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When a hydrogen atom emits a photon of energy 12.1eV, the orbital angular momentum changes by


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Emission of photon of 12.1eV corresponds to the transition from n=3 to n=1
$\therefore$ Change in angular momentum
$=(n_2 - n_1)\large\frac{h}{2\pi}$
Hence the answer is (b)
answered Feb 7, 2014 by sharmaaparna1

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