In the Bohr's model of hydrogen-like atom the force between the nucleus and the electron is modified as $$F = \frac{e^2}{4\pi\epsilon_0}\left(\frac{1}{r^2} + \frac{\beta}{r^3}\right)$$, where $$\beta$$ is a constant. For this atom, the radius of the $$n^{th}$$ orbit in terms of the Bohr radius $$\left(a_0 = \frac{\epsilon_0 h^2}{m\pi e^2}\right)$$ is :

In Bohr's model, the electron moves in a circular orbit around the nucleus. The centripetal force required for this motion is provided by the electrostatic force between the electron and the nucleus. The given force is:

$$ F = \frac{e^2}{4\pi\epsilon_0} \left( \frac{1}{r^2} + \frac{\beta}{r^3} \right) $$

For circular motion, the centripetal force is $$ \frac{m v^2}{r} $$, where $$ m $$ is the mass of the electron and $$ v $$ is its orbital speed. Setting this equal to the given force:

$$ \frac{m v^2}{r} = \frac{e^2}{4\pi\epsilon_0} \left( \frac{1}{r^2} + \frac{\beta}{r^3} \right) $$

Multiply both sides by $$ r $$:

$$ m v^2 = \frac{e^2}{4\pi\epsilon_0} \left( \frac{1}{r} + \frac{\beta}{r^2} \right) $$

Bohr's quantization condition states that the angular momentum is quantized: $$ m v r = n \frac{h}{2\pi} $$, where $$ n $$ is the quantum number and $$ h $$ is Planck's constant. Solving for $$ v $$:

$$ v = \frac{n h}{2\pi m r} $$

Square both sides:

$$ v^2 = \frac{n^2 h^2}{4\pi^2 m^2 r^2} $$

Substitute this expression for $$ v^2 $$ into the force equation:

$$ m \left( \frac{n^2 h^2}{4\pi^2 m^2 r^2} \right) = \frac{e^2}{4\pi\epsilon_0} \left( \frac{1}{r} + \frac{\beta}{r^2} \right) $$

Simplify the left side:

$$ \frac{n^2 h^2}{4\pi^2 m r^2} = \frac{e^2}{4\pi\epsilon_0} \left( \frac{1}{r} + \frac{\beta}{r^2} \right) $$

Multiply both sides by $$ r^2 $$ to eliminate denominators:

$$ \frac{n^2 h^2}{4\pi^2 m} = \frac{e^2}{4\pi\epsilon_0} \left( r + \beta \right) $$

Now, solve for $$ r $$. Multiply both sides by $$ 4\pi^2 m $$:

$$ n^2 h^2 = 4\pi^2 m \cdot \frac{e^2}{4\pi\epsilon_0} (r + \beta) $$

Simplify the right side:

$$ n^2 h^2 = \pi m \cdot \frac{e^2}{\epsilon_0} (r + \beta) $$

Divide both sides by $$ \pi m \frac{e^2}{\epsilon_0} $$:

$$ r + \beta = \frac{n^2 h^2 \epsilon_0}{\pi m e^2} $$

The Bohr radius is given as $$ a_0 = \frac{\epsilon_0 h^2}{m\pi e^2} $$. Notice that:

$$ \frac{h^2 \epsilon_0}{\pi m e^2} = a_0 $$

Therefore:

$$ r + \beta = n^2 a_0 $$

Solving for $$ r $$:

$$ r = n^2 a_0 - \beta $$

Thus, the radius of the $$ n^{\text{th}} $$ orbit is $$ r_n = a_0 n^2 - \beta $$.

Comparing with the options:

A. $$ r_n = a_0 n - \beta $$

B. $$ r_n = a_0 n^2 + \beta $$

C. $$ r_n = a_0 n^2 - \beta $$

D. $$ r_n = a_0 n + \beta $$

Hence, the correct answer is Option C.