1.

The atom of hydrogen absorbs 12.75 eV of energy in ground state. Then what will be the change in orbital angular momentum of the electron in it.

Answer»

`(h)/(2pi)`
`(h)/(pi)`
`(2h)/(pi)`
`(3H)/(2pi)`

SOLUTION :`DeltaE=E_(n)-E_(1)`
`12.75=-(13.6)/(n^(2))+(13.6)/(1^(2))`
`:. (13.6)/(n^(2))=0.85`
`:.n^(2)=(13.6)/(0.85) :. n^(2)=16 :.n=4`
`:.`Angular momentum in `4^(th)` orbit `l_(4)=(4H)/(2pi)`
`:.` Change in angular momentum `=l_(4)-l_(1)`
`=(4h)/(2pi-(h)/(2pi)`
`=(3h)/(2pi)`


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