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energy E of a hydrogen atom with princip...

energy E of a hydrogen atom with principal 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 `n=2` state of hydrogen is approximately

A

1.5 eV

B

0.85 eV

C

3.4 eV

D

1.9 eV

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The correct Answer is:
D
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