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(hle) where n is an integer, h is Planck’s 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)

The correct answer is:
$\frac{he}{2\pi m}$

Step-by-step Explanation:

1. Magnetic Flux Quantization:
The magnetic flux $\Phi$ passing through a circular orbit of radius $r$ in a uniform magnetic field $B$ is given by:
$\Phi =B·A=B\left(\pi r^2\right)$

According to the given model, the flux is quantized as:
$\Phi =n\left(\frac{h}{e}\right)$
where $n$ is an integer, $h$ is Planck's constant, and $e$ is the magnitude of the electron's charge.

For the lowest energy state, we set $n=1$:
$B\left(\pi r^2\right)=\frac{h}{e}$
From this, we can express the radius squared as:
$r^2=\frac{h}{e\pi B}$

2. Circular Motion in a Magnetic Field:
For an electron of mass $m$ and speed $v$ moving in a circular orbit perpendicular to a uniform magnetic field $B$, the magnetic force provides the necessary centripetal force:
$evB=\frac{m{v}^2}{r}$
Simplifying this gives:
$v=\frac{eBr}{m}$

3. Magnetic Moment:
The magnetic moment $\mu$ of a circulating electron is the product of the equivalent current $I$ and the area of the orbit $A$:
$\mu =I·A$
The equivalent current is the charge divided by the orbital period $T=\frac{2\pi r}{v}$:
$I=\frac{e}{T}=\frac{ev}{2\pi r}$
Substituting $I$ and $A=\pi r^2$ into the equation for $\mu$:
$\mu =\left(\frac{ev}{2\pi r}\right)·\left(\pi r^2\right)=\frac{evr}{2}$

Substitute the expression for $v$ from Step 2 into this equation:
$\mu =\frac{e}{2}\left(\frac{eBr}{m}\right)r=\frac{e^{2}B{r}^2}{2m}$

4. Substitute the Quantized Radius:
Now substitute the expression for $r^2$ obtained in Step 1:
$\mu =\frac{e^{2}B}{2m}\left(\frac{h}{e\pi B}\right)$
Simplifying the terms by canceling $e$ and $B$:
$\mu =\frac{he}{2\pi m}$