If the total electromagnetic energy falling on a surface is `U` then the total momentum delivered (for complate absorption) is

Options

A. `U//c`
B. `cU`
C. `U//c^(2)`
D. `c^(2)U`

Answer

To solve the problem of finding the total momentum delivered when the total electromagnetic energy falling on a surface is \( U \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We are given that a total electromagnetic energy \( U \) falls on a surface, and we need to find the total momentum delivered to that surface upon complete absorption of this energy.

2. **Using the Relativistic Energy-Momentum Relation**: According to Einstein's theory of relativity, the energy-momentum relation is given by:
\[ E^2 = P^2c^2 + m^2c^4 \]
where:
- \( E \) is the energy,
- \( P \) is the momentum,
- \( m \) is the mass of the particle,
- \( c \) is the speed of light.

3. **Considering Massless Particles**: Electromagnetic waves (photons) are massless particles, which means \( m = 0 \). Therefore, the equation simplifies to:
\[ E^2 = P^2c^2 \]

4. **Substituting Given Energy**: Since the total energy falling on the surface is \( U \), we can substitute \( E \) with \( U \):
\[ U^2 = P^2c^2 \]

5. **Solving for Momentum**: Rearranging the equation to solve for momentum \( P \):
\[ P^2 = \frac{U^2}{c^2} \]
Taking the square root of both sides gives:
\[ P = \frac{U}{c} \]

6. **Conclusion**: Therefore, the total momentum delivered for complete absorption of the electromagnetic energy \( U \) is:
\[ P = \frac{U}{c} \]

### Final Answer:
The total momentum delivered (for complete absorption) is \( P = \frac{U}{c} \).

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