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

Options

A. H
B. `He^(+)`
C. `B^(4+)`
D. `Li^(2+)`

Answer

B

Explanation

`I.E. =54.4 eV`
`:. E_(1)` (energy of electron in first orbit)
`= 54.4 eV`
`E_(4) = (-13.6 Z^(2))/(n^(2)) eV "atom"^(-1)`
`:. -54.4 = -13.6 Z^(2)` (Here n=1)
`Z^(2) = 4 rArr Z = 2`
`:.` One electron species in `He^(+) (Z = 2)`