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If a hydrogen atom emit a photon of ener...

If a hydrogen atom emit a photon of energy `12.1 eV` , its orbital angular momentum changes by `Delta L. then`Delta L` equals

A

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

B

`211xx10^(-34)`J-s

C

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

D

`4.22xx10^(-34)`J-s

Text Solution

Verified by Experts

The correct Answer is:
B
`DeltaE=-13.6eV(1/n_(1)^(2)-1/n_(2)^(2)), DeltaL=h/(2pi)(n_(1)-n_(2)`
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