Let p and E denote the linear momentum and the energy of a photon. For another photon of smaller wavelength (in same medium)

Options

A. Both p and E increase
B. p increases and E decreases
C. p decreases and E increases
D. Both p and E decreases

Explanation

To solve the question regarding the relationship between the linear momentum (p) and energy (E) of a photon when considering a photon of smaller wavelength, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formulas**:
- The momentum (p) of a photon is given by the formula:
\[ p = \frac{h}{\lambda} \]
- The energy (E) of a photon is given by the formula:
\[ E = \frac{hc}{\lambda} \]
Here, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the photon.

2. **Analyze the Effect of Wavelength**:
- When the wavelength \( \lambda \) decreases (i.e., for a photon of smaller wavelength), we need to examine how this affects both momentum and energy.
- Since \( h \) and \( c \) are constants, both the momentum and energy are inversely proportional to the wavelength \( \lambda \).

3. **Determine the Relationship**:
- As \( \lambda \) decreases, \( \frac{1}{\lambda} \) increases. This implies:
\[ p \text{ increases as } \lambda \text{ decreases} \]
\[ E \text{ increases as } \lambda \text{ decreases} \]

4. **Conclusion**:
- Therefore, for a photon of smaller wavelength, both the momentum \( p \) and the energy \( E \) increase. The correct answer to the question is that both \( p \) and \( E \) increase.

### Final Answer:
Both \( p \) and \( E \) increase.

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