Question

The ratio of magnetic dipole moment to that of the angular momentum of an electron of mass $m\text{kg}$ and charge $e\text{C}.$ In Bohr's orbit of the hydrogen atom is-

• A. $\frac{e}{m}$
• B. $\frac{2e}{m}$
• C. $\frac{e}{2m}$
• D. ${\left(\frac{e}{m}\right)}^2$

Answer 1

The correct option is C $\frac{e}{2m}$

The magnetic dipole moment of an electron is,

${\mu }_B=iA=\left(\frac{ev}{2\pi r}\right)\left(\pi r^2\right)=\frac{1}{2}evr......\left(1\right)$

The angular momentum of the electron is,

$L=mvr\Rightarrow vr=\frac{L}{m}........\left(2\right)$


From (1) $\&$ (2) we get,

${\mu }_B=\frac{1}{2}\frac{eL}{m}$

$\frac{{\mu }_B}{L}=\frac{e}{2m}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.