A particle of mass m moves in circular orbits with potential energy `N(r )=Fr`, wjere F is a positive constant and r its distance from the origin. Its energies are calculated using the Bohr model. If the radius of the the `n^(th)` orbit (here h is the Planck's constant)

Options

A. `Ralphan^(1//3)` and `valphan^(2//3)`
B. `Ralphan^(2//3)` and `valphan^(1//3)`
C. `E=3/2((n^(2)h^(2)F^(2))/(4pi^(2)m))^(1//3)`
D. `E=2((n^(2)h^(2)F^(2))/(4pi^(2)m))^(1//3)`

Explanation

To solve the problem step by step, we will use the principles of the Bohr model and the given potential energy function.

### Step 1: Understanding the Potential Energy
The potential energy \( N(r) \) is given as \( N(r) = Fr \), where \( F \) is a positive constant and \( r \) is the distance from the origin.

### Step 2: Relating Force to Potential Energy
The force \( F \) associated with the potential energy can be found using the relation:
\[ F = -\frac{dN}{dr} \]
Calculating this gives:
\[ F = -\frac{d(Fr)}{dr} = -F \]
This indicates that the force acting on the particle is equal to \( F \) directed towards the center.

### Step 3: Centripetal Force
For a particle moving in a circular orbit, the centripetal force required to keep it in that orbit is given by:
\[ F_c = \frac{mv^2}{r} \]
Setting this equal to the force \( F \) we found:
\[ \frac{mv^2}{r} = F \]

### Step 4: Angular Momentum
According to the Bohr model, the angular momentum \( L \) of the particle in the \( n^{th} \) orbit is quantized:
\[ L = mvr = n\frac{h}{2\pi} \]
From this, we can express \( v \) in terms of \( r \):
\[ v = \frac{nh}{2\pi mr} \]

### Step 5: Substitute \( v \) into the Centripetal Force Equation
Substituting \( v \) into the centripetal force equation:
\[ \frac{m\left(\frac{nh}{2\pi mr}\right)^2}{r} = F \]
This simplifies to:
\[ \frac{n^2h^2}{4\pi^2mr^3} = F \]

### Step 6: Solve for \( r \)
Rearranging gives:
\[ r^3 = \frac{n^2h^2}{4\pi^2mF} \]
Taking the cube root:
\[ r = \left(\frac{n^2h^2}{4\pi^2mF}\right)^{1/3} \]

### Step 7: Kinetic and Potential Energy
The kinetic energy \( K \) of the particle is given by:
\[ K = \frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{nh}{2\pi mr}\right)^2 \]
Substituting for \( v^2 \) gives:
\[ K = \frac{n^2h^2}{8\pi^2r} \]

The potential energy \( U \) at radius \( r \) is:
\[ U = Fr \]

### Step 8: Total Energy
The total energy \( E \) is the sum of kinetic and potential energy:
\[ E = K + U = \frac{n^2h^2}{8\pi^2r} + Fr \]

### Step 9: Substitute \( r \) into Total Energy
Substituting \( r \) from step 6 into the total energy equation will yield:
\[ E = \frac{n^2h^2}{8\pi^2\left(\frac{n^2h^2}{4\pi^2mF}\right)^{1/3}} + F\left(\frac{n^2h^2}{4\pi^2mF}\right)^{1/3} \]

### Step 10: Simplifying the Energy Expression
After simplification, we arrive at:
\[ E = -\frac{3}{2}\frac{n^2h^2F^{2/3}}{4\pi^2m^{1/3}} \]

### Final Answer
Thus, the energy of the particle in the \( n^{th} \) orbit is:
\[ E = -\frac{3}{2}\frac{n^2h^2F^{2/3}}{4\pi^2m^{1/3}} \]

---