A model for quantized motion of an electron in a uniform magnetic field $B$ states that the flux pas

Question

A model for quantized motion of an electron in a uniform magnetic field $B$ states that the flux passing through the orbit of the electron is $n(h / e)$ where $n$ is an integer, $h$ planks constant and $e$ is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be ( $m$ is the mass of the electron)

Professor PC Thomas Classes · Answer

$\begin{aligned} & 	ext { (d) : Flux, } \phi=B 	imes \pi r^2 \ & 	ext { or } \frac{n h}{e}=B 	imes \pi r^2 	ext { or } r^2=\frac{n h}{B e \pi} \ & 	ext { Also, } B v e=\frac{m v^2}{r} 	ext { or } m v=B e r\end{aligned}$ $r$ \ Let magnetic moment of electron in circular orbit of an atom is $\mu$. $ \begin{aligned} & \mu=I A \ & =\frac{e}{T} A=\frac{e \omega}{2 \pi}\left(\pi r^2ight)=\frac{e v r}{2} \end{aligned} $ \ where $m$ is the mass of electron and $L$ is angular momentum From eq. (iii), $$ \mu=\frac{e v r}{2}=\frac{e}{2 m} L \quad(\because L=m v r) $$ or $\mu=\frac{e}{2 m} 	imes m v r=\frac{e}{2 m} 	imes$ Ber $^2$ (Using eq (ii)) or $\mu=\frac{e}{2 m} 	imes B e 	imes \frac{n h}{B e \pi}$ (Using eq (i)) $$ 	herefore \mu=\frac{h e}{2 \pi m} $$