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One electron species having ionization e...

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

A

H

B

`He^(+)`

C

`B^(4+)`

D

`Li^(2+)`

Text Solution

Verified by Experts

The correct Answer is:
B
`I.E. = (13.6 Z^(2))/(n^(2)) eV`
`= 13.6 Z^(2)` for one-electron species
`:. 13.6 Z^(2) = 54.4 or Z^(2) = 4 or Z = 2, " i.e., " He^(+)`
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