Question

A particle of mass m is moving in a circular orbit under the influence of the central force\( f(r) = −kr\), corresponding to the potential energy \(v(r) = \frac{kr^2}{2}\), where \(k\) is a positive force constant and \( r\) is the radial distance from the origin. According to the Bohr’s quantization rule, the angular momentum of the particle is given by \(L= nℏ\), where\( ℏ = ℎ/(2pi), ℎ\) is the Planck’s constant, and \(n\) a positive integer. If \(v\) and \(E\) are the speed and total energy of the particle, respectively, then which of the following expression(s) is(are) correct?

• A. \(r^2\)=\(nh\sqrt{\frac{1}{mk}}\)
• B. \(v^2\)=\(nh\sqrt{\frac{k}{m^3}}\)
• C. \(\frac{L}{mr^2}\)=\(\sqrt{\frac{k}{m}}\)
• D. \(E\)=\(\frac{nh}{2}\sqrt{\frac{k}{m}}\)

Answer 1

The Correct Option is A, B, C

Analysis of Potential Energy and Force 

1. Given Potential Energy

The given potential energy function is:

\[ V(r) = \frac{k r^2}{2}. \]

2. Force Calculation

The force is obtained as:

\[ F = -\frac{dV}{dr} = -kr. \]

3. Circular Motion Condition

For circular motion, the **centripetal force** is provided by the central force:

\[ \frac{m v^2}{r} = kr \Rightarrow m v^2 = k r^2. \]

(Equation 1)

4. Applying Bohr’s Quantization Rule

Using **Bohr’s quantization rule**, \( L = n\hbar \):

\[ L = m v r = n \hbar \Rightarrow v = \frac{n\hbar}{m r}. \]

(Equation 2)

5. Substituting \( v \) into Equation (1)

Substituting \( v \) from Equation (2) into Equation (1):

\[ m \cdot \frac{n\hbar}{m r} \cdot \frac{n\hbar}{m r} = k r^2. \]

Rearranging:

\[ \frac{n^2 \hbar^2}{m r^2} = k r^2. \]

Solving for \( r \):

\[ r^4 = \frac{n^2 \hbar^2}{m k} \Rightarrow r^2 = \frac{n\hbar}{\sqrt{m k}}. \]

**(A) is correct.** ✅

6. Velocity Calculation

From \( v = \frac{n\hbar}{m r} \), squaring both sides:

\[ v^2 = \frac{n\hbar}{m r} \cdot \frac{k r}{m} = \frac{n\hbar k}{m^3 r}. \]

**(B) is correct.** ✅

7. Angular Momentum Per Unit Mass Per Unit Radius

\[ \frac{L}{m r^2} = \frac{n\hbar}{m r^2}. \]

Using \( r^2 = \frac{n\hbar}{\sqrt{m k}} \):

\[ \frac{L}{m r^2} = \frac{\sqrt{k}}{\sqrt{m}}. \]

**(C) is correct.** ✅

8. Total Energy Calculation

The total energy is the sum of kinetic and potential energy:

\[ E = K + U = \frac{1}{2} m v^2 + \frac{1}{2} k r^2. \]

Substituting \( v^2 \) and \( r^2 \):

\[ E = \frac{1}{2} \cdot \frac{k r^2}{m} + \frac{1}{2} k r^2. \]

Simplifying:

\[ E = \frac{n\hbar}{\sqrt{m k}} k. \]

Since this does not match the expected total energy formula, **(D) is not correct.** ❌

Final Answer:

• (A) Correct ✅
• (B) Correct ✅
• (C) Correct ✅
• (D) Incorrect ❌