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

`1.05 times 10^(-34)Js`

`2.11 times 10^(-34)Js`

`3.16 times 10^(-34)Js`

`4.22 times 10^(-34)Js`

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The correct Answer is:

Since `E=-13.6/n^(2)eV`
`E_(1)=-13.6 eV, E_(2)=-3.4 eV`
`E_(3)=-1.50 eV, E_(4)=-0.85 eV`
From above, we can see that `E_(3)-E_(1)=12.1 eV`
i.e., the electron must be making a transition from n = 3 to n = 1 level.
`DeltaL=(3-1)h/(2pi)=h/pi=2.11 times 10^(-34)` J s

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

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