SATHEE: Atoms And Nuclei Question 327

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) $M = \frac{e h}{4 π m}, I = \frac{e V}{2 π r}$

B) $M = \frac{2 e h}{5 π m}, I = \frac{e V}{4 π r}$

C) $M = \frac{h}{π m}, I = \frac{e}{π r}$

D) $M = \frac{e h}{π m}, I = \frac{e V}{π r}$

Show Answer

Answer:

Correct Answer: A

Solution:

$I = \frac{e}{T} = \frac{e V}{2 π r}$ so,

$M = \frac{e v}{2 π r} \times π r^{2} = \frac{e v r}{2}$

According to Bohr’s theory angular momentum $m v r = \frac{n h}{2 π} or v r = \frac{n h}{2 π m} so, M= \frac{n e h}{4 π m}$ For the ground state $n = 1, so, M = \frac{e h}{4 π m}$

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Atoms And Nuclei Question 327

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