A light beam consists of two types of photons. In one type, each photon has the energy 2 eV and in another type, each photon has energy 3 eV. The light beam is incident on a photoelectric material of work function 1 eV. The maximum kinetic enregy of emitted photoelectron is

Options

A. 2 eV
B. 3 eV
C. 4 eV
D. 1 eV

Answer

To solve the problem, we need to find the maximum kinetic energy of the emitted photoelectrons when a light beam consisting of two types of photons (with energies of 2 eV and 3 eV) strikes a photoelectric material that has a work function of 1 eV.

### Step-by-Step Solution:

1. **Understanding the Work Function**:
The work function (ϕ) of the material is the minimum energy required to remove an electron from the surface of the material. In this case, ϕ = 1 eV.

2. **Finding the Kinetic Energy for Each Type of Photon**:
According to the photoelectric effect, the maximum kinetic energy (K.E.) of the emitted photoelectron can be calculated using the formula:
\[ K.E. = E - \phi \]
where \(E\) is the energy of the incident photon and \(\phi\) is the work function.

3. **Case 1: Photon Energy of 2 eV**:
For the first type of photon with energy \(E_1 = 2 \, \text{eV}\):
\[ K.E._1 = E_1 - \phi = 2 \, \text{eV} - 1 \, \text{eV} = 1 \, \text{eV} \]

4. **Case 2: Photon Energy of 3 eV**:
For the second type of photon with energy \(E_2 = 3 \, \text{eV}\):
\[ K.E._2 = E_2 - \phi = 3 \, \text{eV} - 1 \, \text{eV} = 2 \, \text{eV} \]

5. **Determining the Maximum Kinetic Energy**:
The maximum kinetic energy of the emitted photoelectron will be the higher value obtained from the two cases. Thus:
\[ K.E. = \max(K.E._1, K.E._2) = \max(1 \, \text{eV}, 2 \, \text{eV}) = 2 \, \text{eV} \]

### Final Answer:
The maximum kinetic energy of the emitted photoelectron is **2 eV**.