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CBSE Class 12 Physics 2019 Outside Delhi Set 1 Paper

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Question : 16 of 27
Marks: +1, -0
Prove that the magnetic moment of the electron revolving around a nucleus in an orbit of radius rr with orbital speed vv is equal to evr2\frac{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   evr2\;\frac{e v r}{2}
Deducing expression of the magnetic moment of hydrogen atom
The magnetic moment is
m=IAm = I A
But current is I=  eT=  ev2πrI = \;\frac{e}{T} = \;\frac{e v}{2 \pi r}
where, T=  2πrvT = \;\frac{2 \pi r}{v} and the area, A=πr2A = \pi r^2
m=  ev2πrπr2=  evr2m = \;\frac{e v}{2 \pi r} \pi r^2 = \;\frac{e v r}{2}
But from Bohr's second postulate
  mevr=  nh2π=  h2π\;m_e v r = \;\frac{n h}{2 \pi} = \;\frac{h}{2 \pi}    for   n=1\;\text{ for }\; n = 1
  vr=  nh2πme\;v r = \;\frac{n h}{2 \pi m_e}
Hence the magnetic moment is
m=  e2  h2πme=  eh4πmem = \;\frac{e}{2} \;\frac{h}{2 \pi m_e} = \;\frac{e h}{4 \pi m_e} (Here n=1n = 1 )
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