A) \[\frac{5h}{2\pi }\]
B) \[\frac{h}{2\pi }\]
C) \[\frac{h}{\pi }\]
D) \[\frac{2h}{3\pi }\]
Correct Answer: C
Solution :
Energy of electron in nrt orbit of hydrogen atom is given by \[\tan \theta =\frac{AD}{OD}\] \[\Rightarrow \] \[\tan {{60}^{\circ }}=\frac{2}{a}\] \[\Rightarrow \] \[a=\frac{I}{2\sqrt{3}}=\frac{9\times {{10}^{-2}}}{2\sqrt{3}}\] \[\text{B=3 }\!\!\times\!\!\text{ }\frac{\text{4 }\!\!\pi\!\!\text{ }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-7}}}\text{ }\!\!\times\!\!\text{ 2}}{\text{4 }\!\!\pi\!\!\text{ }\!\!\times\!\!\text{ }\frac{\text{9 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}}{\text{2}\sqrt{\text{3}}}}\text{ }\!\![\!\!\text{ sin6}{{\text{0}}^{\text{o}}}\text{+sin6}{{\text{0}}^{\text{o}}}\text{ }\!\!]\!\!\text{ }\] n = 2 The angular momentum of electron is given by \[=\frac{4\sqrt{3}}{9}\times {{10}^{-5}}\left( \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} \right)\]
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