if we assume only gravitational attraction between proton and electron in hydrogen atom and the Bohr quantizaton rule to be followed, then the expression for the ground state energy of the atom will be (the mass of proton is M and that of electron is m.)

Options

A. increases 4 times
B. decreases 4 times
C. increases 8 times
D. decreases 8 times

Explanation

To derive the expression for the ground state energy of a hydrogen atom considering only gravitational attraction between the proton and electron, we can follow these steps:

### Step 1: Write the gravitational force equation
According to Newton's law of gravitation, the gravitational force (F) between the proton and electron can be expressed as:
\[ F = \frac{G M m}{r^2} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the proton,
- \( m \) is the mass of the electron,
- \( r \) is the distance between the proton and electron.

### Step 2: Write the centripetal force equation
For an electron in circular motion around the proton, the centripetal force required to keep the electron in orbit is given by:
\[ F = \frac{m v^2}{r} \]
where:
- \( v \) is the velocity of the electron.

### Step 3: Set the gravitational force equal to the centripetal force
Since the gravitational force provides the necessary centripetal force for the electron's circular motion, we can set the two forces equal to each other:
\[ \frac{G M m}{r^2} = \frac{m v^2}{r} \]

### Step 4: Simplify the equation
We can simplify this equation by multiplying both sides by \( r \) and dividing by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{G M}{r} = v^2 \]
Thus, we have:
\[ v^2 = \frac{G M}{r} \]

### Step 5: Apply Bohr's quantization condition
According to Bohr's quantization rule, the angular momentum of the electron is quantized and given by:
\[ M v r = \frac{n h}{2 \pi} \]
For the ground state of hydrogen, \( n = 1 \):
\[ M v r = \frac{h}{2 \pi} \]

### Step 6: Solve for \( v \)
From the angular momentum equation, we can express \( v \) as:
\[ v = \frac{h}{2 \pi M r} \]

### Step 7: Substitute \( v \) into the expression for \( v^2 \)
Now, substitute \( v \) from the angular momentum equation into the equation \( v^2 = \frac{G M}{r} \):
\[ \left(\frac{h}{2 \pi M r}\right)^2 = \frac{G M}{r} \]
Squaring the left side gives:
\[ \frac{h^2}{4 \pi^2 M^2 r^2} = \frac{G M}{r} \]

### Step 8: Rearranging to find \( r \)
Multiply both sides by \( r^2 \):
\[ h^2 = 4 \pi^2 M^2 G M r \]
Rearranging gives:
\[ r = \frac{h^2}{4 \pi^2 G M^2} \]

### Step 9: Calculate the total energy
The total energy \( E \) of the electron in the hydrogen atom is the sum of its kinetic energy \( (KE) \) and potential energy \( (PE) \):
\[ E = KE + PE \]
The kinetic energy is given by:
\[ KE = \frac{1}{2} m v^2 = \frac{1}{2} m \left(\frac{G M}{r}\right) \]
The potential energy due to gravitational attraction is:
\[ PE = -\frac{G M m}{r} \]
Thus, the total energy becomes:
\[ E = \frac{1}{2} m \left(\frac{G M}{r}\right) - \frac{G M m}{r} \]
\[ E = \frac{G M m}{2r} - \frac{G M m}{r} = -\frac{G M m}{2r} \]

### Step 10: Substitute \( r \) back into the energy equation
Substituting \( r = \frac{h^2}{4 \pi^2 G M^2} \) into the energy expression:
\[ E = -\frac{G M m}{2} \cdot \frac{4 \pi^2 G M^2}{h^2} \]
\[ E = -\frac{2 \pi^2 G^2 M^3 m}{h^2} \]

### Final Expression
Thus, the expression for the ground state energy of the hydrogen atom, assuming only gravitational attraction, is:
\[ E = -\frac{2 \pi^2 G^2 M^3 m}{h^2} \]