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

Options

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

Answer

C

Explanation

According to Bohr' theory,
Ionization enegry `=- Z^(2)E_(1)`
`= 13.6 eV Z^(2)`
Thus,
`13.6eV Z^(2) = 54.4 eV`
`Z^(2) = (54.4)/(13.6)=4`
`Z = 2`
This implies `He^(+)`ion.