A) \[\frac{h}{\pi }\]
B) \[\frac{2h}{\pi }\]
C) \[\frac{5h}{2\pi }\]
D) \[\frac{7h}{2\pi }\]
Correct Answer: C
Solution :
By using \[E=-\frac{13.6}{{{n}^{2}}}\,eV\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(for\,\,\,{{H}_{2}}\,atom)\] \[\Rightarrow \,\,-0.544\,\,=\,\,\frac{13.6}{{{n}^{2}}}\,\,\Rightarrow \,\,{{n}^{2}}=25\Rightarrow n=5\] \[\therefore \,\,Angular\,\,momentum\,\,=\,\,n\frac{h}{2\pi }\,\,=\,\,\frac{5h}{2\pi }\,\]
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