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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_n=-(13.6)/n^2eV`. The energy of a photon ejected when the electron jumps from n=3 state to 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:
A
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