logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
0 votes

The orbital magnetic dipole moment of electron in hydrogen atom is

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

Can you answer this question?
 
 

1 Answer

0 votes
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
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...