The frequency and the intensity of a beam of light falling on the surface of a photoelectric material are increased by a factor of two. This will

Options

A. increase the maximum kinetic energy, the photoelectrons,
as well as photoelectric current by a factor of 2
B. increase the maximum kinetic energy of the photoelectrons and
would increase the photoelectric current by a factor of 2
C. increase the maximum kinetic energy of the photoelectrons by a factor of 2 and will gave no effect on the magnitude of the photoelectric current produced
D. not produce any effect on the kinetic energy of the emitted electrins but will increase the photoelectric current by a factor of 2

Explanation

To solve the problem, we will analyze the effects of increasing the frequency and intensity of light on the photoelectric effect, specifically focusing on the maximum kinetic energy of emitted electrons and the photoelectric current.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The photoelectric effect describes the emission of electrons from a material when it is exposed to light (photons). The maximum kinetic energy (K.E.) of the emitted electrons can be expressed using Einstein's photoelectric equation:
\[ K.E. = E_{photon} - \phi \]
where \(E_{photon} = h \nu\) (energy of the photon), \(h\) is Planck's constant, \(\nu\) is the frequency of light, and \(\phi\) is the work function of the material.

2. **Initial Conditions**:
Let the initial frequency of light be \(\nu\). The initial maximum kinetic energy of the emitted electrons is:
\[ K.E._{max} = h \nu - \phi \]

3. **Increasing Frequency**:
The problem states that the frequency of the light is increased by a factor of 2. Thus, the new frequency becomes:
\[ \nu' = 2\nu \]
The energy of the photon at this new frequency is:
\[ E_{photon}' = h \nu' = h (2\nu) = 2h \nu \]
The new maximum kinetic energy of the emitted electrons is:
\[ K.E.'_{max} = E_{photon}' - \phi = 2h \nu - \phi \]

4. **Comparing Maximum Kinetic Energies**:
We can express the new maximum kinetic energy in terms of the initial maximum kinetic energy:
\[ K.E.'_{max} = 2h \nu - \phi = 2(K.E._{max} + \phi) - \phi = 2K.E._{max} + \phi \]
This shows that the new maximum kinetic energy is greater than twice the initial maximum kinetic energy because of the additional \(\phi\).

5. **Increasing Intensity**:
The intensity of light is also increased by a factor of 2. Intensity is directly proportional to the number of photons incident on the surface. Therefore, if the intensity is doubled, the number of photons is also doubled, which means the number of emitted electrons (assuming 100% quantum efficiency) will also double.

6. **Conclusion**:
- The maximum kinetic energy of the photoelectrons increases by more than a factor of 2 due to the increase in frequency.
- The photoelectric current, which is proportional to the number of emitted electrons, increases by a factor of 2 due to the increase in intensity.

Thus, the correct answer is that the maximum kinetic energy of the photoelectrons increases and the photoelectric current increases by a factor of 2.

### Final Answer:
- The maximum kinetic energy of the photoelectrons increases by more than a factor of 2.
- The photoelectric current increases by a factor of 2.