An electron is moving on a circular path around a stationary neutron under gravitational interaction Masses of the neutron and electron are `M` and `m` respectively. If Bohr's quantum condition holds here the minimum permissible de Broglie wavelength associated with the electron will be -

Options

A. `(h^(2))/(2piGMm)`
B. `(h^(2))/(2piGMm^(2))`
C. `(h^(2))/(2piGM^(2)m)`
D. `(h^(2))/(4pi GM^(2)m)`

Answer

To solve the problem of finding the minimum permissible de Broglie wavelength associated with an electron moving in a circular path around a stationary neutron under gravitational interaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Forces**:
The electron is moving in a circular path around the neutron due to gravitational attraction. The gravitational force provides the necessary centripetal force for the electron's circular motion.

2. **Setting Up the Equation**:
The centripetal force required for circular motion is given by:
\[ \frac{mv^2}{r} \]
where \( m \) is the mass of the electron, \( v \) is its velocity, and \( r \) is the radius of the circular path.

The gravitational force between the electron and the neutron is given by:
\[ F_g = \frac{G M m}{r^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the neutron, and \( m \) is the mass of the electron.

3. **Equating the Forces**:
Setting the centripetal force equal to the gravitational force:
\[ \frac{mv^2}{r} = \frac{G M m}{r^2} \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ mv^2 = \frac{G M m}{r} \]
Rearranging gives:
\[ v^2 = \frac{G M}{r} \]

4. **Using Bohr's Quantum Condition**:
According to Bohr's quantum condition, the angular momentum \( L \) of the electron is quantized:
\[ L = mvr = n \frac{h}{2\pi} \]
where \( n \) is a positive integer and \( h \) is Planck's constant.

5. **Expressing Velocity**:
From the angular momentum equation, we can express \( v \):
\[ v = \frac{n h}{2 \pi m r} \]

6. **Substituting for Velocity**:
We can substitute this expression for \( v \) into the equation for \( v^2 \):
\[ \left(\frac{n h}{2 \pi m r}\right)^2 = \frac{G M}{r} \]
Simplifying gives:
\[ \frac{n^2 h^2}{4 \pi^2 m^2 r^2} = \frac{G M}{r} \]
Multiplying both sides by \( r^2 \):
\[ n^2 h^2 = 4 \pi^2 m^2 G M r \]

7. **Solving for \( r \)**:
Rearranging gives:
\[ r = \frac{n^2 h^2}{4 \pi^2 m^2 G M} \]

8. **Finding the de Broglie Wavelength**:
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{mv} \]
Substituting \( v \) from the angular momentum equation:
\[ \lambda = \frac{h}{m \left(\frac{n h}{2 \pi m r}\right)} = \frac{2 \pi r}{n} \]

9. **Substituting for \( r \)**:
Now substituting \( r \) back into the wavelength equation:
\[ \lambda = \frac{2 \pi \left(\frac{n^2 h^2}{4 \pi^2 m^2 G M}\right)}{n} = \frac{n h^2}{2 \pi m^2 G M} \]

10. **Finding the Minimum Wavelength**:
For the minimum permissible de Broglie wavelength, we set \( n = 1 \):
\[ \lambda_{min} = \frac{h^2}{2 \pi m^2 G M} \]

### Final Answer:
The minimum permissible de Broglie wavelength associated with the electron is:
\[ \lambda_{min} = \frac{h^2}{2 \pi G M m^2} \]