Question

The radius of the orbit of an electron, in an hydrogen atom is $0.5\mathring{A}$. If it makes $16$ revolutions per second, then the magnetic moment associated with orbital motion of the electron is

• A. $1.25\times {10}^{-23}A{m}^2$
• B. $0.6\times {10}^{-23}A{m}^2$
• C. $2.25\times {10}^{+23}A{m}^2$
• D. None of the above

Answer 1

The correct option is A $1.25\times {10}^{-23}A{m}^2$

Given, $r=0.5\mathring{A}=0.5\times {10}^{-10}m$
We know the constant
Current, $I=\frac{q}{t}=\frac{ne}{t}(\because q=ne)$
$=\frac{{10}^{16}\times 1.6\times {10}^{-19}}{1}=1.6\times {10}^{-3}A$
We have, $M=i\times A=i\times \pi r^2=1.6\times {10}^{-3}\times 3.14\times {(0.5\times {10}^{-10})}^2=1.25\times {10}^{-23}A-m^2$