Step 1: Understanding the Concept:
An electron orbiting a nucleus behaves like a tiny current loop, which produces a magnetic dipole moment.
The orbiting electron also possesses angular momentum.
The ratio of these two quantities is a fundamental constant for an orbiting charged particle.
Step 2: Key Formulas or Approach:
Angular momentum of a particle in circular motion: \( L = mvr \).
Magnetic dipole moment of a current loop: \( m_{\text{orb}} = I \times A \).
Current produced by an orbiting electron: \( I = \frac{e}{T} \), where \( T \) is the time period.
Step 3: Detailed Explanation:
Consider an electron of mass \( m \) and charge \( e \) moving in a circular orbit of radius \( r \) with speed \( v \).
The angular momentum is \( L = mvr \).
The time period of revolution is \( T = \frac{2\pi r}{v} \).
The equivalent current is \( I = \frac{e}{T} = \frac{e}{2\pi r / v} = \frac{ev}{2\pi r} \).
The area of the circular loop is \( A = \pi r^2 \).
The magnetic dipole moment magnitude is:
\[ m_{\text{orb}} = I \cdot A = \left(\frac{ev}{2\pi r}\right) \cdot (\pi r^2) \]
Simplify the expression:
\[ m_{\text{orb}} = \frac{evr}{2} \]
We are asked to find the ratio of angular momentum \( L \) to magnetic dipole moment \( m_{\text{orb}} \).
\[ \text{Ratio} = \frac{L}{m_{\text{orb}}} \]
Substitute the expressions for \( L \) and \( m_{\text{orb}} \):
\[ \text{Ratio} = \frac{mvr}{\frac{evr}{2}} \]
Cancel the common terms \( v \) and \( r \):
\[ \text{Ratio} = \frac{m}{\frac{e}{2}} \]
\[ \text{Ratio} = \frac{2m}{e} \]
Note: The inverse ratio, \( \frac{m_{\text{orb}}}{L} = \frac{e}{2m} \), is known as the gyromagnetic ratio.
Step 4: Final Answer:
The ratio of angular momentum to magnetic dipole moment is \( \frac{2m}{e} \).