MHT CET · Physics · Atomic Physics
An electron in the ground state of hydrogen atom is revolving in a circular orbit of
radius \(\mathrm{R}\). The orbital magnetic moment of the electron is
\((\mathrm{m}=\) mass of electron, \(\mathrm{h}=\) Planck's constant, \(\mathrm{e}=\) electronic charge \()\)
- A \(\frac{-\mathrm{eh}}{\pi \mathrm{m}}\)
- B \(\frac{\mathrm{eh}}{2 \pi \mathrm{m}}\)
- C \(\frac{2 \mathrm{eh}}{\pi \mathrm{m}}\)
- D \(\frac{\mathrm{eh}}{4 \pi \mathrm{m}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{eh}}{4 \pi \mathrm{m}}\)
Step-by-step Solution
Detailed explanation
(A)
In ground state \((\mathrm{n}=1)\) according to Bohr's theory:
\(\mathrm{mvR}=\frac{\mathrm{h}}{2 \pi}\) or \(\mathrm{v}=\frac{\mathrm{h}}{2 \pi \mathrm{mR}}\)
Now time period, \(\mathrm{T}=\frac{2 \pi \mathrm{R}}{\mathrm{v}}=\frac{2 \pi \mathrm{R}}{\mathrm{h} / 2 \pi \mathrm{mR}}=\frac{4 \pi^{2} \mathrm{mR}^{2}}{\mathrm{~h}}\)
Magnetic moment, \(\mathrm{M}=\mathrm{iA}\)
Where, \(\mathrm{I}=\frac{\text { charge }}{\text { time period }}=\frac{\mathrm{e}}{\frac{4 \pi^{2} \mathrm{mR}^{2}}{\mathrm{~h}}}=\frac{\mathrm{eh}}{4 \pi^{2} \mathrm{mR}^{2}}\)
and \(\mathrm{A}=\pi \mathrm{R}^{2}\)
\(\therefore \mathrm{M}=\left(\pi \mathrm{R}^{2}\right)\left(\frac{\text { eh }}{4 \pi^{2} \mathrm{mR}^{2}}\right)\) or \(\mathrm{M}=\frac{\text { eh }}{4 \pi \mathrm{m}}\)
Direction of magnetic moment \(\vec{M}\) is perpendicular to the plane of orbit.
In ground state \((\mathrm{n}=1)\) according to Bohr's theory:
\(\mathrm{mvR}=\frac{\mathrm{h}}{2 \pi}\) or \(\mathrm{v}=\frac{\mathrm{h}}{2 \pi \mathrm{mR}}\)
Now time period, \(\mathrm{T}=\frac{2 \pi \mathrm{R}}{\mathrm{v}}=\frac{2 \pi \mathrm{R}}{\mathrm{h} / 2 \pi \mathrm{mR}}=\frac{4 \pi^{2} \mathrm{mR}^{2}}{\mathrm{~h}}\)
Magnetic moment, \(\mathrm{M}=\mathrm{iA}\)
Where, \(\mathrm{I}=\frac{\text { charge }}{\text { time period }}=\frac{\mathrm{e}}{\frac{4 \pi^{2} \mathrm{mR}^{2}}{\mathrm{~h}}}=\frac{\mathrm{eh}}{4 \pi^{2} \mathrm{mR}^{2}}\)
and \(\mathrm{A}=\pi \mathrm{R}^{2}\)
\(\therefore \mathrm{M}=\left(\pi \mathrm{R}^{2}\right)\left(\frac{\text { eh }}{4 \pi^{2} \mathrm{mR}^{2}}\right)\) or \(\mathrm{M}=\frac{\text { eh }}{4 \pi \mathrm{m}}\)
Direction of magnetic moment \(\vec{M}\) is perpendicular to the plane of orbit.
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