In a photoelectric effect experimetent, a metallic surface of work function 2.2 eV is illuminated with a light of wavelenght 400 nm. Assume that an electron makes two collisions before being emitted and in each collision 10% additional energy is lost.
Q. Find the kinetic energy of this electron as it comes out of the metal.

Options

A. 0.46 eV
B. 0.31 eV
C. 0.23 eV
D. none of these

Explanation

To solve the problem step by step, we will follow the outlined procedure:

### Step 1: Calculate the energy of the incoming photon
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant, \( 6.626 \times 10^{-34} \, \text{Js} \)
- \( c \) is the speed of light, \( 3 \times 10^8 \, \text{m/s} \)
- \( \lambda \) is the wavelength of the light, \( 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \)

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{400 \times 10^{-9} \, \text{m}} \]
Calculating this gives:
\[ E \approx 4.97 \times 10^{-19} \, \text{J} \]
To convert this energy into electron volts (1 eV = \( 1.6 \times 10^{-19} \, \text{J} \)):
\[ E \approx \frac{4.97 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.1 \, \text{eV} \]

### Step 2: Calculate the energy lost after the first collision
The electron loses 10% of its energy in the first collision:
\[ \text{Energy lost after first collision} = 0.1 \times 3.1 \, \text{eV} = 0.31 \, \text{eV} \]
The remaining energy after the first collision is:
\[ \text{Remaining energy} = 3.1 \, \text{eV} - 0.31 \, \text{eV} = 2.79 \, \text{eV} \]

### Step 3: Calculate the energy lost after the second collision
The electron again loses 10% of its remaining energy:
\[ \text{Energy lost after second collision} = 0.1 \times 2.79 \, \text{eV} = 0.279 \, \text{eV} \]
The remaining energy after the second collision is:
\[ \text{Remaining energy} = 2.79 \, \text{eV} - 0.279 \, \text{eV} = 2.511 \, \text{eV} \]

### Step 4: Calculate the kinetic energy of the emitted electron
The kinetic energy of the emitted electron can be found using the work function of the metal:
\[ \text{Kinetic Energy} = \text{Remaining energy} - \text{Work function} \]
Given that the work function is \( 2.2 \, \text{eV} \):
\[ \text{Kinetic Energy} = 2.511 \, \text{eV} - 2.2 \, \text{eV} = 0.311 \, \text{eV} \]

### Final Answer
The kinetic energy of the electron as it comes out of the metal is approximately \( 0.311 \, \text{eV} \).

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