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If the ionization energy of He^(+) is 19...

If the ionization energy of `He^(+)` is `19.6xx10^(-18) J` per atom then the energy of `Be^(3+)` ion in the second stationary state is `:`

A

`-4.9xx10^(-18)J`

B

`-44.1xx10^(-18)J`

C

`-11.025xx10^(-18)J`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
D
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