Question

Bohr model is applied to a particle of mass 'm' and charge 'q' moving in a plane under the influence of a transverse magnetic field B. The energy of the charged particle in the \(n^{th}\) level will be (h=Planck's constant)

• A. \(\frac{2nhqB}{πm}\)
• B. \(\frac{nhqB}{πm}\)
• C. \(\frac{nhqB}{2πm}\)
• D. \(\frac{nhqB}{4πm}\)

Hint:

Quantise angular momentum m v r = n h / 2 pi and use q v B = m v^2 / r.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The particle moves in a circle in the plane perpendicular to the field. The magnetic force supplies the centripetal force: \(qvB=\dfrac{mv^2}{r}\), so \(mv=qBr\).

Step 2: Apply the Bohr quantisation:
Angular momentum: \(mvr=\dfrac{nh}{2\pi}\). Using \(mv=qBr\), we get \(qBr^2=\dfrac{nh}{2\pi}\), so \(r^2=\dfrac{nh}{2\pi qB}\).

Step 3: Find the energy:
All the energy is kinetic: \(E=\dfrac12mv^2=\dfrac12m\left(\dfrac{qBr}{m}\right)^2=\dfrac{q^2B^2r^2}{2m}\).
\[ E=\dfrac{q^2B^2}{2m}\cdot\dfrac{nh}{2\pi qB}=\dfrac{nhqB}{4\pi m} \]
Option D.

Step 4: Why the other options are wrong.
Options A, B and C are larger by factors of 8, 4 and 2. They come from dropping the factor \(\dfrac12\) in kinetic energy or the factor \(2\pi\) in the quantisation rule.

Final Answer:
The energy is n h q B / (4 pi m). \[ \boxed{\text{(D) }\dfrac{nhqB}{4\pi m}} \]