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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

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

C

`3.16 xx 10^(-34)J-s`

D

`4.22 xx 10^(-34)J-s`

Text Solution

Verified by Experts

The correct Answer is:
B
`Delta E =-13.6eV ((1)/(n_(1)^(2))-(1)/(n_(2)^(n))),Delta L =(h)/(2pi)(n_1 -n_(2))`
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