Atoms And Nuclei Question 327
Question: If the electron revolving around the nucleus in a radius ‘r’ with orbital speed ‘v’ has magnetic moment evr/2. Hence, using Bohr’s postulate of the quantization of angular momentum obtain the magnetic moment (M) of hydrogen atom in its ground state and current (I) due to revolution of electron.
Options:
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} $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ 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} $
BETA
