Question

An electron projected perpendicular to a uniform magnetic field \( B \) moves in a circle. If Bohr’s quantization is applicable, then the radius of the electronic orbit in the first excited state is:

• A. \( \sqrt\frac{2h}{{\pi e B}} \)
• B. \( \sqrt\frac{4h}{{\pi e B}} \)
• C. \( \sqrt\frac{h}{{\pi e B}} \)
• D. \(\sqrt \frac{h}{{2\pi e B}} \)

Hint:

In uniform magnetic fields: - Charged particles move in circular paths due to the Lorentz force. - Bohr’s quantization provides discrete angular momentum states: \( mvr = n \frac{h}{2\pi} \). - The radius of the orbit depends on the quantum number \( n \) and the magnetic field strength \( B \).

Answer 1

The Correct Option is D

Step 1: Apply the quantization condition. From Bohr’s quantization rule, the angular momentum of the electron is quantized: \[ m v r = n \frac{h}{2\pi}. \] For the first excited state, \( n = 2 \), so: \[ m v r = 2 \frac{h}{2\pi} = \frac{h}{\pi}. \] Step 2: Equating with the Lorentz force. The centripetal force is provided by the Lorentz force: \[ \frac{m v^2}{r} = e v B. \] Rearranging for \( r \): \[ r = \frac{m v}{e B}. \] Step 3: Substituting momentum. From Bohr’s condition: \[ m v = \frac{h}{\pi}, \] substituting in the equation for \( r \): \[ r = \frac{h}{\pi e B}. \] For \( n = 2 \), the radius is: \[ r = \frac{h}{\sqrt{2\pi e B}}. \] Thus, the answer is \( \boxed{\sqrt\frac{h}{{2\pi e B}}} \).