Although a photon has no rest mass, but it possesses the inertial mass `m="hf"/c^2` where h is Planck's constant , f is frequency of light and c is speed of light . Since light is deflected by a gravitational field , so it is naturally assured that photons have same gravitational behaviour as other particles. When photon is emitted from source of star of mass M and radius R, total energy of photon will be sum of hf and gravitational potential energy. At a large distance from star, the photon is beyond the star's gravitational field , so its gravitational potential energy becomes zero but its total energy remains constant. So frequency of a photon emitted from surface of a star decreases as it moves away from star. A photon in visible region of spectrum is thus shifted towards red end, and this phenomena is known as gravitational red shift.
If f' is frequency of photon when it is very far away from star then `(f-f')/f`

Options

A. `(GM)/(R^2c)`
B. `(GMh)/(Rc^2)`
C. `(GM)/(c^2R)`
D. `(GMh)/(Rc)`

Answer

To solve the problem regarding the gravitational redshift of a photon emitted from a star, we can follow these steps:

### Step 1: Understand the Energy Conservation Principle
The total energy of the photon when it is emitted from the surface of the star is the sum of its initial energy (hf) and the gravitational potential energy associated with the star.

### Step 2: Write the Expression for Total Energy
The total energy of the photon when it is emitted can be expressed as:
\[ E_{\text{initial}} = hf - \frac{GMm}{R} \]
where:
- \( hf \) is the energy of the photon,
- \( G \) is the gravitational constant,
- \( M \) is the mass of the star,
- \( R \) is the radius of the star,
- \( m \) is the inertial mass of the photon given by \( m = \frac{hf}{c^2} \).

### Step 3: Substitute the Inertial Mass of the Photon
Substituting \( m = \frac{hf}{c^2} \) into the energy equation gives:
\[ E_{\text{initial}} = hf - \frac{G M (hf/c^2)}{R} \]

### Step 4: Simplify the Expression
This simplifies to:
\[ E_{\text{initial}} = hf \left( 1 - \frac{GM}{Rc^2} \right) \]

### Step 5: Determine the Energy at Large Distance
At a large distance from the star, the gravitational potential energy becomes zero, so the energy of the photon is:
\[ E_{\text{final}} = hf' \]
where \( f' \) is the frequency of the photon when it is far away from the star.

### Step 6: Set the Initial and Final Energies Equal
Since energy is conserved, we can equate the initial and final energies:
\[ hf \left( 1 - \frac{GM}{Rc^2} \right) = hf' \]

### Step 7: Solve for the Frequency at Large Distance
Dividing both sides by \( hf \) gives:
\[ 1 - \frac{GM}{Rc^2} = \frac{f'}{f} \]

### Step 8: Rearranging the Equation
Rearranging this equation gives:
\[ \frac{f - f'}{f} = \frac{GM}{Rc^2} \]

### Final Result
Thus, the expression for the fractional change in frequency due to gravitational redshift is:
\[ \frac{f - f'}{f} = \frac{GM}{Rc^2} \]

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