JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Modern Physics

A small particle of mass

m

moves in such a way that its potential energy

U = \frac{1}{2} m ω^{2} r^{2}

where

ω

is constant and

r

is the distance of the particle from origin. Assuming Bohr's quantization of momentum and circular orbit, the radius of

n^{th}

orbit will be proportional to.

(A) $n$ ✓ Correct
(B)n
(C) $n^{2}$
(D) $\frac{1}{n}$

Answer: (A)

Solution

The force is

F = - \frac{d U}{d r} = - m ω^{2} r

, and for a circular orbit it supplies the centripetal force

\frac{m v ^{2}}{r} = m ω^{2} r

, giving

v = ω r

. Bohr's quantization

m v r = \frac{nh}{2 π}

then becomes

mω r^{2} = \frac{nh}{2 π}

, so

r^{2} \propto n

. The radius of the

n^{th}

orbit is proportional to

n

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Single correct option

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