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

$(a)\;1.05\times10^{-34}Joule.sec\qquad(b)\;2.11\times10^{-34}Joule.sec\qquad(c)\;3.16\times10^{-34}Joule.sec\qquad(d)\;4.22\times10^{-34}Joule.sec$

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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
$=(n_2 - n_1)\large\frac{h}{2\pi}$
$=(3-1)\large\frac{h}{2\pi}$
$=\large\frac{h}{\pi}$
$=\large\frac{6.626\times10^{-34}}{3.14}$
$=2.11\times10^{-34}JSec$
Hence the answer is (b)
answered Feb 7, 2014 by sharmaaparna1
 

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