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A hydrogen atom in its ground state abso...

A hydrogen atom in its ground state absorbs `10.2 eV` of energy. The orbital angular momentum is increased by

A

`1.05xx10^(-34) J-sec`

B

`2.11xx10^(-34) J-sec`

C

`3.16xx10^(-34) J-sec`

D

`4.22x10^(-34) J-sec`

Text Solution

Verified by Experts

The correct Answer is:
A
After absorbing 10.2 eV, electron goes to n=2 from n=1
`DeltaL=(n_(2)-n_(1))(h)/(2pi)=(2-1)(h)/(2pi)`
`=(6.6xx10^(-34))/(2xx3.14)=1.05xx10^(-34) J.sec`
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