A radiation of energy `E` falls normally on a perfectly reflecting surface . The momentum transferred to the surface is

Options

A. `E/C`
B. `(2E)/(C )`
C. `(2E)/(C^(2))`
D. `E/(C^(2))`

Answer

To solve the problem of calculating the momentum transferred to a perfectly reflecting surface when radiation of energy \( E \) falls normally on it, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Energy of Radiation:**
The energy \( E \) of the radiation can be expressed in terms of its frequency \( f \) using the equation:
\[ E = h \cdot f \]
where \( h \) is Planck's constant.

2. **Relating Energy to Momentum:**
The momentum \( p \) of a photon (which is a quantum of radiation) can be expressed as:
\[ p = \frac{E}{c} \]
where \( c \) is the speed of light. This can also be derived from the relationship:
\[ p = \frac{h}{\lambda} \]
where \( \lambda \) is the wavelength of the radiation.

3. **Calculating Initial Momentum:**
When the radiation falls on the surface, it has an initial momentum \( p_i \):
\[ p_i = \frac{E}{c} \]

4. **Reflection of Radiation:**
Since the surface is perfectly reflecting, the radiation will reflect back with the same magnitude of momentum but in the opposite direction. Thus, the final momentum \( p_f \) after reflection is:
\[ p_f = -\frac{E}{c} \]

5. **Calculating Change in Momentum:**
The change in momentum \( \Delta p \) (which is the momentum transferred to the surface) can be calculated as:
\[ \Delta p = p_f - p_i \]
Substituting the values of \( p_f \) and \( p_i \):
\[ \Delta p = -\frac{E}{c} - \frac{E}{c} = -\frac{2E}{c} \]

6. **Final Result:**
The magnitude of the momentum transferred to the surface is:
\[ |\Delta p| = \frac{2E}{c} \]

### Conclusion:
The momentum transferred to the perfectly reflecting surface is \( \frac{2E}{c} \).