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The orbital magnetic dipole moment of electron in hydrogen atom is

$(a)\;\mu=\large\frac{e h n}{\pi m}\qquad(b)\;\mu=\large\frac{e h n}{2 \pi m}\qquad(c)\;\mu=\large\frac{e h n}{4 \pi m}\qquad(d)\;\mu=\large\frac{ 2e h n}{\pi m}$

1 Answer

Answer : (b) $\;\mu=\large\frac{e h n}{ 4\pi m}$
Explanation :
The orbiting electron behaves as a current loop. The equivalent current
$i=\large\frac{net\;charge}{time\;of\;revolution }=\large\frac{e}{T}=e f_{rot}$
Now , magnetic dipole moment
$\mu=n i A$
n=no. of turns of loop =1 , A= area of loop = $\;n r^2$
$\mu=(1)\;(e f_{rot})\;(\pi r^2) \quad \;=> \mu=\pi e r^2 f_{rot}$
Putting the values of $\;f_{rot}\;$ and r we obtain ,
$\mu=\pi e \;(\large\frac{n^2 h^2 \varepsilon }{n m e^2 z})^{2}\;(\large\frac{m z^2 e^4}{4 \varepsilon^2 h^3 n^3})$
$\mu=\large\frac{e h n}{4 \pi m}$
answered Feb 21, 2014 by yamini.v

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