Animated Solution for Physics - Atoms and Nuclei: A particle of mass $m$ moves in a circular orbit in a central potential field $U(r)=\frac{1}{2}k{r}^2$. If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number $n$ as

Bohr's Model in a Harmonic Potential

When we first learn about Bohr's model of the atom, it is almost exclusively in the context of the hydrogen atom, where the electron is bound by the Coulombic electrostatic force. But the true beauty of Bohr's postulates lies in their versatility. What happens if we change the rules of the game? What if the binding force isn't Coulombic, but something entirely different?

In this fascinating problem, we are presented with a particle of mass $m$ moving in a circular orbit, but the central potential field is given by $U(r)=\frac{1}{2}k{r}^2$. This is the potential of a 3D isotropic harmonic oscillator. Let's embark on a journey to see how Bohr's quantization conditions adapt to this new environment.

From Potential to Force

Before we can talk about orbits and velocities, we need to understand the force acting on the particle. In classical mechanics, a conservative force is the negative spatial gradient of the potential energy.

Mathematically, this is expressed as:

Substituting our given potential $U(r)=\frac{1}{2}k{r}^2$ into this equation, we perform a simple differentiation:

The negative sign simply indicates that the force is attractive, pulling the particle towards the center. The magnitude of this central force is $|F|=kr$. Notice how this differs from the Coulomb force, which falls off as $1/r^2$. Here, the force actually increases linearly with distance, much like a spring!

Balancing Forces

For the particle to maintain a stable circular orbit of radius $r$, this central attractive force must provide the exact centripetal force required for circular motion.

We equate the centripetal force to our derived central force:

By multiplying both sides by $r$, we obtain a highly useful intermediate relation:

This equation is a goldmine. Not only does it relate velocity to radius, but it also gives us a direct shortcut to the kinetic energy, since $KE=\frac{1}{2}m{v}^2$.

Applying Bohr's Condition

Now we introduce the quantum realm. Bohr's second postulate states that the angular momentum of the particle in a stable orbit must be quantized in integral multiples of $h/2\pi$.

For the $n^{\text{th}}$ orbit, this is written as:

We need to find how the radius $r_n$ depends on the principal quantum number $n$. To do this, we isolate the velocity $v_n$ from Bohr's condition:

Now, we substitute this quantum expression for velocity back into our classical force balance equation ($m{v}_n^2=k{r}_n^2$):

Expanding the square, we get:

To find the proportionality, we group all the constants ($h,\pi ,m,k$) and isolate $r_n$. Cross-multiplying gives us:

This tells us that $r_n^4\propto n^2$. Taking the square root of both sides twice, we arrive at our first major conclusion:

Unveiling the Energy

Finally, we need to determine how the total energy $E_n$ scales with $n$. The total energy is the sum of the potential and kinetic energies:

Remember our golden relation from earlier? We found that $m{v}_n^2=k{r}_n^2$. Let's substitute this directly into the kinetic energy term. This is a brilliant shortcut that saves us from messy algebra!

Adding them up, the total energy simplifies beautifully to:

We already established that $r_n\propto \sqrt{n}$, which means $r_n^2\propto n$. Substituting this proportionality into our energy equation yields our final result:

By stepping outside the familiar territory of the hydrogen atom and applying fundamental principles, we've successfully deduced that for a harmonic potential, the orbital radius scales as $\sqrt{n}$ and the total energy scales linearly with $n$.