A particle of mass m is moving in a circular orbit under the influence of the central force F(r) = -

Q: Why did we ignore the negative sign in the force $F(r) = -kr$?

The negative sign simply indicates that the force is attractive (directed towards the center). When equating it to the centripetal force $mv^2/r$, we only use the magnitude $kr$ because the centripetal force itself is already defined as being directed towards the center.

Q: How did we know the potential energy is $kr^2/2$?

The problem explicitly states $V(r) = kr^2/2$. However, even if it wasn't given, you can find it by integrating the force: $V(r) = -\int F(r) dr = -\int (-kr) dr = kr^2/2$.

Q: Why is the kinetic energy equal to the potential energy here?

This is a specific consequence of the force being proportional to $r$. From $mv^2/r = kr$, we get $mv^2 = kr^2$. Since kinetic energy is $mv^2/2$, it becomes $kr^2/2$, which happens to be exactly the same expression as the potential energy for this specific force field.

Q: Can we use the standard Bohr model formulas for radius and velocity here?

No! The standard Bohr model assumes an electrostatic force which is proportional to $1/r^2$. Here, the force is proportional to $r$. You must derive the expressions from scratch using the given force law.

Q: What does the $n$ represent in these equations?

The integer $n$ is the principal quantum number. It dictates the allowed, discrete orbits the particle can occupy. As $n$ increases, the particle moves to higher, more energetic orbits further from the center.

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 $L = n\hbar$, where $\hbar = \frac{h}{2\pi}$, $h$ 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?

Options:

(A) $r^2 = n\hbar \sqrt{\frac{1}{mk}}$

(B) $v^2 = n\hbar \sqrt{\frac{k}{m^3}}$

(C) $\frac{L}{mr^2} = \sqrt{\frac{k}{m}}$

(D) $E = \frac{n\hbar}{2} \sqrt{\frac{k}{m}}$

Sigma Percentile · Accepted Answer

Imagine a tiny particle of mass $m$ caught in a cosmic dance, whirling in a perfect circular orbit. But what keeps it tethered? It's a central force, pulling it inward, described by $F(r) = -kr$. The negative sign simply means it's attractive, directed towards the origin.

### The Classical Foundation: Balancing the Forces

For any object to maintain a circular path, it requires a centripetal force. In our scenario, this role is played entirely by the central force. By equating the required centripetal force to the magnitude of our central force, we establish our first crucial relationship:

$$ \frac{mv^2}{r} = kr $$

This elegant equation is the bridge between the particle's speed $v$ and its orbital radius $r$. With a quick rearrangement, we can express the velocity explicitly:

$$ v^2 = \frac{k}{m}r^2 \implies v = r\sqrt{\frac{k}{m}} $$

This tells us that the further the particle is from the center, the faster it must travel to avoid spiraling inward.

### The Quantum Leap: Bohr's Quantization

Now, we inject a dose of quantum mechanics into our classical model. According to Bohr's quantization rule, the angular momentum $L$ of the particle isn't just any random value; it's restricted to integer multiples of the reduced Planck's constant, $\hbar$.

$$ L = mvr = n\hbar $$

This is where the magic happens. We now have a system of equations linking the classical mechanics of circular motion with the quantum rules of angular momentum.

### Decoding the Radius and Velocity

Let's put our equations to work and test the given options. We'll start by finding an expression for $r^2$. By substituting our velocity expression into the angular momentum equation, we get:

$$ m \left( r\sqrt{\frac{k}{m}} \right) r = n\hbar $$

$$ r^2 \sqrt{mk} = n\hbar $$

$$ r^2 = n\hbar \sqrt{\frac{1}{mk}} $$

This perfectly matches option (A)! Now, what about the velocity squared? We already know that $v^2 = \frac{k}{m}r^2$. Substituting our newly found $r^2$ into this relation yields:

$$ v^2 = \frac{k}{m} \left( n\hbar \sqrt{\frac{1}{mk}} \right) = n\hbar \sqrt{\frac{k}{m^3}} $$

And just like that, option (B) is also proven correct.

What about the ratio $\frac{L}{mr^2}$? Since $L = mvr$, this ratio simplifies beautifully:

$$ \frac{L}{mr^2} = \frac{mvr}{mr^2} = \frac{v}{r} $$

From our very first step, we know that $\frac{v}{r} = \sqrt{\frac{k}{m}}$. Thus, option (C) stands true as well.

### The Energy Finale

Finally, let's evaluate the total energy $E$ of the system. The total energy is the sum of kinetic ($K$) and potential ($V$) energies.

$$ E = K + V = \frac{1}{2}mv^2 + \frac{1}{2}kr^2 $$

Recall our force balance equation: $mv^2 = kr^2$. This means the kinetic energy is exactly equal to the potential energy!

$$ E = \frac{1}{2}kr^2 + \frac{1}{2}kr^2 = kr^2 $$

Now, we substitute our expression for $r^2$:

$$ E = k \left( n\hbar \sqrt{\frac{1}{mk}} \right) = n\hbar \sqrt{\frac{k^2}{mk}} = n\hbar \sqrt{\frac{k}{m}} $$

Looking at option (D), it claims $E = \frac{n\hbar}{2} \sqrt{\frac{k}{m}}$. It has an extra factor of $\frac{1}{2}$, making it the only incorrect option in the bunch.

And there you have it! By seamlessly blending classical mechanics with quantum rules, we've successfully navigated through the properties of this fascinating orbital system.

Select Answer:

Visualized Solution

$F_{centripetal} = F_{central}$

$v = r \frac{k}{m}$

$L = n ℏ$

$Finding r^{2}$

$Finding v^{2}$

$Finding \frac{L}{m r ^{2}}$

$Finding Total Energy E$

The Sigma Insight: Bohr's Atomic Model and Energy Levels

The Classical Foundation

The Quantum Leap

Decoding the Radius and Velocity

The Energy Finale

Similar Questions

A particle of mass $m$ moves in a circular orbit in a central potential field $U (r) = \frac{1}{2} k r^{2}$ . If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number $n$ as

A particle of mass $m$ moves in a circular orbit in a central potential field $U (r) = U_{0} r^{4}$ . If Bohr's quantisation conditions are applied, radii of possible orbitals $r_{n}$ vary with $n^{1/ α}$ , where $α$ is .................

A diatomic molecule is made of two masses $m_{1}$ and $m_{2}$ which are separated by a distance $r$ . If we calculate its rotational energy by applying Bohr's rule of angular momentum quantisation, its energy will be given by ( $n$ is an integer)

If an electron is moving in the $n$ th orbit of the hydrogen atom, then its velocity $v_{n}$ for the $n$ th orbit is given as

In the Bohr model of the hydrogen atoms

In the Bohr model of the hydrogen atom, the ratio of the kinetic energy to the total energy of the electron in a quantum state $n$ is ........... .

A diatomic molecule has moment of inertia $I$ . By Bohr's quantization condition its rotational energy in the $n$ th level ( $n = 0$ is not allowed) is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fcentripetal​=Fcentral​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=rmk​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L=nℏ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding r2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding v2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding mr2L​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Total Energy E