A) \[\frac{1}{\sqrt{r}}\]
B) \[\frac{1}{r}\]
C) \[\sqrt{r}\]
D) \[{{r}^{2}}\]
Correct Answer: C
Solution :
Angular Momentum \[=L=\frac{nh}{2\pi }\] The radius of the orbit \[=r={{a}_{0}}{{n}^{2}}\Rightarrow n=\sqrt{r/{{a}_{0}}}\] Where \[{{a}_{0}}=\]radius of first orbit of hydrogen atom. \[\therefore \] \[L=\sqrt{\frac{r}{{{a}_{0}}}}\times \frac{h}{2\pi }\Rightarrow L\alpha \sqrt{r}\] Hence, the correction option is(c).You need to login to perform this action.
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