One electron species having ionization enegry of `54.4 eV` is

Options

A. `Be^(+2)`
B. `Be^(+3)`
C. He^(+)`
D. `H`

Answer

C

Explanation

`He^(+)`
Ionization potential ` = E_(oo) - E_1`
`54.4 = 0 - E_1 or E_1 = - 54.4eV`
But `E_1 = 13.6 xx z^2/((1)^2) eV or - 54.4 eV = -13.6 xxZ^2`
`or Z = 2, ` So `He^(+)` ion.