Animated Solution for Physics - Atoms and Nuclei: A particle of mass $m$ moves in a circular orbit in a central potential field $U(r)=U_0{r}^4$. If Bohr's quantisation conditions are applied, radii of possible orbitals $r_n$ vary with $n^{1/\alpha }$, where $\alpha$ is .................

Imagine you are observing a tiny particle of mass $m$ gracefully orbiting a central point. But there is a fascinating twist here. This particle is not bound by the familiar gravitational pull of a planet, nor is it held by the standard electrostatic attraction of a nucleus.

Instead, it is trapped in a unique, hypothetical central potential field described by the equation $U(r)=U_0{r}^4$. Our mission is to uncover how the radius of its allowed quantum orbits depends on the principal quantum number $n$.

To understand the dynamics of this particle, we must first discover the invisible force guiding its path. In physics, potential energy is like a landscape of hills and valleys, and force is the natural tendency to roll downhill. Mathematically, for a conservative field, the force is simply the negative spatial gradient of the potential energy.

Let's perform a quick differentiation on our given potential function. Taking the derivative of $U_0{r}^4$ with respect to $r$ yields $4U_0{r}^3$.

The negative sign is a beautiful indicator that the force is attractive, constantly pulling the particle towards the center of its orbit.

Since our particle is maintaining a perfect circular orbit, this attractive central force must be the very thing providing the necessary centripetal force. We can set up a dynamic equilibrium by equating the magnitude of our central force to the standard centripetal force formula, $\frac{m{v}^2}{r}$.

This equation is our golden key to unlocking the relationship between the particle's velocity and its orbital radius. By rearranging the terms, we can isolate the velocity squared.

Since $\frac{4U_0}{m}$ is just a constant, we can strip away the equals sign and reveal the pure proportionality: $v^2\propto r^4$. Taking the square root of both sides gives us a beautifully simple relationship.

The Quantum Leap

Now, we bring in the genius of Niels Bohr. Even though Bohr originally designed his model for the hydrogen atom, his core postulate about the quantization of angular momentum is a universal tool. He stated that the angular momentum of an orbiting particle must be an integral multiple of $\frac{h}{2\pi }$.

In this equation, $n$ is the principal quantum number, an integer that dictates the specific, allowed orbits. Because $m$, $h$, and $2\pi$ are all constants, we can distill this profound quantum condition into a simple proportionality.

Finding Alpha

We are now at the final crossroads. We have two powerful proportionalities: $v\propto r^2$ from our classical mechanics analysis, and $vr\propto n$ from our quantum mechanics postulate. Let's merge them by substituting our velocity relationship into the quantum condition.

This immediately simplifies to a cubic relationship.

To find how the radius $r$ scales with $n$, we simply take the cube root of both sides.

The problem states that the radius of the $n^{\text{th}}$ orbital varies as $n^{1/\alpha }$. By comparing this given form with our derived result, the answer reveals itself with absolute clarity.

And there we have it! By seamlessly blending classical mechanics with quantum postulates, we've solved the mystery of the non-Coulombic potential.