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Energy E of a hydrogen atom with princip...

Energy `E` of a hydrogen atom with principle quantum number `n` is given by `E = (-13.6)/(n^(2)) eV`. The energy of a photon ejected when the electron jumps from `n = 3` state to `n = 2` state of hydrogen is approximately

A

0. 85 eV

B

3.4 eV

C

1.9 eV

D

1.5 eV

Text Solution

Verified by Experts

The correct Answer is:
C
Energy of photon =`E_3-E_2 ="-13.6"/9=("-13.6"/4)=5/36xx13.6=1.9` eV
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