Answer

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**Hint :**The answer to this problem can be found by equating the given potential energy of the particle and the kinetic energy of the particle. We have to find the radius of the nth orbit so equate the angular momentum of the particle and Bohr’s angular momentum and get the value equate it with the kinetic energy.

**Complete step by step answer**

Given, The potential energy of the small particle, $ U = \dfrac{1}{2}m{\omega ^2}{r^2}{\text{ }} \to {\text{1}} $

Where, U is the potential energy of the small particle

$ \omega $ is a constant

$ r $ is the distance of the particle from the origin

$ m $ is the mass of the small particle

Then the kinetic energy of the small particle, $ K.E = \dfrac{1}{2}m{v^2}{\text{ }} \to {\text{2}} $

Where,

K.E is the kinetic energy of the small particle

$ m $ is the mass of the small particle

V is the velocity of the particle with which it moves

There is a hint given in the question itself to use the equations of Bohr's model of quantisation of angular momentum and circular orbits

The angular momentum of a particle in nth orbit is,

$ L = mvr{\text{ }} \to 3 $

L is the angular momentum of a particle

$ m $ is the mass of the small particle

V is the velocity of the particle with which it moves

$ r $ is the distance of the particle from the origin

By Bohr’s first postulate, the angular momentum of the electron

$ L = \dfrac{{nh}}{{2\pi }}{\text{ }} \to 4 $

L is the angular momentum of a particle

n is the orbit in which it revolves

h is the Planck constant

Equating 3 and 4 we get

$ mvr = \dfrac{{nh}}{{2\pi }} $

$ mv = \dfrac{{nh}}{{2\pi r}}{\text{ }} \to 5 $

Substitute equation 5 in equation 2

$ K.E = \dfrac{1}{2}{\left( {\dfrac{{nh}}{{2\pi r}}} \right)^2} $

$ K.E = \dfrac{1}{4}\dfrac{{{n^2}{h^2}}}{{{\pi ^2}{r^2}}}{\text{ }} \to {\text{6}} $

We know that,

$ Kinetic{\text{ }}energy{\text{ }} = \dfrac{1}{2}{}potential{\text{ }}energy $

Then, from equation 1 and equation 6

$ \dfrac{1}{4}\dfrac{{{n^2}{h^2}}}{{{\pi ^2}{r^2}}} = \dfrac{1}{2}\left( {\dfrac{1}{2}m{\omega ^2}{r^2}} \right) $

$ \dfrac{{{n^2}{h^2}}}{{{\pi ^2}{r^2}}} = m{\omega ^2}{r^2} $

$ \dfrac{{{n^2}{h^2}}}{{{\pi ^2}{r^2}m{\omega ^2}}} = {r^2} $

$ \dfrac{{{n^2}{h^2}}}{{{\pi ^2}m{\omega ^2}}} = {r^2} \times {r^2} $

$ {r^4} = \dfrac{{{n^2}{h^2}}}{{{\pi ^2}m{\omega ^2}}} $

From above equation we get

$ {r^4} \propto {n^2} $

$ r \propto \sqrt n $

The radius, $ r $ of orbit is proportional to $ \sqrt n $ (square root of n)

**Hence the correct answer is option (B) $ \sqrt n $ .**

**Note**

It is an indirect question since we have to find the relation between the radius and the nth orbit, we are using the equations having n (nth orbit) and r (radius) to relate them. It is given in the question to Assume Bohr's model of quantisation of angular momentum and circular orbits.

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