The ratio of the energy of the electron in ground state of hydrogen to the electron in first excited state of `Be^(3+)` is :

Options

A. `1 : 4`
B. `1 : 8`
C. `1 : 16`
D. `16 : 1`

Answer

A

Explanation

(a) `E_n = - (13.6 Z^2)/(n^2) eV`, where
`Z = "atomic number" = 4`
`N = "number of orbits"`
`Be^(3 +) (n = 2)` (first excited state)
`E_1 = -13.6 eV, E_2 - 13.6 xx 16//4`.