MHT CET · Physics · Atomic Physics
An electron moving around the nucleus with an angular momentum \(L\) has a magnetic moment ( \(e=\) charge on electron, \(m=\) mass of electron \()\)
- A \(\frac{2 e}{m} L\)
- B \(\frac{e}{m} L\)
- C \(\frac{e}{2 m} L\)
- D \(\frac{e}{2 \pi m} L\)
Answer & Solution
Correct Answer
(C) \(\frac{e}{2 m} L\)
Step-by-step Solution
Detailed explanation
The magnetic moment of a current carrying coil is given by: \(\mu=i A=i \pi r^2\)
The moving electron in an orbit is equivalent to a coil with current \(i=\frac{e}{T}\), where \(T\) is the period of rotation around the nucleus.
Therefore, \(\mu=\frac{e}{T} \pi r^2\)
Using definition of angular momentum \(L=m v r\) or \(r=\frac{L}{m v}\) and the velocity of the electron is given by \(v=\frac{2 \pi r}{T}\) or \(\frac{1}{T}=\frac{v}{2 \pi r}\)
We can substitute \(r\) and \(\frac{1}{\mathrm{~T}}\) into the expression for magnetic moment,
\(\mu=e\left(\frac{v}{2 \pi r}\right) \pi r^2=\frac{e v}{2} r=\frac{e v}{2} \frac{L}{m v}=\left(\frac{e}{2 m}\right) L\)
The moving electron in an orbit is equivalent to a coil with current \(i=\frac{e}{T}\), where \(T\) is the period of rotation around the nucleus.
Therefore, \(\mu=\frac{e}{T} \pi r^2\)
Using definition of angular momentum \(L=m v r\) or \(r=\frac{L}{m v}\) and the velocity of the electron is given by \(v=\frac{2 \pi r}{T}\) or \(\frac{1}{T}=\frac{v}{2 \pi r}\)
We can substitute \(r\) and \(\frac{1}{\mathrm{~T}}\) into the expression for magnetic moment,
\(\mu=e\left(\frac{v}{2 \pi r}\right) \pi r^2=\frac{e v}{2} r=\frac{e v}{2} \frac{L}{m v}=\left(\frac{e}{2 m}\right) L\)
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