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An electron is revolving in the 2^(nd) o...

An electron is revolving in the `2^(nd)` orbit of `He^+` ion. To this if 12.1 eV of energy supplied. Then to which orbit it will be excited.

A

6

B

8

C

4

D

2

Text Solution

Verified by Experts

The correct Answer is:
A
`E=- 13.6xx(Z^2)/(n^2) eV//"atom"`
for `He^+ (Z=2) n=2 thererore E = 0 13.6 eV`
Supplied energy = 12.1 eV
`=-1.5 eV` (i.e. energy of `6^(th)` orbit)
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