Suppose an electron is attracted towards the origin by a force k/r, where k is a constant and r is t

Q: Why do we equate the attractive force to $mv^2/r$?

In circular motion, there must be a net inward force acting as the centripetal force. Here, the given attractive force $k/r$ is the only force acting on the electron, so it must provide the required $mv^2/r$.

Q: Why is the kinetic energy independent of $n$?

When we equate $k/r$ to $mv^2/r$, the radius $r$ cancels out completely! This leaves $mv^2 = k$, meaning the kinetic energy ($mv^2/2$) is just a constant $k/2$, regardless of which orbit the electron is in.

Q: Can we use the standard Bohr radius formula $r_n \propto n^2$ here?

No! The standard formula $r_n \propto n^2$ is derived specifically for the Coulomb force, which is proportional to $1/r^2$. Since the force here is $1/r$, the entire derivation changes, leading to a different result.

Q: What does Bohr's quantization condition actually mean physically?

It means that an electron cannot just orbit at any random distance. It can only exist in specific "allowed" orbits where its angular momentum is an exact multiple of $h/2\pi$. It's like walking on stairs instead of a ramp.

Q: If the force was $F = kr$, how would the radius depend on $n$?

If $F = kr$, equating it to $mv^2/r$ gives $mv^2 = kr^2$. Using $v \propto 1/r$, we'd get $1/r^2 \propto r^2$, meaning $r^4 \propto n^2$, so $r \propto \sqrt{n}$. This is the classic harmonic oscillator case!

Question

Suppose an electron is attracted towards the origin by a force $k/r$, where $k$ is a constant and $r$ is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the $n$th orbital of the electron is found to be $r_n$ and the kinetic energy of the electron to be $T_n$. Then, which of the following is true?

Options:

(A) $T_n \propto \frac{1}{n^2}, r_n \propto n^2$

(B) $T_n$ is independent of $n, r_n \propto n$

(C) $T_n \propto \frac{1}{n}, r_n \propto n$

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

Sigma Percentile · Accepted Answer

### The Non-Standard Bohr Atom: A Thought Experiment

Imagine you are a physicist in the early 20th century. Niels Bohr has just proposed his revolutionary model of the hydrogen atom, where an electron orbits a proton under the influence of the Coulomb force, which follows an inverse-square law ($F \propto 1/r^2$).

But what if the universe played by different rules? What if the attractive force between the nucleus and the electron wasn't an inverse-square force, but instead varied simply as $1/r$? This is exactly the thought experiment we are diving into today. It's a beautiful exercise that tests whether you truly understand the **principles** of Bohr's model, rather than just memorizing its final formulas.

### Phase 1: The Force Balance and a Surprising Cancellation

For any object to move in a perfect circle, there must be a centripetal force pulling it towards the center. In our hypothetical atom, this role is played by the given attractive force, $F = \frac{k}{r_n}$.

Let's set up the fundamental equation of circular motion. We equate the required centripetal force to our attractive force:

$$ \frac{mv^2}{r_n} = \frac{k}{r_n} $$

Now, look closely at this equation. Something magical happens. The radius $r_n$ appears in the denominator on both sides. When we simplify, it completely cancels out!

$$ mv^2 = k $$

This is a profound result. It tells us that the quantity $mv^2$ is a constant, completely independent of the size of the orbit.

### Phase 2: The Secret of the Kinetic Energy

We are asked to find how the kinetic energy, $T_n$, depends on the principal quantum number $n$. We know the formula for kinetic energy:

$$ T_n = \frac{1}{2}mv^2 $$

Since we just discovered that $mv^2 = k$, we can substitute this directly into our kinetic energy equation:

$$ T_n = \frac{k}{2} $$

Because $k$ is a given constant, $T_n$ is also a constant. It does not contain the variable $n$. Therefore, **the kinetic energy is completely independent of the orbit number $n$**. This immediately narrows down our options!

### Phase 3: Bringing in Bohr's Masterstroke

To find the radius, we need the second pillar of Bohr's theory: the quantization of angular momentum. Bohr postulated that an electron can only exist in orbits where its angular momentum ($mvr_n$) is an integer multiple of $\frac{h}{2\pi}$.

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

From this, we can isolate the velocity $v$:

$$ v = \frac{nh}{2\pi m r_n} $$

This equation links the velocity, the radius, and the quantum number $n$.

### Phase 4: Solving for the Radius

Now, let's take this expression for velocity and substitute it back into our earlier finding, $mv^2 = k$.

$$ m \left( \frac{nh}{2\pi m r_n} \right)^2 = k $$

Let's carefully expand the square. Don't rush this step; it's where silly mistakes happen!

$$ m \frac{n^2 h^2}{4\pi^2 m^2 r_n^2} = k $$

One mass term ($m$) cancels out from the numerator and denominator:

$$ \frac{n^2 h^2}{4\pi^2 m r_n^2} = k $$

Our goal is to find how $r_n$ depends on $n$. Let's rearrange the equation to isolate $r_n^2$:

$$ r_n^2 = \left( \frac{h^2}{4\pi^2 m k} \right) n^2 $$

The term in the parentheses is entirely made of constants ($h, \pi, m, k$). Therefore, we can clearly see the proportionality:

$$ r_n^2 \propto n^2 $$

Taking the square root of both sides, we arrive at our final, elegant conclusion:

$$ r_n \propto n $$

### The Final Verdict

By applying the core principles of circular motion and angular momentum quantization to a non-standard force, we discovered two key facts:
1. The kinetic energy $T_n$ is independent of $n$.
2. The orbital radius $r_n$ is directly proportional to $n$.

This perfectly matches option (b). This problem is a fantastic reminder that in physics, understanding the *process* is infinitely more powerful than memorizing the *result*.

Select Answer:

Visualized Solution

The Physical Setup

Centripetal Force Condition

Kinetic Energy Analysis

Independence of Kinetic Energy

Bohr's Quantization Postulate

Expressing Velocity

Substituting Velocity

Solving for Radius

Radius Dependence on $n$

Final Conclusion

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

The Non-Standard Bohr Atom

Phase 1

Phase 2

Phase 3

Phase 4

The Final Verdict

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