The frequency and intensity of a light source are both doubled. Consider the following statements
A. The saturation photocurrent remains almost the same
B. The maximum kinetic energy of the photoelectrons is double

Options

A. Both A and B are true
B. A is true but B is false
C. A is false but B is true
D. Both A and B are false

Explanation

To analyze the question regarding the effects of doubling the frequency and intensity of a light source on the saturation photocurrent and the maximum kinetic energy of photoelectrons, we can break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
- The photoelectric effect states that when light of sufficient frequency strikes a metal surface, it can eject electrons from that surface. The energy of the incoming photons is given by \( E = h\nu \), where \( h \) is Planck's constant and \( \nu \) is the frequency of the light.

2. **Effect of Doubling Frequency**:
- When the frequency of the light is doubled, the energy of each photon also doubles. Thus, if the original frequency is \( \nu \), the new frequency is \( 2\nu \), leading to photon energy \( E' = h(2\nu) = 2h\nu \).

3. **Effect on Saturation Photocurrent**:
- The saturation photocurrent is directly related to the number of photoelectrons emitted per second. The number of emitted photoelectrons is proportional to the number of incident photons.
- If the intensity of light is doubled, it does not mean that the number of photons emitted per second doubles if the frequency is also doubled. The increase in intensity due to doubling the frequency does not increase the number of photons emitted because the energy per photon has increased. Therefore, the saturation photocurrent remains almost the same.

4. **Effect on Maximum Kinetic Energy**:
- The maximum kinetic energy of the emitted photoelectrons can be calculated using Einstein's photoelectric equation:
\[ K_{max} = h\nu - \phi \]
where \( \phi \) is the work function of the metal.
- If the frequency is doubled, the new maximum kinetic energy becomes:
\[ K'_{max} = h(2\nu) - \phi = 2h\nu - \phi \]
- Comparing the new kinetic energy with the original:
\[ K'_{max} = 2(h\nu) - \phi = 2K_{max} + \phi - \phi = 2K_{max} \]
- This shows that the maximum kinetic energy is indeed more than double the original kinetic energy, thus the statement that the maximum kinetic energy of the photoelectrons is double is incorrect.

5. **Conclusion**:
- Statement A: The saturation photocurrent remains almost the same - **True**.
- Statement B: The maximum kinetic energy of the photoelectrons is double - **False**.

### Final Answer:
- Statement A is true, and Statement B is false.