A particle of mass m moves in circular orbits with potential energy V(r) = Fr, where F is a positive

Q: Why do we use $F = -dV/dr$?

Because the conservative force is the negative gradient of the potential energy. This force provides the necessary centripetal acceleration for circular motion.

Q: Can we apply Bohr's model to any potential?

Bohr's quantization of angular momentum ($L = nh/2\pi$) is a postulate that can be applied to circular orbits in any central force field, not just the Coulomb force.

Q: Why is the kinetic energy not $-E$ like in the hydrogen atom?

The relation $K = -E$ (or $K = -U/2$) is specific to inverse-square forces (where $V \propto -1/r$). For a general potential $V \propto r^k$, the Virial theorem states $2K = k V$. Here $k=1$, so $2K = V$, meaning $K = V/2$.

Q: How did we get $E = \frac{3}{2}FR$?

Total energy $E = K + V$. Since $K = \frac{1}{2}mv^2$ and the centripetal force $mv^2/R = F$, we have $K = \frac{1}{2}FR$. Given $V = FR$, $E = \frac{1}{2}FR + FR = \frac{3}{2}FR$.

Q: What if the potential was $V(r) = kr^2$?

Then the force would be $2kr$, leading to a harmonic oscillator potential. The dependencies of $R$ and $E$ on $n$ would change accordingly based on the same two fundamental equations.

Question

A particle of mass $m$ moves in circular orbits with potential energy $V(r) = Fr$, where $F$ is a positive constant and $r$ is its distance from the origin. Its energies are calculated using the Bohr model. If the radius of the particle's orbit is denoted by $R$ and its speed and energy are denoted by $v$ and $E$, respectively, then for the $n^{	ext{th}}$ orbit (here $h$ is the Planck's constant)-

Options:

(A) $R \propto n^{1/3}$ and $v \propto n^{2/3}$

(B) $R \propto n^{2/3}$ and $v \propto n^{1/3}$

(C) $E = \frac{3}{2} \left( \frac{n^2 h^2 F^2}{4\pi^2 m} ight)^{1/3}$

(D) $E = 2 \left( \frac{n^2 h^2 F^2}{4\pi^2 m} ight)^{1/3}$

Sigma Percentile · Accepted Answer

### The Setup: A New Kind of Atom

Imagine you are standing at the center of a microscopic system, watching a particle of mass $m$ orbit around you. But this isn't your standard hydrogen atom where the force is governed by Coulomb's law. Instead, we are handed a completely different rulebook: the potential energy is given by $V(r) = Fr$, where $F$ is a positive constant.

This simple linear relationship changes everything. Our goal is to figure out how the radius, velocity, and total energy of this particle depend on its quantum state, $n$. I know this might look intimidating because it breaks away from the familiar $1/r$ potential, but let's take a breath. The beauty of physics is that the fundamental laws remain exactly the same.

### The Master Equations: Dynamics and Quantization

First, we need to understand the physical force driving this circular motion. We know that any conservative force is the negative gradient of its potential energy. By differentiating our potential, $F_{	ext{rad}} = -\frac{dV}{dr} = -F$, we find that a constant attractive force of magnitude $F$ is pulling the particle inward.

For the particle to maintain a stable circular orbit, this inward pull must perfectly match the required centripetal force. This gives us our first master equation: 
$$F = \frac{mv^2}{R}$$

Now, the problem explicitly instructs us to use Bohr's model. Bohr's genius wasn't just about hydrogen; his postulate that angular momentum is quantized applies universally to central forces. So, we write down our second master equation:
$$mvR = \frac{nh}{2\pi}$$

### Solving for the Orbit's Radius and Velocity

We now have a system of two equations with two unknowns ($v$ and $R$). Let's isolate the velocity from the quantization condition: $v = \frac{nh}{2\pi m R}$.

Next, we substitute this expression for $v$ directly into our force equation. This strategic move eliminates velocity entirely, allowing us to solve for the radius:
$$F = \frac{m}{R} \left( \frac{nh}{2\pi m R} ight)^2 = \frac{n^2 h^2}{4\pi^2 m R^3}$$

Rearranging this to solve for $R^3$, we get $R^3 = \frac{n^2 h^2}{4\pi^2 m F}$. Taking the cube root reveals the dependency:
$$R = \left( \frac{n^2 h^2}{4\pi^2 m F} ight)^{1/3}$$
This clearly shows that **$R \propto n^{2/3}$**.

What about the velocity? Since $v \propto \frac{n}{R}$, we can substitute our new proportionality for $R$ to find $v \propto \frac{n}{n^{2/3}} = n^{1/3}$. This confirms that **option (B) is correct**.

### The Energy Landscape

Finally, let's map out the energy of this system. The total energy $E$ is the sum of kinetic ($K$) and potential ($V$) energies.

From our very first equation, we know that $mv^2 = FR$. Therefore, the kinetic energy is simply $K = \frac{1}{2}mv^2 = \frac{1}{2}FR$.

Adding the given potential energy $V = FR$, the total energy becomes:
$$E = \frac{1}{2}FR + FR = \frac{3}{2}FR$$

To get the final expression, we substitute the exact value of $R$ we derived earlier:
$$E = \frac{3}{2} F \left( \frac{n^2 h^2}{4\pi^2 m F} ight)^{1/3}$$

By bringing the $F$ inside the cube root (where it becomes $F^3$), we arrive at our elegant final answer:
$$E = \frac{3}{2} \left( \frac{n^2 h^2 F^2}{4\pi^2 m} ight)^{1/3}$$
This perfectly matches **option (C)**.

### Conclusion

This problem is a phenomenal exercise in generalizing quantum principles. By anchoring ourselves to the fundamental definitions of force and Bohr's quantization, we successfully navigated a non-standard potential. Always remember: when the potential changes, the physics doesn't—only the algebra does!

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Visualized Solution

Visualizing the Potential $V (r) = F r$

Force from Potential $F_{rad} = - \frac{d V}{d r}$

Calculating the Force $F_{rad} = - F$

Centripetal Force $F = \frac{m v ^{2}}{R}$

Bohr's Quantization $m v R = \frac{nh}{2 π}$

Velocity in terms of $R$

Substituting Velocity

Solving for Radius $R$

Radius Dependence $R \propto n^{2/3}$

Velocity Dependence $v \propto n^{1/3}$

Kinetic Energy $K = \frac{1}{2} F R$

Total Energy $E = \frac{3}{2} F R$

Final Energy Expression

The Way Forward: Virial Theorem

The Sigma Insight: Bohr's Atomic Model and Energy Levels

The Setup

The Master Equations

Solving for the Orbit's Radius and Velocity

The Energy Landscape

Conclusion

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Visualized Solution

Visualizing the Potential .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V(r)=Fr

Force from Potential .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Frad​=−drdV​

Calculating the Force .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Frad​=−F

Centripetal Force .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F=Rmv2​

Bohr's Quantization .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } mvR=2πnh​

Velocity in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R