Question

A particle is moving in a circular orbit of radius $r$ under the influence of an attractive central force. Assume the Bohr quantization condition to hold. It is found that the frequency of revolution of the particle is independent of the quantum number $n$. Then, the force is proportional to

• A. $1/r^3$
• B. $1/r^2$
• C. $r$
• D. $r^2$

Answer 1

Answer: (C)

r

Answer 2

To solve this problem, we will use the principles of classical mechanics for circular motion and Bohr's quantization condition.

1. Centripetal Force: For a particle moving in a circular orbit of radius $r$ with speed $v$ under the influence of a central force $F$, the force provides the necessary centripetal acceleration: $F = \frac{mv^2}{r} \quad (1)$

2. Bohr Quantization Condition: According to Bohr's quantization condition, the angular momentum ($L$) of the particle is quantized: $L = mvr = n\hbar \quad (2)$ where $m$ is the mass of the particle, $n$ is the principal quantum number ($n=1, 2, 3, \ldots$), and $\hbar = \frac{h}{2\pi}$ (Planck's constant divided by $2\pi$). From equation (2), we can express the speed $v$: $v = \frac{n\hbar}{mr} \quad (3)$

3. Substitute $v$ into the Force Equation: Substitute the expression for $v$ from equation (3) into equation (1): $F = \frac{m}{r} \left(\frac{n\hbar}{mr}\right)^2$ $F = \frac{m}{r} \frac{n^2\hbar^2}{m^2r^2}$ $F = \frac{n^2\hbar^2}{mr^3} \quad (4)$

4. Frequency of Revolution: The frequency of revolution $f$ is given by the ratio of the speed to the circumference of the orbit: $f = \frac{v}{2\pi r} \quad (5)$

5. Substitute $v$ into the Frequency Equation: Substitute the expression for $v$ from equation (3) into equation (5): $f = \frac{1}{2\pi r} \left(\frac{n\hbar}{mr}\right)$ $f = \frac{n\hbar}{2\pi mr^2} \quad (6)$

6. Apply the Condition: $f$ is independent of $n$: We are given that the frequency of revolution $f$ is independent of the quantum number $n$. This means $f$ is a constant value, let's call it $f_0$. From equation (6): $f_0 = \frac{n\hbar}{2\pi mr^2}$ We can rearrange this to find the relationship between $n$ and $r$: $n = \frac{2\pi m f_0}{\hbar} r^2 \quad (7)$ Since $2\pi$, $m$, $f_0$, and $\hbar$ are constants, this implies that $n$ is proportional to $r^2$ (i.e., $n \propto r^2$).

7. Substitute the relationship of $n$ into the Force Equation: Substitute the expression for $n$ from equation (7) into equation (4): $F = \frac{1}{mr^3} \left(\frac{2\pi m f_0}{\hbar} r^2\right)^2 \hbar^2$ $F = \frac{1}{mr^3} \frac{(2\pi m f_0)^2}{\hbar^2} r^4 \hbar^2$ $F = \frac{4\pi^2 m^2 f_0^2}{m} r^{4-3}$ $F = (4\pi^2 m f_0^2) r$

Since $4\pi^2 m f_0^2$ is a collection of constants, we can denote it as a single constant $K$. $F = Kr$

Thus, the force is proportional to $r$.