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Ratio of  magnetic dipole moment to the angular momentum for hydrogen like atoms is ( e and m are electronic charge and mass respectively)
Asked by samairakhanna185 | 20 Mar, 2019, 10:22: PM

Let v the velocity of electron in its orbit of radius r, charge on electron is taken as e and mass of electron is m

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 $L=mvr$

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

Answered by Ankit K | 21 Mar, 2019, 09:02: AM
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