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: Enforce quantization of angular momentum. Based on Bohr's quantization rule, the electron's angular momentum is discrete:\[m v r = n \frac{h}{2\pi}.\]For the first excited state, \( n = 2 \), which simplifies to:\[m v r = 2 \frac{h}{2\pi} = \frac{h}{\pi}.\]Step 2: Set angular momentum quantization equal to the Lorentz force. The centripetal force is supplied by the Lorentz force:\[\frac{m v^2}{r} = e v B.\]Solving for the radius \( r \):\[r = \frac{m v}{e B}.\]Step 3: Substitute the quantized momentum. From Bohr's condition, we have:\[m v = \frac{h}{\pi},\]Substitute this into the equation for \( r \):\[r = \frac{h}{\pi e B}.\]For \( n = 2 \), the specific radius is:\[r = \frac{h}{\sqrt{2\pi e B}}.\]Therefore, the solution is \( \boxed{\sqrt\frac{h}{{2\pi e B}}} \).