Question

The orbital speed of electron, orbiting around a nucleus in a circular orbit of radius $50pm$ is $2.2\times {10}^{6}m{s}^{-1}$. Then the magnetic dipole moment of an electron is

• A. $1.6\times {10}^{-19}A{m}^2$
• B. $5.3\times {10}^{-21}A{m}^2$
• C. $8.8\times {10}^{-24}A{m}^2$
• D. $8.8\times {10}^{-26}A{m}^2$

Answer 1

Answer: (C)

$8.8\times {10}^{-24}A{m}^2$

Magnetic dipole moment

$m=iA=\frac{e}{T}\times \pi r^2=\frac{e}{\left(2\pi r/v\right)}\times \pi r^2=\frac{erv}{2}$

$=\frac{1.6\times {10}^{-19}\times 50\times {10}^{-12}\times 2.2\times {10}^6}{2}$

$=8.8\times {10}^{-24}A{m}^2$