CBSE Class 12 Physics 2019 Outside Delhi Set 1 Paper

Question : 16 of 27

Marks: +1, -0

Prove that the magnetic moment of the electron revolving around a nucleus in an orbit of radius

r

with orbital speed

v

is equal to

e v r \/ 2

. Hence using Bohr's postulate of quantization of angular momentum, deduce the expression for the magnetic moment of hydrogen atom in the ground state.

Solution:

Proving magnetic moment as

\;{e v r}/{2}

Deducing expression of the magnetic moment of hydrogen atom
The magnetic moment is

m=I A

But current is

I=\;{e}/{T}=\;{e v}/{2 π r}

where,

T=\;{2 π r}/{ {\v}}

and the area,

A=π r^2

m=\;{e v}/{2 π r} π r^2=\;{e v r}/{2}

But from Bohr's second postulate

\;m_e v r=\;{n h}/{2 π}=\;{h}/{2 π}

\;\text " for "\; n=1

\;v r=\;{n h}/{2 π m_e}

Hence the magnetic moment is

m=\;{e}/{2} \;{h}/{2 π m_e}=\;{e h}/{4 π m_e}

(Here

n=1

)

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