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Question

The energy of an excited H-atom is -3.4eV. The angular momentum of e in the given orbit is:

A
h/π
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B
h/2π
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C
2h/π
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D
h/3π
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Solution

The correct option is A h/π
As we know,
E=13.61n2;

3.4=13.6n2; n=2
So, angular momentum of e in the given orbit is mvr=nh2π=hπ as n is 2 here.

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