If in Bohr's atomic model, it is assumed that force between electron and proton varies inversely as `r^(4)` , energy of the system will be proportional to

Options

A. `n^(2)`
B. `n^(4)`
C. `n^(6)`
D. `n^(8)`

Answer

To solve the problem, we need to analyze how the energy of the system changes when the force between the electron and proton varies inversely as \( r^4 \) instead of the usual \( r^2 \).

### Step-by-Step Solution:

1. **Understanding the Force**:
The force between the electron and proton in a hydrogen atom according to Coulomb's law is given by:
\[ F = \frac{k \cdot Z \cdot e^2}{r^2} \]
where \( k \) is Coulomb's constant, \( Z \) is the atomic number (which is 1 for hydrogen), \( e \) is the charge of the electron, and \( r \) is the distance between the electron and proton.

In this case, we are given that the force varies as:
\[ F \propto \frac{1}{r^4} \]

2. **Centripetal Force**:
For an electron in a circular orbit, the centripetal force is provided by the electrostatic force:
\[ F_{\text{centripetal}} = \frac{m v^2}{r} \]
Setting this equal to the modified force gives:
\[ \frac{m v^2}{r} = \frac{k \cdot e^2}{r^4} \]

3. **Rearranging for Velocity**:
Rearranging this equation, we find:
\[ m v^2 = \frac{k \cdot e^2}{r^3} \]
Thus, we can express \( v^2 \) as:
\[ v^2 = \frac{k \cdot e^2}{m r^3} \]

4. **Angular Momentum**:
The angular momentum \( L \) of the electron is quantized and given by:
\[ L = m v r = n \frac{h}{2\pi} \]
Squaring both sides gives:
\[ m^2 v^2 r^2 = \left(n \frac{h}{2\pi}\right)^2 \]

5. **Substituting for \( v^2 \)**:
Substitute \( v^2 \) from step 3 into the angular momentum equation:
\[ m^2 \left(\frac{k \cdot e^2}{m r^3}\right) r^2 = \left(n \frac{h}{2\pi}\right)^2 \]
Simplifying gives:
\[ k \cdot e^2 \cdot m = \left(n \frac{h}{2\pi}\right)^2 \cdot \frac{1}{r} \]

6. **Finding \( r \)**:
Rearranging for \( r \):
\[ r = \frac{(n \frac{h}{2\pi})^2}{k \cdot e^2 \cdot m} \]

7. **Energy of the System**:
The total energy \( E \) of the electron in the orbit is the sum of kinetic and potential energy. The kinetic energy \( K \) is given by:
\[ K = \frac{1}{2} m v^2 \]
Substituting \( v^2 \):
\[ K = \frac{1}{2} m \left(\frac{k \cdot e^2}{m r^3}\right) = \frac{k \cdot e^2}{2 r^3} \]

The potential energy \( U \) is given by:
\[ U = -\frac{k \cdot e^2}{r} \]

Therefore, the total energy \( E \) is:
\[ E = K + U = \frac{k \cdot e^2}{2 r^3} - \frac{k \cdot e^2}{r} \]

8. **Proportionality of Energy**:
Since we have \( r \propto n^2 \) from earlier steps, we can substitute this into the energy expression. The energy will be proportional to:
\[ E \propto -\frac{1}{n^4} \]
Hence, the energy of the system will be proportional to \( n^{-6} \).

### Final Answer:
The energy of the system will be proportional to \( n^{-6} \).