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) = kr2/2, 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 F are the speed and total energy of the particle, respectively, then which of the following expression(s) is(are) correct? (A) r2 = nℏ√ 1/mk (B) v2 = nℏ√k/m3 (C) L/mr2 = √k/m (D) E = nℏ/ 2 √k/m

 The Quantum Harmonic Oscillator: Bohr’s Quantization Meets Central Forces

---

Rewritten Question

A particle of mass $m$ orbits under the central force $F(r)=-kr$ (potential $V(r)=\frac{1}{2}k{r}^2$) and follows Bohr’s quantization rule $L=n\mathrm{\hslash }$. Which of the following are correct?
(A) $r^2=n\mathrm{\hslash }\sqrt{\frac{1}{mk}}$
(B) $v^2=n\mathrm{\hslash }\sqrt{\frac{k}{m^3}}$
(C) $\frac{L}{m{r}^2}=\sqrt{\frac{k}{m}}$
(D) $E=\frac{n\mathrm{\hslash }}{2}\sqrt{\frac{k}{m}}$

---

Historical Context

Niels Bohr’s 1913 model quantized angular momentum ($L=n\mathrm{\hslash }$) to explain atomic spectra. While originally for Coulomb forces, this problem adapts it to a harmonic oscillator force ($F=-kr$), akin to springs or molecular vibrations. Such systems bridge classical mechanics and quantum theory.

---

Key Physics and Equations

1. Force Balance:
   Centripetal force = Central force:

   $$\begin{array}{cccc}& \frac{m{v}^2}{r}=kr⟹v^2=\frac{k{r}^2}{m}\mathrm{.}& & \text{(1)}\end{array}$$

2. Bohr Quantization:
   Angular momentum

---

Step-by-Step Analysis

Option (A): $r^2=n\mathrm{\hslash }\sqrt{\frac{1}{mk}}$

Substitute $v$ from (2) into (1):

$${\left(\frac{n\mathrm{\hslash }}{mr}\right)}^2=\frac{k{r}^2}{m}⟹\frac{n^2{\mathrm{\hslash }}^2}{m^2{r}^2}=\frac{k{r}^2}{m}⟹r^4=\frac{n^2{\mathrm{\hslash }}^2}{mk}⟹r^2=\frac{n\mathrm{\hslash }}{\sqrt{mk}}\mathrm{.}$$

Conclusion: Correct ✅.

---

Option (B): $v^2=n\mathrm{\hslash }\sqrt{\frac{k}{m^3}}$

From $r^2=\frac{n\mathrm{\hslash }}{\sqrt{mk}}$, substitute into $v^2=\frac{k{r}^2}{m}$:

$$v^2=\frac{k}{m}\cdot \frac{n\mathrm{\hslash }}{\sqrt{mk}}=\frac{n\mathrm{\hslash }k}{m\sqrt{mk}}=n\mathrm{\hslash }\sqrt{\frac{k}{m^3}}\mathrm{.}$$

Conclusion: Correct ✅.

---

Option (C): $\frac{L}{m{r}^2}=\sqrt{\frac{k}{m}}$

From $L=n\mathrm{\hslash }$ and $r^2=\frac{n\mathrm{\hslash }}{\sqrt{mk}}$:

$$\frac{L}{m{r}^2}=\frac{n\mathrm{\hslash }}{m\cdot \frac{n\mathrm{\hslash }}{\sqrt{mk}}}=\frac{\sqrt{mk}}{m}=\sqrt{\frac{k}{m}}\mathrm{.}$$

Conclusion: Correct ✅.

---

Option (D): $E=\frac{n\mathrm{\hslash }}{2}\sqrt{\frac{k}{m}}$

Total energy $E=\text{KE}+\text{PE}$:

$$E=\frac{1}{2}m{v}^2+\frac{1}{2}k{r}^2\mathrm{.}$$

From $m{v}^2=k{r}^2$ (see Equation 1):

$$E=\frac{1}{2}k{r}^2+\frac{1}{2}k{r}^2=k{r}^2\mathrm{.}$$

Substitute $r^2=\frac{n\mathrm{\hslash }}{\sqrt{mk}}$:

$$E=k\cdot \frac{n\mathrm{\hslash }}{\sqrt{mk}}=n\mathrm{\hslash }\sqrt{\frac{k}{m}}\mathrm{.}$$

The given option $\frac{n\mathrm{\hslash }}{2}\sqrt{\frac{k}{m}}$ is half the correct value.
Conclusion: Incorrect ❌.

---

Final Answer

Correct Options: (A), (B), (C)

$$\boxed{{\displaystyle A}},\boxed{{\displaystyle B}},\boxed{{\displaystyle C}}$$

Note: This problem illustrates how Bohr’s quantization elegantly adapts to non-Coulomb forces, revealing universal principles in quantum systems.