(st)` excited state.

Options

A. `3.4 eV`
B. `10.2 eV`
C. `12.1 eV`
D. `1.5 eV`

Answer

B

Explanation

This excitation means the transition of electron is from `n_(i) =1` to `n_(f) = 2`. According to Bohr's theory,
`DeltaE = Z^(2)E_(1)((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))`
Since `Z^(2)E =-` Ionization energy
`DeltaE=- 13.6eV ((1)/(2^(2))-(1)/(1^(2))) = (-13.6eV) ((-3)/(4)) = 10.2eV`