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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

`3^(2) xx 21.7 xx 10^(-19) J`

B

`21. 79 xx 10^(-19)J`

C

`(1)/(3) xx 21.79 xx 10^(-19)`J

D

`(1)/(3^(2)) xx 21.79 xx 10^(-19)` J

Text Solution

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The correct Answer is:
B
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