If the frequency of light in a photoelectric experiment is double then maximum kinetic energy of photoelectron

Options

A. It becomes more then double
B. It becomes less then double
C. It becomes exactly double
D. It does not change

Answer

To solve the problem, we need to understand the relationship between the frequency of light and the maximum kinetic energy of the photoelectrons emitted during the photoelectric effect.

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect**:
The photoelectric effect occurs when light shines on a metal surface, causing electrons to be emitted. The energy of the incident photons is absorbed by the electrons, and if this energy is greater than the work function of the metal, electrons are emitted.

2. **Energy of Photons**:
The energy (E) of a photon is given by the equation:
\[ E = h \nu \]
where \(h\) is Planck's constant and \(\nu\) is the frequency of the light.

3. **Maximum Kinetic Energy of Photoelectrons**:
The maximum kinetic energy (K.E.) of the emitted photoelectrons can be expressed as:
\[ K.E. = h \nu - \phi \]
where \(\phi\) is the work function (the minimum energy required to remove an electron from the metal).

4. **Doubling the Frequency**:
If the frequency of light is doubled, we have:
\[ \nu' = 2\nu \]
The energy of the new photons becomes:
\[ E' = h \nu' = h (2\nu) = 2h \nu \]

5. **Calculating New Maximum Kinetic Energy**:
The new maximum kinetic energy of the photoelectrons when the frequency is doubled is:
\[ K.E.' = h \nu' - \phi = 2h \nu - \phi \]

6. **Comparing Kinetic Energies**:
The original maximum kinetic energy was:
\[ K.E. = h \nu - \phi \]
Now, we can compare the two:
\[ K.E.' = 2(h \nu - \phi) + \phi = 2K.E. + \phi - \phi = 2K.E. \]
Thus, when the frequency is doubled, the maximum kinetic energy of the photoelectrons also doubles.

### Final Answer:
The maximum kinetic energy of the photoelectrons will be doubled.