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JEE Advance - Physics (2024 - Paper 1 Online - No. 5)

A particle of mass $m$ is moving in a circular orbit under the influence of the central force $F(r)=-k r$, corresponding to the potential energy $V(r)=k r^2 / 2$, where $k$ is a positive force constant and $r$ is the radial distance from the origin. According to the Bohr's quantization rule, the angular momentum of the particle is given by $L=n \hbar$, where $\hbar=h /(2 \pi), h$ is the Planck's constant, and $n$ a positive integer. If $v$ and $E$ are the speed and total energy of the particle, respectively, then which of the following expression(s) is(are) correct?
$r^2=n \hbar \sqrt{\frac{1}{m k}}$
$v^2=n \hbar \sqrt{\frac{k}{m^3}}$
$\frac{L}{m r^2}=\sqrt{\frac{k}{m}}$
$E=\frac{n \hbar}{2} \sqrt{\frac{k}{m}}$

Explanation

The central force will provide necessary centripetal force

$$\Rightarrow \mathrm{kr}=\frac{\mathrm{mv}^2}{\mathrm{r}}$$

or, $$\mathrm{kr}^2=\mathrm{mv}^2\quad \text{... (1)}$$

By quantisation rule $$\mathrm{n} \hbar=\mathrm{mvr}$$

or, $$\frac{\mathrm{n} \hbar}{\mathrm{r}}=\mathrm{mv}\quad \text{... (2)}$$

$$\begin{aligned} & \frac{(1)}{(2)^2} \Rightarrow \frac{\mathrm{kr}^2}{\frac{\mathrm{n}^2 \hbar^2}{\mathrm{r}^2}}=\frac{\mathrm{mv}^2}{\mathrm{~m}^2 \mathrm{v}^2} \\ & \Rightarrow \frac{\mathrm{k}}{\mathrm{n}^2 \hbar^2} \mathrm{r}^4=\frac{1}{\mathrm{~m}} \\ & \Rightarrow \mathrm{r}=\left(\frac{\mathrm{n}^2 \hbar^2}{\mathrm{~km}}\right)^{\frac{1}{4}} \Rightarrow \mathrm{r}^2=\frac{\mathrm{n} \hbar}{\sqrt{\mathrm{mk}}} \end{aligned}$$

(B) Using (1), $$\mathrm{K} \cdot \frac{\mathrm{n} \hbar}{\sqrt{\mathrm{mk}}}=\mathrm{mv}^2$$

$$\Rightarrow \mathrm{v}^2=\mathrm{n} \hbar \sqrt{\frac{\mathrm{k}}{\mathrm{m}^3}}$$

(C) $$\frac{\mathrm{L}}{\mathrm{mr}^2}=\frac{\mathrm{mvr}}{\mathrm{mr}^2}=\frac{\mathrm{v}}{\mathrm{r}}=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$$ from (1)

(D) $$\mathrm{E}=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{kr}^2=\frac{\mathrm{n} \hbar}{2} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}+\frac{1}{2} \mathrm{k} \frac{\mathrm{n} \hbar}{\sqrt{\mathrm{mk}}}$$

$$\mathrm{E}=\mathrm{n} \hbar \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$$

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