A) \[M=\frac{eh}{4\pi m},I=\frac{eV}{2\pi r}\]
B) \[M=\frac{2eh}{5\pi m},I=\frac{eV}{4\pi r}\]
C) \[M=\frac{h}{\pi m},I=\frac{e}{\pi r}\]
D) \[M=\frac{eh}{\pi m},I=\frac{eV}{\pi r}\]
Correct Answer: A
Solution :
[a] \[I=\frac{e}{T}=\frac{eV}{2\pi r}\] so, \[M=\frac{ev}{2\pi r}\times \pi {{r}^{2}}=\frac{evr}{2}\] According to Bohr's theory angular momentum \[mvr=\frac{nh}{2\pi }\text{ or }vr=\frac{nh}{2\pi m}\text{ so, M=}\frac{neh}{4\pi m}\] For the ground state \[n=1,\text{ so, }M=\frac{eh}{4\pi m}\]You need to login to perform this action.
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