MHT CET · Physics · Magnetic Effects of Current
An electron(e) is revolving in a circular orbit of radius 'r' in hydrogen atom. The angular momentum of the electron is (M = magnetic dipole moment associated with it and \(\mathrm{m}=\) mass of electron \()\)
- A \(\frac{4 \mathrm{mM}}{\mathbf{C}}\)
- B \(\frac{2 m M}{e}\)
- C \(\frac{3 \mathrm{Mmm}}{\mathbf{C}}\)
- D \(\frac{\mathrm{mM}}{\mathrm{e}}\)
Answer & Solution
Correct Answer
(B) \(\frac{2 m M}{e}\)
Step-by-step Solution
Detailed explanation
The angular momentum of electron is given as \(L\)
\(
\begin{array}{l}
=m v r=m(r \omega) \cdot r=m \omega r^{2} \\
\Rightarrow \omega r^{2}=\frac{L}{m} \quad \ldots(i)
\end{array}
\)
Magnetic dipole moment,
\(
\begin{array}{l}
M=i A=\frac{e}{T} \cdot \pi r^{2}=\frac{e}{2 \pi / \omega} \cdot \pi r^{2}=\frac{e}{2} \cdot \omega r^{2} \\
\Rightarrow M=\frac{e}{2} \cdot \frac{L}{m} \text { [From Eq. (i)] } \\
\Rightarrow L=\frac{2 m M}{e}
\end{array}
\)
\(
\begin{array}{l}
=m v r=m(r \omega) \cdot r=m \omega r^{2} \\
\Rightarrow \omega r^{2}=\frac{L}{m} \quad \ldots(i)
\end{array}
\)
Magnetic dipole moment,
\(
\begin{array}{l}
M=i A=\frac{e}{T} \cdot \pi r^{2}=\frac{e}{2 \pi / \omega} \cdot \pi r^{2}=\frac{e}{2} \cdot \omega r^{2} \\
\Rightarrow M=\frac{e}{2} \cdot \frac{L}{m} \text { [From Eq. (i)] } \\
\Rightarrow L=\frac{2 m M}{e}
\end{array}
\)
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