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in Physics by (121k points)
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. `1.5` eV
B. `0.85` eV
C. `3.4` eV
D. `1.9` eV

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1 Answer

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by (121k points)
Correct Answer - D

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