When photons of energy `hv` fall on an aluminium plate (of work function `E_(0)`), photoelectrons of maximum kinetic energy `K` are ejected . If the frequency of the radiation is doubled , the maximum kinetic energy of the ejected photoelectrons will be

Options

A. K + `E_(0)`
B. 2K
C. K
D. K+hv

Explanation

To solve the problem, we need to use the photoelectric effect equation, which relates the energy of the incoming photons to the work function of the material and the kinetic energy of the emitted photoelectrons.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- 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 radiation.
- The work function of the aluminium plate is denoted as \( E_0 \).
- The maximum kinetic energy \( K \) of the emitted photoelectrons can be expressed as:
\[ K = h\nu - E_0 \]

2. **Doubling the Frequency:**
- When the frequency of the radiation is doubled, the new frequency becomes \( 2\nu \).
- The energy of the incoming photons at this new frequency is:
\[ E' = h(2\nu) = 2h\nu \]

3. **Calculating the New Maximum Kinetic Energy:**
- The new maximum kinetic energy \( K' \) of the emitted photoelectrons can be calculated using the same photoelectric effect equation:
\[ K' = E' - E_0 \]
- Substituting the new photon energy into the equation gives:
\[ K' = 2h\nu - E_0 \]

4. **Relating the New Kinetic Energy to the Old Kinetic Energy:**
- We know from the initial conditions that \( K = h\nu - E_0 \).
- We can express \( E_0 \) in terms of \( K \):
\[ E_0 = h\nu - K \]
- Substituting this into the expression for \( K' \):
\[ K' = 2h\nu - (h\nu - K) = 2h\nu - h\nu + K = h\nu + K \]

5. **Final Result:**
- Thus, the new maximum kinetic energy \( K' \) can be expressed as:
\[ K' = K + h\nu \]

### Conclusion:
The maximum kinetic energy of the ejected photoelectrons when the frequency of the radiation is doubled is \( K + h\nu \).