A plane electromagnetic wave is incident on a material surface. The wave delivers momentum p and energy E.

Options

A. `p = 0, E!= 0.`
B. `p!=0, E=0.`
C. `p!=0, E!=0.`
D. `p=0, E=0.`

Answer

To solve the problem, we need to analyze the relationship between the momentum \( P \) and energy \( E \) of a plane electromagnetic wave incident on a material surface.

### Step-by-Step Solution:

1. **Understanding the Definitions**:
- The energy \( E \) of a photon (quantum of electromagnetic wave) is given by the equation:
\[ E = h \nu \]
where \( h \) is Planck's constant and \( \nu \) is the frequency of the wave.
- The momentum \( P \) of a photon can be expressed as:
\[ P = \frac{E}{c} \]
where \( c \) is the speed of light.

2. **Relating Energy and Momentum**:
- From the above equations, we can express momentum in terms of frequency:
\[ P = \frac{h \nu}{c} \]
- This shows that if the frequency \( \nu \) is non-zero, then \( P \) must also be non-zero.

3. **Analyzing the Options**:
- **Option 1**: \( P = 0, E \neq 0 \) - This is not possible because if \( E \) is non-zero, then \( P \) must also be non-zero.
- **Option 2**: \( P \neq 0, E = 0 \) - This is also not possible because if \( P \) is non-zero, then \( E \) cannot be zero.
- **Option 3**: \( P \neq 0, E \neq 0 \) - This is possible and consistent with our analysis.
- **Option 4**: \( P = 0, E = 0 \) - This is not possible in the context of an incident electromagnetic wave, as it would imply no energy or momentum is delivered.

4. **Conclusion**:
- The only valid option is **Option 3**: \( P \neq 0 \) and \( E \neq 0 \). This means that when an electromagnetic wave strikes a surface, it delivers both momentum and energy.