Question

Ratio of magnetic dipole moment to the angular momentum for hydrogen like atoms in nth orbit

Answer 1

Answer:

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

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

Answer 2

Answer:

here is your answer

Explanation: