Question

Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the \(n^{\mathrm{th}}\) orbital will therefore be proportional to :

• A. \(n^2\)
• B. \(n\)
• C. \(n^{1/2}\)
• D. \(n^{1/4}\)

Hint:

Combine \(2\pi r = n\lambda\) with \(\lambda = h/p\) and \(p = qBr\) for circular motion in a magnetic field.

Answer 1

The Correct Option is C

Condition: \(2\pi r = n\lambda = \frac{nh}{mv}\).
In the magnetic field, \(\frac{mv^2}{r} = qvB\), so \(mv = qBr\).
Substituting: \(2\pi r = \frac{nh}{qBr}\), so \(r^2 = \frac{nh}{2\pi qB}\).
\(r = \sqrt{\frac{nh}{2\pi qB}} \propto n^{1/2}\).
Answer: (3) \(n^{1/2}\)