A hypothetical atom consists of a particle of mass m moving in a...

The potential energy of the particle is given by $U(r) = \alpha r^4$.

The conservative force acting on the particle is $F = -\dfrac{dU}{dr} = -4\alpha r^3$.

The magnitude of this force provides the necessary centripetal force for circular motion:

$\dfrac{mv^2}{r} = 4\alpha r^3$

$mv^2 = 4\alpha r^4$

$v = \sqrt{\dfrac{4\alpha}{m}} r^2$

According to Bohr's quantization postulate, the angular momentum is quantized:

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

Substituting the expression for $v$ into the quantization condition:

$m \left( \sqrt{\dfrac{4\alpha}{m}} r^2 \right) r = \dfrac{nh}{2\pi}$

$\sqrt{4\alpha m} r^3 = \dfrac{nh}{2\pi}$

$r^3 = \dfrac{nh}{2\pi \sqrt{4\alpha m}}$

From this, we can see that the radius of the $n^{\text{th}}$ orbit depends on $n$ as:

$r_n \propto n^{1/3}$

The kinetic energy of the particle is given by $K = \dfrac{1}{2}mv^2$. Using $mv^2 = 4\alpha r^4$, we get:

$K_n = \dfrac{1}{2}(4\alpha r_n^4) = 2\alpha r_n^4$

Substituting $r_n \propto n^{1/3}$ into the kinetic energy expression:

$K_n \propto (n^{1/3})^4$

$K_n \propto n^{4/3}$

Answer: $r_n \propto n^{1/3}$ ; $K_n \propto n^{4/3}$