Given potential

The central potential is $U(r)=U_0 r^4.$

We need how the allowed orbital radius $r_n$ depends on $n$ using Bohr quantization.

Force due to the potential

For a central potential, $F(r)=-\frac{dU}{dr}=-4U_0 r^3.$

Its magnitude provides the centripetal force for circular motion: $\frac{mv^2}{r}=4U_0 r^3.$

Therefore, $mv^2=4U_0 r^4. \qquad (1)$

Bohr quantization condition

Angular momentum is quantized as $mvr=n\hbar. \qquad (2)$

From (2), $v=\frac{n\hbar}{mr}.$

Substitute into the force equation

Using $v=\dfrac{n\hbar}{mr}$ in (1): $m\left(\frac{n\hbar}{mr}\right)^2=4U_0 r^4.$

Simplify: $\frac{n^2\hbar^2}{mr^2}=4U_0 r^4.$

$n^2\hbar^2=4U_0 m\, r^6.$

Hence, $r^6 \propto n^2.$

So, $r_n \propto n^{2/6}=n^{1/3}.$

Compare with the required form

Given, $r_n \propto n^{1/\alpha}.$

Since $r_n \propto n^{1/3},$ we get $\frac{1}{\alpha}=\frac{1}{3} \Rightarrow \alpha=3.$

Comparison with stored answer

Stored correct answer: $3$

Derived answer: $3$

They agree.