Question

A particle of mass $m$ moves along a circular orbit in a centro-symmetrical potential field $U(r)=U_0r^2$ where $U_0$ is a constant and $r$ is the distance of the particle from the centre of the potential field. Use Bohr quantization condition. $E_n$ and $r_n$ respectively are the energy and radius of the $n^{th}$ state. Then

• A. $r_n \propto n^2$
• B. $r_n \propto n^{\frac{1}{2}}$
• C. $E_n = \frac{3U_0r^2}{2}$
• D. $E_n = 2U_0r^2$

Answer 1

Answer: (B)

B

Answer 2

The particle of mass $m$ moves in a centro-symmetrical potential field $U(r) = U_0r^2$. The force acting on the particle is $F(r) = -\frac{dU}{dr} = -\frac{d}{dr}(U_0r^2) = -2U_0r$. The magnitude of the force is $F(r) = 2U_0r$. This force is attractive and acts towards the center, providing the centripetal force for the circular orbit. For a circular orbit of radius $r$, the centripetal force is $\frac{mv^2}{r}$, where $v$ is the speed of the particle. So, $\frac{mv^2}{r} = 2U_0r$. This gives $mv^2 = 2U_0r^2$.

The kinetic energy of the particle is $K = \frac{1}{2}mv^2 = \frac{1}{2}(2U_0r^2) = U_0r^2$. The potential energy is given as $U(r) = U_0r^2$. The total energy of the particle in a circular orbit of radius $r$ is $E = K + U = U_0r^2 + U_0r^2 = 2U_0r^2$. So, for the $n^{th}$ state with radius $r_n$ and energy $E_n$, the energy is related to the radius by $E_n = 2U_0r_n^2$.

Now, we apply the Bohr quantization condition for angular momentum, which states that the angular momentum $L$ is quantized: $L = mvr = n\frac{h}{2\pi}$, where $n$ is a positive integer ($n=1, 2, 3, \dots$) and $h$ is Planck's constant. From the force equation, $mv^2 = 2U_0r^2$, we get $v^2 = \frac{2U_0r^2}{m}$, so $v = r\sqrt{\frac{2U_0}{m}}$. Substitute this expression for $v$ into the Bohr quantization condition:

$m \left(r\sqrt{\frac{2U_0}{m}}\right) r = n\frac{h}{2\pi}$

$mr^2 \sqrt{\frac{2U_0}{m}} = n\frac{h}{2\pi}$

$r^2 \sqrt{m^2 \frac{2U_0}{m}} = n\frac{h}{2\pi}$

$r^2 \sqrt{2mU_0} = n\frac{h}{2\pi}$

Let $r_n$ be the radius of the $n^{th}$ state.

$r_n^2 = n \frac{h}{2\pi\sqrt{2mU_0}}$

$r_n = \left(\frac{h}{2\pi\sqrt{2mU_0}}\right)^{1/2} n^{1/2}$.

Since $\frac{h}{2\pi\sqrt{2mU_0}}$ is a constant, we have $r_n \propto n^{1/2}$.