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

To solve the problem, we analyze the motion of a particle of mass \(m\) moving in a circular orbit under a central force \(f(r) = -kr\), with potential energy \(V(r) = \frac{kr^2}{2}\). Using Bohr's quantization rule \(L = n\hbar\), we find expressions for \(r^2\), \(v^2\), \(\frac{L}{mr^2}\), and \(E\).

1. Relation Between Force and Circular Motion:
For circular motion of radius \(r\), the centripetal force is provided by the restoring force:
\[ \frac{mv^2}{r} = kr \] which gives:
\[ v^2 = \frac{kr^2}{m} \]

2. Angular Momentum Quantization:
The angular momentum \(L\) is:
\[ L = mvr = n\hbar \] Using \(v = \frac{L}{mr}\), substitute \(v\) in the centripetal force equation:
\[ \frac{m}{r} \left(\frac{L}{mr}\right)^2 = kr \implies \frac{L^2}{m r^3} = k r \implies L^2 = m k r^4 \] Rearranging:
\[ r^4 = \frac{L^2}{m k} \] Substitute \(L = n \hbar\):
\[ r^4 = \frac{n^2 \hbar^2}{m k} \] Taking square root:
\[ r^2 = n \hbar \sqrt{\frac{1}{m k}} \] This confirms the first expression is correct.

3. Expression for \(v^2\):
Using \(v = \frac{L}{mr}\) and the expression for \(r^2\):
\[ v^2 = \left(\frac{L}{mr}\right)^2 = \frac{L^2}{m^2 r^2} \] Substitute \(L = n \hbar\) and \(r^2 = n \hbar \sqrt{\frac{1}{m k}}\):
\[ v^2 = \frac{n^2 \hbar^2}{m^2 \cdot n \hbar \sqrt{\frac{1}{m k}}} = \frac{n \hbar}{m^2} \sqrt{m k} = n \hbar \sqrt{\frac{k}{m^3}} \] This confirms the second expression is correct.

4. Expression for \(\frac{L}{m r^2}\):
\[ \frac{L}{m r^2} = \frac{n \hbar}{m r^2} \] Substitute \(r^2 = n \hbar \sqrt{\frac{1}{m k}}\):
\[ \frac{L}{m r^2} = \frac{n \hbar}{m \cdot n \hbar \sqrt{\frac{1}{m k}}} = \frac{1}{m \sqrt{\frac{1}{m k}}} = \sqrt{\frac{k}{m}} \] This confirms the third expression is correct.

5. Expression for Total Energy \(E\):
The total energy is kinetic plus potential energy:
\[ E = \frac{1}{2} m v^2 + \frac{1}{2} k r^2 \] Using \(v^2 = \frac{k r^2}{m}\), we have:
\[ E = \frac{1}{2} m \frac{k r^2}{m} + \frac{1}{2} k r^2 = k r^2 \] Substitute \(r^2 = n \hbar \sqrt{\frac{1}{m k}}\):
\[ E = k \cdot n \hbar \sqrt{\frac{1}{m k}} = n \hbar \sqrt{\frac{k}{m}} \] The expression given is \(\frac{n h}{2} \sqrt{\frac{k}{m}}\). Note that \(\hbar = \frac{h}{2\pi}\), so:
\[ E = n \hbar \sqrt{\frac{k}{m}} = n \frac{h}{2\pi} \sqrt{\frac{k}{m}} \neq \frac{n h}{2} \sqrt{\frac{k}{m}} \] So the fourth expression is not correct.

Final Answer:
The correct expressions are:
\[ r^2 = n \hbar \sqrt{\frac{1}{m k}}, \quad v^2 = n \hbar \sqrt{\frac{k}{m^3}}, \quad \frac{L}{m r^2} = \sqrt{\frac{k}{m}} \] The expression for energy \(E = \frac{n h}{2} \sqrt{\frac{k}{m}}\) is incorrect.

Answer 2

The Bohr quantization rule states that the angular momentum is quantized: \(L = n\hbar = n\frac{h}{2\pi}\)

Since \(L = mvr\), we have \(mvr = n\hbar\)

The force is \(F = -kr\), which means the centripetal force is \(kr = \frac{mv^2}{r}\), so \(kr^2 = mv^2\).

Then \(v = \sqrt{\frac{k}{m}} r\)

Substituting into the angular momentum equation, \(mr \sqrt{\frac{k}{m}} r = n\hbar\)

\(r^2 \sqrt{mk} = n\hbar\)

\(r^2 = n\hbar \sqrt{\frac{1}{mk}}\), which is answer (A).

\(v = \frac{n\hbar}{mr} = \frac{n\hbar}{m \sqrt{n\hbar} (mk)^{-1/4}} = \frac{n\hbar}{m(n\hbar)^{1/2}m^{-1/4}k^{-1/4}} = \sqrt{\frac{n\hbar \sqrt{k}}{m^{3/2}}} = \sqrt{n \hbar}\sqrt{\frac{\sqrt{k}}{\sqrt{m^3}}} = \sqrt{n \hbar \sqrt{\frac{k}{m^3}}}\) \(v^2 = n \hbar \sqrt{\frac{k}{m^3}}\), which is answer (B)

\(\frac{L}{mr^2} = \frac{n \hbar}{m (n\hbar \sqrt{\frac{1}{mk}})} = \frac{\hbar}{\sqrt{\frac{n^2\hbar^2}{mk}}m} = \sqrt{\frac{mk}{n^2 \hbar^2 m^2}} = \sqrt{\frac{k}{m}} \frac{1}{m \sqrt{r^2}}=\sqrt{\frac{k}{m}}\) \(\frac{L}{mr^2} = \sqrt{\frac{k}{m}}\) , which is answer (C)

The total energy is \(E = T + V = \frac{1}{2}mv^2 + \frac{1}{2}kr^2 = \frac{1}{2}(kr^2) + \frac{1}{2}(kr^2) = kr^2\)

Since \(r^2 = n\hbar \sqrt{\frac{1}{mk}}\), we have \(E = k(n\hbar \sqrt{\frac{1}{mk}}) = n\hbar \sqrt{\frac{k^2}{mk}} = n\hbar \sqrt{\frac{k}{m}}\), which doesn’t correspond to the options given in (D).

Answer: (A, B, C)