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When a hydrogen atoms emits a photon of ...

When a hydrogen atoms emits a photon of energy `12.1 eV` , its orbital angular momentum changes by (where h os Planck's constant)

A

`(h)/(2pi)`

B

`(3h)/(2h)`

C

`(h)/(pi)`

D

`(2h)/(pi)`

Text Solution

Verified by Experts

The correct Answer is:
C
`Delta E = E_3 - E_1 = 12.1 eV`
Electron jumps from 3rd orbit to 1st orbit
`mvr = (3h)/(2pi) - (h)/(2pi)`
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