In a hydrogen atom, if energy atom, if energy of an electron in group state is `- 13.6 eV`, then that in the `2^(nd)` excited state is :

Options

A. `6.04 eV`
B. `3.4 eV`
C. `1.51 eV`
D. `13.6 eV`

Answer

C

Explanation

According to Bohr' theory, the enegry of any state is claculated as follows:
`E_(n) = (Z^(2)E_(1))/(n^(2))`
where `E_(1)`is `-13.6eV`.
Ground state implies `n =1, 1st`excited state is `n = 2`, and `2nd` excited state sui `n = 3`. Thus,
`E_(3) = ((1)^(2)(-13.6eV))/((3)^(2)) =- 1.51 eV`