The electric potential between a proton and an electron is given by V = V_0 ln(r/r_0), where r_0 is

Q: Why did the radius $r$ cancel out when finding the velocity?

This is a unique property of the logarithmic potential given in this specific problem. Because the force turns out to be inversely proportional to $r$ (i.e., $F \propto 1/r$), and the centripetal force is also proportional to $1/r$, the $r$ terms cancel out, making the velocity constant for all orbits!

Q: Can we always use Bohr's quantization condition $mvr = nh/2\pi$?

The problem explicitly states "Assuming Bohr's model to be applicable". Bohr's quantization of angular momentum is a fundamental postulate of his model, so we must use it whenever we are analyzing orbits within the framework of Bohr's theory, regardless of the specific force law.

Q: What if the potential was the standard Coulomb potential?

For a standard Coulomb potential, $V \propto 1/r$, the force is $F \propto 1/r^2$. Equating this to $mv^2/r$ gives $v \propto 1/\sqrt{r}$. Substituting this into Bohr's condition yields $r_n \propto n^2$, which is the familiar result for the hydrogen atom.

Q: Does the constant $r_0$ affect the final answer?

No, it doesn't! When we differentiate $\ln(r/r_0)$, which is $\ln(r) - \ln(r_0)$, the constant term $\ln(r_0)$ disappears. This means the force, and consequently the radius dependence on $n$, is completely independent of $r_0$.

Question

The electric potential between a proton and an electron is given by $V = V_0 \ln \frac{r}{r_0}$, where $r_0$ is a constant. Assuming Bohr's model to be applicable, write variation of $r_n$ with $n$. Here, $n$ is the principal quantum number.

Options:

(A) $r_n \propto n$

(B) $r_n \propto \frac{1}{n}$

(C) $r_n \propto n^2$

(D) $r_n \propto \frac{1}{n^2}$

Sigma Percentile · Accepted Answer

### A Non-Standard Potential

In standard physics problems, we are incredibly used to the Coulomb potential, where $V \propto \frac{1}{r}$. But what happens when the universe throws a curveball and gives us a logarithmic potential?

In this fascinating problem, the electric potential between a proton and an electron is given by $V = V_0 \ln \left(\frac{r}{r_0}\right)$. Our mission is to find out how the radius of the $n$-th orbit, $r_n$, depends on the principal quantum number $n$, assuming Bohr's model still holds true.

### Finding the Force

The first step in analyzing any orbit is to understand the force keeping the particle in that orbit. In electrostatics, the electric field is the negative gradient of the potential. Therefore, the magnitude of the electrostatic force $F$ on the electron is simply the charge $e$ multiplied by the magnitude of the potential gradient.

$$|F| = e \left| \frac{dV}{dr} \right|$$

Let's take the derivative of our given potential:

$$|F| = e \frac{d}{dr} \left[ V_0 \ln \left(\frac{r}{r_0}\right) \right]$$

Using the chain rule, the derivative of $\ln(r/r_0)$ is just $1/r$. So, the force simplifies beautifully to:

$$|F| = \frac{eV_0}{r}$$

### The Centripetal Connection

For the electron to maintain a stable circular orbit, this electrostatic force must provide the exact centripetal force required. We equate the two:

$$\frac{mv^2}{r} = \frac{eV_0}{r}$$

**Look closely at this equation!** The radius $r$ is present in the denominator on both sides. This means it cancels out completely!

$$mv^2 = eV_0$$

$$v = \sqrt{\frac{eV_0}{m}}$$

This is a profound result. It tells us that for this specific logarithmic potential, the orbital velocity of the electron is a constant, completely independent of which orbit it is in!

### Applying Bohr's Quantization

Now, we bring in the heavy machinery: Bohr's second postulate. Bohr stated that the angular momentum of an orbiting electron must be quantized in integral multiples of $\frac{h}{2\pi}$.

$$mvr_n = \frac{nh}{2\pi}$$

We already know that the velocity $v$ is a constant. Let's substitute our expression for $v$ into Bohr's condition:

$$m \left( \sqrt{\frac{eV_0}{m}} \right) r_n = \frac{nh}{2\pi}$$

Now, we simply rearrange the terms to isolate the radius of the $n$-th orbit, $r_n$:

$$r_n = \frac{nh}{2\pi \sqrt{meV_0}}$$

### The Final Verdict

Take a step back and look at our final expression for $r_n$. The mass $m$, the elementary charge $e$, the potential constant $V_0$, and Planck's constant $h$ are all fixed values. The only variable on the right side is the principal quantum number $n$.

Therefore, we can conclude that the radius of the $n$-th orbit is directly proportional to $n$.

**$$r_n \propto n$$**

This is a stark contrast to the standard hydrogen atom where $r_n \propto n^2$. It beautifully demonstrates how changing the fundamental force law alters the quantum mechanical structure of the atom!

Select Answer:

Visualized Solution

Visualizing the Orbit

Electrostatic Force

Centripetal Force

Orbital Velocity

Bohr's Quantization Condition

Radius of $n$ -th Orbit

Final Proportionality

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

A Non-Standard Potential

Finding the Force

The Centripetal Connection

Applying Bohr's Quantization

The Final Verdict

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