JEE Advanced 2024 — Paper 1 — PHY-II solution

JEE Advanced2024Paper 1PHY-II

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 $L = n ℏ$ , where $ℏ = h / (2 π)$ , $h$ 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?

A
$r^{2} = n ℏ \frac{1}{mk}$
B
$v^{2} = n ℏ \frac{k}{m ^{3}}$
C
$\frac{L}{m r ^{2}} = \frac{k}{m}$
D
$E = \frac{n ℏ}{2} \frac{k}{m}$

Solution

For uniform circular motion under the central force $F = - k r$ , the inward force supplies the centripetal acceleration:

$\frac{m v ^{2}}{r} = k r \Rightarrow m v^{2} = k r^{2} \Rightarrow m v = mk r$ .

Multiplying by $r$ : $m v r = mk r^{2}$ . Using Bohr quantization $L = m v r = n ℏ$ :

$r^{2} = \frac{n ℏ}{mk} = n ℏ \frac{1}{mk}$ . Option (A) is correct.

From $v^{2} = \frac{k r ^{2}}{m} = \frac{k}{m} \cdot \frac{n ℏ}{mk} = n ℏ \frac{k}{m ^{3}}$ . Option (B) is correct.

Angular speed: $ω = \frac{v}{r}$ , and $\frac{L}{m r ^{2}} = ω$ . From $m v^{2} = k r^{2}$ , $v / r = k / m$ , so $\frac{L}{m r ^{2}} = \frac{k}{m}$ . Option (C) is correct.

Total energy: $E = \frac{1}{2} m v^{2} + \frac{1}{2} k r^{2} = k r^{2} = k \cdot \frac{n ℏ}{mk} = n ℏ \frac{k}{m}$ , not $\frac{n ℏ}{2} k / m$ . Option (D) is incorrect.

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