Sankalp NEET Full Test-3 Question-30 Solution

Question: 30. A particle of mass m is moving around the origin with a constant force F pulling it towards the origin. If Bohr model is used to describe its motion, the radius r of the n^{\text {th }} orbit and the particle’s speed v in the orbit depend on n as

(1) r \propto n^{1 / 3} ; v \propto n^{1 / 3}

(2) r \propto n^{1 / 3} ; v \propto n^{2 / 3}

(3) r \propto n^{2 / 3} ; v \propto n^{1 / 3}

(4) r \propto n^{4 / 3} ; v \propto n^{-1 / 3}

Answer: Option (3)

Explanation:

The particle moves in a circular orbit under a constant central

force F directed towards the origin.

This force provides the required centripetal force for circular motion.

Hence,

\frac{m v^{2}}{r} = F

From this equation, the speed of the particle is related to the radius by:

v^{2} = \frac{F r}{m} v \propto r^{1/2}

According to Bohr’s quantization condition,

the angular momentum of the particle is quantized as:

m v r = n \hbar

So,

v = \frac{n \hbar}{m r}

Now substitute this expression for v into the force equation:

\frac{m}{r} \left(\frac{n \hbar}{m r}\right)^{2} = F

Simplifying,

\frac{n^{2} \hbar^{2}}{m r^{3}} = F

Rearranging,

r^{3} = \frac{n^{2} \hbar^{2}}{m F}

Taking cube root,

r \propto n^{2/3}

Now use this result to find how

v depends on n.

From v \propto r^{1/2},

v \propto \left(n^{2/3}\right)^{1/2} = n^{1/3}

Thus, the radius of the n^{\text{th}} orbit varies

as r \propto n^{2/3} and the speed of the particle varies

as v \propto n^{1/3}.

Therefore, Option (3) is the correct answer.