The ratio of magnetic dipole moment of an electron of charge e and mass m in the Bohr orbit in hydrogen to the angular momentum of the electron in the orbit is:

  • Option 1)

    \frac{e}{m}

  • Option 2)

    \frac{e}{2m}

  • Option 3)

    \frac{m}{e}

  • Option 4)

    \frac{2m}{e}

 

Answers (1)

 

Bohr quantisation principle -

mvr=\frac{nh}{2\pi } \\2\pi r= n\lambda

- wherein

Angular momentum of an electron in  stationary orbit is quantised.

 

magnetic dipole moment, (\mu ) =IA                 

=\left ( \frac{e}{T} \right )\cdot \pi r^{2}= \frac{e}{\left ( \frac{2\pi r}{v} \right )}\cdot \pi r^{2} = \frac{evr}{2}

angular momentum of electron L=mvr

\therefore \frac{\mu }{L} = \frac{evr}{2\cdot mvr} = \frac{e}{2m}

 


Option 1)

\frac{e}{m}

This solution is incorrect.

Option 2)

\frac{e}{2m}

This solution is correct.

Option 3)

\frac{m}{e}

This solution is incorrect.

Option 4)

\frac{2m}{e}

This solution is incorrect.

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