A) \[2\pi {{r}_{0}}\]
B) \[4\pi {{r}_{0}}\]
C) \[6\pi {{r}_{0}}\]
D) \[\pi {{r}_{0}}\]
Correct Answer: C
Solution :
\[mvr\,=\frac{nh}{2\pi }\] ?(1) \[p=\frac{h}{\lambda }=mv\] ?(2) Placing the value of mv from Eq. (2) into Eq. (1) for 3rd orbit \[\frac{h}{\lambda }{{r}_{3}}=\frac{3h}{2\pi }\] \[\lambda =\frac{2\pi {{r}_{3}}}{3}\] \[{{r}_{3}}\,={{n}^{2}}{{r}_{0}}\,=9{{r}_{0}}\] So, \[\lambda =\,\frac{2\pi .9{{r}_{0}}}{3}=6\pi {{r}_{0}}\]
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