A particle of mass is moving in a circular orbit under the influence of the central force corresponding to the potential energy where is a positive force constant and is the radial distance from the origin. [2024]

Question

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 $V (r) = \frac{k r^{2}}{2},$ 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 $ℏ = \frac{h}{(2 π)},$ $h$ 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? [2024]

Smart Achievers · Accepted Answer

(1, 2, 3)

Mass $F (r) = - k r,$ is moving in a circular orbit under the influence of the central force $F (r) = - k r$

which provides the necessary centripetal force.

$∴$ $F (r) = F (c)$ or, $k r = \frac{m v^{2}}{r}$

or, $k r^{2} = m v^{2}$ $\dots (i)$

Using the quantisation rule, $n ℏ = m v r$

or, $\frac{n ℏ}{r} = m v$ $\dots (i i)$

Squaring both sides of Eq. (ii), $\frac{n^{2} ℏ^{2}}{r^{2}} = m^{2} v^{2}$ $\dots (i i i)$

Dividing Eq. (i) by Eq. (iii),

$\frac{k r^{2}}{\frac{n^{2} ℏ^{2}}{r^{2}}} = \frac{m v^{2}}{m^{2} v^{2}}$

$\Rightarrow \frac{k}{n^{2} ℏ^{2}} r^{4} = \frac{1}{m}$

$\Rightarrow r = {(\frac{n^{2} ℏ^{2}}{k m})}^{\frac{1}{4}}$ $\Rightarrow r^{2} = n ℏ \sqrt{\frac{1}{m k}}$

Hence, option (1) is correct.

Now using Eq. (i),

$k \cdot \frac{n ℏ}{\sqrt{m k}} = m v^{2}$ $\Rightarrow v^{2} = n ℏ \sqrt{\frac{k}{m^{3}}}$

Hence, option (2) is correct.

Again using Eq. (i),

$\frac{L}{m r^{2}} = \frac{m v r}{m r^{2}} = \frac{v}{r} = \sqrt{\frac{k}{m}}$

Hence, option (3) is correct.

Total energy, $E = K + E$ or,

$E = \frac{1}{2} m v^{2} + \frac{1}{2} k r^{2}$ $= \frac{n ℏ}{2} \sqrt{\frac{k}{m}} + \frac{1}{2} k \cdot \frac{n ℏ}{\sqrt{m k}}$

$∴$ $E = n ℏ \sqrt{\frac{k}{m}}$

Hence, option (4) is incorrect.

Solution

1 $r^{2} = n ℏ \sqrt{\frac{1}{m k}}$

2 $v^{2} = n ℏ \sqrt{\frac{k}{m^{3}}}$

3 $\frac{L}{m r^{2}} = \sqrt{\frac{k}{m}}$

4 $E = \frac{n ℏ}{2} \sqrt{\frac{k}{m}}$

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Solution

1 r2=nℏ1mk

2 v2=nℏkm3

3 Lmr2=km

4 E=nℏ2km

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1 $r^{2} = n ℏ \sqrt{\frac{1}{m k}}$

2 $v^{2} = n ℏ \sqrt{\frac{k}{m^{3}}}$

3 $\frac{L}{m r^{2}} = \sqrt{\frac{k}{m}}$

4 $E = \frac{n ℏ}{2} \sqrt{\frac{k}{m}}$