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 a photon of original frequency f falls through a small height H near the earth's surface, then fractional charge in frequency will be (acceleration due to gravity is g)

Options

A. `"gH"/c`
B. `(gH^2)/c`
C. `(c^2)/"gH"`
D. `"gH"/c^2`

Answer

To solve the problem regarding the fractional change in frequency of a photon falling through a small height \( H \) near the Earth's surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Energy of the Photon**:
The energy of a photon is given by the equation:
\[ E = hf \]
where \( h \) is Planck's constant and \( f \) is the frequency of the photon.

2. **Considering Gravitational Potential Energy**:
When the photon is at a height \( H \) above the Earth's surface, its total energy \( E_1 \) can be expressed as:
\[ E_1 = hf + \text{(gravitational potential energy)} \]
The gravitational potential energy of the photon at height \( H \) is given by:
\[ \text{Gravitational potential energy} = mgh = \frac{hf}{c^2} gH \]
where \( m = \frac{hf}{c^2} \) is the inertial mass of the photon.

3. **Total Energy at Height \( H \)**:
Therefore, the total energy at height \( H \) becomes:
\[ E_1 = hf + \frac{hf}{c^2} gH \]

4. **Energy When Photon Falls to Height 0**:
When the photon falls to the Earth's surface (height = 0), its energy \( E_2 \) is:
\[ E_2 = hf' \]
where \( f' \) is the new frequency of the photon after falling.

5. **Conservation of Energy**:
According to the conservation of energy, the total energy remains constant:
\[ E_1 = E_2 \]
This gives us:
\[ hf + \frac{hf}{c^2} gH = hf' \]

6. **Rearranging the Equation**:
Rearranging the equation to find \( f' \):
\[ hf' = hf + \frac{hf}{c^2} gH \]
\[ hf' - hf = \frac{hf}{c^2} gH \]
\[ h(f' - f) = \frac{hf}{c^2} gH \]

7. **Finding the Fractional Change in Frequency**:
Dividing both sides by \( hf \):
\[ \frac{f' - f}{f} = \frac{gH}{c^2} \]
Therefore, the fractional change in frequency is:
\[ \frac{\Delta f}{f} = \frac{gH}{c^2} \]

### Final Result:
The fractional change in frequency as the photon falls through a height \( H \) near the Earth's surface is given by:
\[ \frac{\Delta f}{f} = \frac{gH}{c^2} \]