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If 13.6 eV energy is required to lionize...

If 13.6 eV energy is required to lionize the hydrogen atm, then energy required to remove an electron from n = 2 is :

A

`10.2 eV`

B

`0 eV`

C

`3.4 eV`

D

`6.8 eV`

Text Solution

Verified by Experts

The correct Answer is:
C
Energy required to lonize a atom from `n^(th)` orbit is
`= +13.6/n^(2) eV`
`E_(2)=(+13.6)/2^(2) eV= +3.4 eV`
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ALLEN-SIMPLE HARMONIC MOTION-Exercise-04 [B]
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