The frequency of incident light falling on a photosensitive metal plate is doubled, the KE of the emitted photoelectrons is

Options

A. double the earlier value
B. unchanged
C. more than doubled
D. less than doubled

Explanation

To solve the problem, we need to analyze the relationship between the frequency of incident light and the kinetic energy of emitted photoelectrons using Einstein's photoelectric equation.

### Step-by-Step Solution:

1. **Understand the Photoelectric Equation**:
The maximum kinetic energy (KE) of the emitted photoelectrons is given by the equation:
\[ KE = h\nu - \phi_0 \]
where \(h\) is Planck's constant, \(\nu\) is the frequency of the incident light, and \(\phi_0\) is the work function of the metal.

2. **Define the Initial Conditions**:
Let’s denote the initial frequency of the incident light as \(\nu\). The kinetic energy of the emitted photoelectrons at this frequency (\(KE_1\)) can be expressed as:
\[ KE_1 = h\nu - \phi_0 \]

3. **Change the Frequency**:
According to the problem, the frequency of the incident light is doubled. Therefore, the new frequency (\(\nu_2\)) is:
\[ \nu_2 = 2\nu \]

4. **Calculate the New Kinetic Energy**:
Using the new frequency in the photoelectric equation, the new kinetic energy (\(KE_2\)) becomes:
\[ KE_2 = h(2\nu) - \phi_0 = 2h\nu - \phi_0 \]

5. **Compare the Two Kinetic Energies**:
Now, we can compare \(KE_2\) and \(KE_1\):
\[ KE_2 = 2h\nu - \phi_0 \]
\[ KE_1 = h\nu - \phi_0 \]
To find the difference between the two kinetic energies, we subtract \(KE_1\) from \(KE_2\):
\[ KE_2 - KE_1 = (2h\nu - \phi_0) - (h\nu - \phi_0) = 2h\nu - h\nu = h\nu \]

6. **Conclusion**:
From the above calculations, we see that when the frequency is doubled, the increase in kinetic energy is equal to \(h\nu\). Thus, the new kinetic energy \(KE_2\) is greater than \(KE_1\) by \(h\nu\). Therefore, the kinetic energy of the emitted photoelectrons is more than double the initial kinetic energy.

### Final Answer:
The kinetic energy of the emitted photoelectrons is more than double when the frequency of the incident light is doubled.

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