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MHT CET · Physics · Atomic Physics

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. A \(\mathrm{m}_{\text {orb }} \propto \mathrm{n}^2\)
  2. B \(\mathrm{m}_{\mathrm{orb}} \propto \frac{1}{\mathrm{n}^2}\)
  3. C \(\mathrm{m}_{\mathrm{orb}} \propto \frac{1}{\mathrm{n}}\)
  4. D \(\mathrm{m}_{\text {orb }} \propto \mathrm{n}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\mathrm{m}_{\text {orb }} \propto \mathrm{n}\)

Step-by-step Solution

Detailed explanation

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.
\(\therefore\text m _{\text{orb} }=\text e \left(\frac{\pi\text r ^2}{\text T}\right)\qquad\ldots(1)\)
Now, we can make use of the Bohr's hypothesis about angular momentum:
\(\text{mvr} =\frac{\text{nh}}{2 \pi}\qquad\ldots(2)\)
and the velocity of the electron doing a period circular motion.
\(\text{vT} =2 \pi\text r\qquad\ldots(3)\)
On dividing equation (2) by (3), and re-arranging,
\(\left(\frac{\pi\text r^2}{\text T}\right)=\frac{\text{nh}}{4 \pi\text m}\qquad\ldots(4)\)
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,
\(\begin{aligned} & \mathrm{m}_{\text {orb }}=\mathrm{n}\left(\frac{\mathrm{eh}}{4 \pi \mathrm{m}}\right) \\ & \therefore \mathrm{m}_{\text {orb }} \propto \mathrm{n}\end{aligned}\)