For a hypothetical hydrogen like atom, the potential energy of the system is given by `U(r)=(-Ke^(2))/(r^(3))`, where r is the distance between the two particles. If Bohr's model of quantization of angular momentum is applicable then velocity of particle is given by:

Options

A. `v=(n^(2)h^(3))/(Ke^(2)8pi^(3)m^(2))`
B. `v=(n^(3)h^(3))/(8Ke^(2)pi^(3)m^(2))`
C. `v=(n^(3)h^(3))/(24Ke^(2)pi^(3)m^(2))`
D. `v=(n^(2)h^(3))/(24Ke^(2)pi^(3)m^(2))`

Answer

To find the velocity of a particle in a hypothetical hydrogen-like atom where the potential energy is given by \( U(r) = -\frac{K e^2}{r^3} \), we can follow these steps:

### Step 1: Determine the Force from Potential Energy
The force \( F \) can be derived from the potential energy \( U(r) \) using the relation:
\[ F = -\frac{dU}{dr} \]
Calculating the derivative:
\[ U(r) = -\frac{K e^2}{r^3} \]
Taking the derivative:
\[ \frac{dU}{dr} = \frac{3K e^2}{r^4} \]
Thus, the force is:
\[ F = -\frac{dU}{dr} = -\left(\frac{3K e^2}{r^4}\right) = \frac{3K e^2}{r^4} \]

### Step 2: Relate Force to Centripetal Force
The force we calculated is the electrostatic force acting between two charged particles. This force provides the necessary centripetal force for circular motion, given by:
\[ F = \frac{mv^2}{r} \]
Setting the two expressions for force equal gives:
\[ \frac{3K e^2}{r^4} = \frac{mv^2}{r} \]

### Step 3: Rearranging the Equation
Rearranging the equation to isolate \( v^2 \):
\[ 3K e^2 = mv^2 r^3 \]
Thus, we can express \( v^2 \) as:
\[ v^2 = \frac{3K e^2}{m} \cdot \frac{1}{r^3} \]

### Step 4: Apply Bohr's Quantization Condition
According to Bohr's model, the angular momentum \( L \) is quantized:
\[ L = mvr = n\frac{h}{2\pi} \]
From this, we can express \( r \) in terms of \( v \):
\[ r = \frac{nh}{2\pi mv} \]

### Step 5: Substitute \( r \) into the Velocity Equation
Substituting \( r \) into the expression for \( v^2 \):
\[ v^2 = \frac{3K e^2}{m} \cdot \frac{1}{\left(\frac{nh}{2\pi mv}\right)^3} \]
This simplifies to:
\[ v^2 = \frac{3K e^2 (2\pi mv)^3}{m n^3 h^3} \]
Rearranging gives:
\[ v^2 = \frac{3 \cdot 8 \pi^3 K e^2 m^2 v^3}{n^3 h^3} \]

### Step 6: Solve for \( v \)
Dividing both sides by \( v \) (assuming \( v \neq 0 \)):
\[ 1 = \frac{3 \cdot 8 \pi^3 K e^2 m^2 v^2}{n^3 h^3} \]
Thus, we can express \( v^2 \):
\[ v^2 = \frac{n^3 h^3}{3 \cdot 8 \pi^3 K e^2 m^2} \]
Taking the square root gives:
\[ v = \frac{n^{3/2} h^{3/2}}{\sqrt{3 \cdot 8 \pi^3 K e^2 m^2}} \]

### Final Answer
The velocity of the particle is:
\[ v = \frac{n^{3/2} h^{3/2}}{\sqrt{24 \pi^3 K e^2 m^2}} \]