Question

Electron is moving with speed of \(2 \times 10^6\) m/s in an orbit of radius \(5.0 \times 10^{-11}\) metre in hydrogen atom. Determine the magnetic moment of rotating electron.

Hint:

The formula \(\mu = \frac{evr}{2}\) is very useful. It can also be related to the orbital angular momentum \(L = mvr\). The ratio \(\frac{\mu}{L} = \frac{evr/2}{mvr} = \frac{e}{2m}\) is called the gyromagnetic ratio, a fundamental constant for an electron's orbital motion.

Answer 1

Step 1: Understanding the Concept:
An electron revolving in a circular orbit is equivalent to a tiny current loop. According to Ampere's law, any current loop produces a magnetic field and has an associated magnetic dipole moment. The magnitude of this magnetic moment depends on the equivalent current and the area of the loop.

Step 2: Key Formula or Approach:
1. The magnetic dipole moment (\(\mu\)) of a current loop is given by \(\mu = iA\), where \(i\) is the current and \(A\) is the area of the loop.
2. The equivalent current \(i\) produced by an electron (charge \(e\)) revolving with a time period \(T\) is \(i = \frac{e}{T}\).
3. The time period of revolution is the circumference of the orbit divided by the speed: \(T = \frac{2\pi r}{v}\).
Combining these formulas, we get an expression for the magnetic moment: \[ \mu = iA = \left(\frac{e}{T}\right)(\pi r^2) = \left(\frac{e}{2\pi r/v}\right)(\pi r^2) = \frac{evr}{2} \]

Step 3: Detailed Explanation:
We are given the following values: \begin{itemize} \item Speed of the electron, \(v = 2 \times 10^6\) m/s. \item Radius of the orbit, \(r = 5.0 \times 10^{-11}\) m. \item Charge of an electron, \(e = 1.6 \times 10^{-19}\) C. \end{itemize} Now, we substitute these values into the derived formula for magnetic moment: \[ \mu = \frac{evr}{2} \] \[ \mu = \frac{(1.6 \times 10^{-19} \text{ C}) \times (2 \times 10^6 \text{ m/s}) \times (5.0 \times 10^{-11} \text{ m})}{2} \] \[ \mu = \frac{1.6 \times 2 \times 5.0}{2} \times 10^{-19 + 6 - 11} \text{ A·m}^2 \] \[ \mu = 8.0 \times 10^{-24} \text{ A·m}^2 \]

Step 4: Final Answer:
The magnetic moment of the rotating electron is \(8.0 \times 10^{-24}\) A·m\(^2\).