Suppose the potential energy between an electron and a proton at a distance `r` is given by `Ke^(2)// 3 r^(3)`. Application of Bohr's theory tohydrogen atom in this case showns that

Options

A. kinetic energy in the nth atom in this orbit is proportional to `n^(6)`
B. kinetic energy is proportional to `m^(-3)` (m=mass of electron)
C. kinetic energy in the nht orbit is proportional to `n^(-2)`
D. kinetic energy is proportional to `m^(3)` (m=mass of electron)

Answer

To solve the problem, we need to analyze the potential energy given and apply Bohr's theory to derive the relationships between the kinetic energy, radius, and quantum number for the hydrogen atom.

### Step-by-Step Solution:

1. **Identify the Potential Energy**:
The potential energy \( U \) between an electron and a proton at a distance \( r \) is given by:
\[ U = -\frac{K e^2}{3 r^3} \]

2. **Calculate the Force**:
The force \( F \) can be derived from the potential energy using the relation:
\[ F = -\frac{dU}{dr} \]
Differentiating the potential energy:
\[ F = -\frac{d}{dr}\left(-\frac{K e^2}{3 r^3}\right) = \frac{K e^2}{3} \cdot \frac{d}{dr}\left(r^{-3}\right) = \frac{K e^2}{3} \cdot (-3 r^{-4}) = -\frac{K e^2}{r^4} \]

3. **Apply Centripetal Force**:
The centripetal force acting on the electron is provided by this electrostatic force:
\[ \frac{m v^2}{r} = \frac{K e^2}{r^4} \]
Rearranging gives:
\[ m v^2 = \frac{K e^2}{r^3} \]

4. **Relate Kinetic Energy**:
The kinetic energy \( K.E. \) of the electron is given by:
\[ K.E. = \frac{1}{2} m v^2 \]
Substituting \( m v^2 \) from the previous step:
\[ K.E. = \frac{1}{2} \cdot \frac{K e^2}{r^3} \]

5. **Determine the Relationship with Quantum Number \( n \)**:
According to Bohr's theory, the angular momentum \( L \) of the electron is quantized:
\[ L = mvr = n\hbar \]
Squaring both sides gives:
\[ m^2 v^2 r^2 = n^2 \hbar^2 \]
From our previous expression for \( m v^2 \):
\[ m^2 \cdot \frac{K e^2}{r^3} \cdot r^2 = n^2 \hbar^2 \]
Simplifying gives:
\[ \frac{K e^2 m^2}{r} = n^2 \hbar^2 \]
Rearranging for \( r \):
\[ r = \frac{K e^2 m^2}{n^2 \hbar^2} \]

6. **Final Relationships**:
From the expression for kinetic energy:
\[ K.E. \propto \frac{1}{r^3} \quad \text{and} \quad r \propto \frac{1}{n^2} \]
Therefore, substituting \( r \):
\[ K.E. \propto n^6 \]
This indicates that:
\[ K.E. \propto \frac{1}{n^3} \]

### Conclusion:
The kinetic energy is inversely proportional to \( n^3 \) and the radius is directly proportional to \( n^2 \). Thus, we can conclude that the relationships derived from Bohr's theory in this case yield the following results.