Question

In an hydrogen atom, the electron is making $6.6\times {10}^{15}rps$. If the radius of the orbit is $0.53\times {10}^{-10}m$, then the equivalent magnetic dipole moment is approximately

• A. ${10}^{-29}A{m}^2$
• B. ${10}^{-27}A{m}^2$
• C. ${10}^{-23}A{m}^2$
• D. ${10}^{-19}A{m}^2$

Answer 1

Answer: (C)

${10}^{-23}A{m}^2$

$\omega =6.6\times {10}^{15}rps$
$r=0.53\times {10}^{-10}$
Now current due to revolution of electron is
Now,$\omega =6.6\times 2\pi \times {10}^{15}$
$\omega =4.14\times {10}^{16}$
Now current = charge x angular velocity
$I=1.6\times {10}^{-19}\times 4.14\times {10}^{16}$
$i=5.35\times {10}^{-3}$
$Magneticmoment=NIS$
$M=5.35\times {10}^{-3}\times \pi \times (0.53\times {10}^{-10}{)}^2$
$M=4.72\times {10}^{-23}$
Therefore, it is almost equal to $M={10}^{-23}A{m}^2$
option(C)