Question

An electron is revolving around a proton in a circular orbit of radius $r$. Using classical physics, prove that the ratio of value of the angular momentum and the value of magnetic moment of electron revolving in the orbit is a constant.

Hint:

Write L = m v r and mu = I A with I = e v / (2 pi r), then divide. The r and v cancel and 2m/e remains.

Answer 1

Step 1: Understanding the Concept:
A revolving electron is a tiny current loop, so it has a magnetic moment. It also has orbital angular momentum. We find both in terms of its speed v and the orbit radius r.

Step 2: Key Formula or Approach:
Angular momentum: \(L = m_e v r\). Magnetic moment: \(\mu = I A\), where \(I = e/T\) and \(A = \pi r^2\).
Time period of the revolution: \(T = 2\pi r/v\).

Step 3: Detailed Explanation:
Current due to the revolving electron:
\[ I = \frac{e}{T} = \frac{ev}{2\pi r} \]
Magnetic moment:
\[ \mu = I A = \frac{ev}{2\pi r} \times \pi r^2 = \frac{evr}{2} \]
Take the ratio of L to \(\mu\):
\[ \frac{L}{\mu} = \frac{m_e v r}{evr/2} = \frac{2m_e}{e} \]
This depends only on the mass and charge of the electron. Both are constants, so the ratio does not depend on r or v.

Final Answer:
The ratio of angular momentum to magnetic moment is a constant (about \(1.14 \times 10^{-11}\) kg/C). \[ \boxed{\frac{L}{\mu} = \frac{2m_e}{e}} \]