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If the 1^(st) ionization energy of H ato...

If the `1^(st)` ionization energy of H atom is 13.6 eV, then the `2^(nd)` ionization energy of He atom is

A

27.2 eV

B

40.8 eV

C

54.4 eV

D

108.8 eV

Text Solution

Verified by Experts

The correct Answer is:
C
`E_(H)=(-13.6)/n^(2)," "(E_(1))_(H)=-13.6eV`
`(E_(1))_(He^(+))=-13.6Z^(2)/n^(2)=-13.6xx2^(2)/1=-54.4eV`
The second I.E. = `E_(oo)-(E_(1))_(He^(+))`
= `0-(-54.4)eV=54.4eV`.
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