The magnetic moment $\left(\mathrm{m}_{\mathrm{orb}}\right)$ of a revolving electron around the nucleus…

The magnetic moment $\left(\mathrm{m}_{\mathrm{orb}}\right)$ of a revolving electron around the nucleus varies with the principal quantum number (n) as
  1. $\mathrm{m}_{\text {orb }} \propto \mathrm{n}^2$
  2. $\mathrm{m}_{\mathrm{orb}} \propto \frac{1}{\mathrm{n}^2}$
  3. $\mathrm{m}_{\mathrm{orb}} \propto \frac{1}{\mathrm{n}}$
  4. $\mathrm{m}_{\text {orb }} \propto \mathrm{n}$

Solution

Orbital magnetic moment can be defined as, $\mathrm{m}_{\text {orb }}=\mathrm{iA}$ where, $\mathrm{i}=\frac{\mathrm{e}}{\mathrm{T}}, \mathrm{r}$ is the radius of the Bohr orbit, $\mathrm{A}=\pi \mathrm{r}^2$ is the area and $\mathrm{T}$ is the time period of uniform circular motion. \(\therefore \mathrm{m}_{\text {orb }}=\mathrm{e}\left(\frac{\pi \mathrm{r}^2}{\mathrm{~T}}\right)\) ...(1) Now, we can make use of the Bohr's hypothesis about angular momentum: \(\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}\) ...(2)
and the velocity of the electron doing a period circular motion.
\(\mathrm{vT}=2 \pi \mathrm{r}\) ...(3)
On dividing equation (2) by (3), and re-arranging,
\(\left(\frac{\pi r^2}{T}\right)=\frac{n h}{4 \pi m}\) ...(4) On plugging in above into equation (1), $\mathrm{m}_{\mathrm{orb}}=\frac{\mathrm{neh}}{4 \pi \mathrm{m}}$ Orbital magnetic moments of an electron in Bohr orbit is given by, $\begin{aligned} & \mathrm{m}_{\text {orb }}=\mathrm{n}\left(\frac{\mathrm{eh}}{4 \pi \mathrm{m}}\right) \\ & \therefore \mathrm{m}_{\text {orb }} \propto \mathrm{n}\end{aligned}$

Asked in: MHT CET 2022 (05 Aug Shift 2)

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