A light beam of wavelength 400 nm is incident on a metal of work- function 2.2 eV. A particular electron absorbs a photon and makes 2 collisions before coming out of the metal
(a) Assuming that 10% of existing energy is lost to the metal in each collision find the final kinetic energy of this electron as it comes out of the metal.
(b) Under the same assumptions find the maximum number of collisions, the electron should suffer before it becomes unable to come out of the metal.

Explanation

To solve the problem step by step, we will break it down into two parts as given in the question.

### Part (a): Finding the Final Kinetic Energy of the Electron

1. **Calculate the Energy of the Incident Photon:**
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h = 4.1357 \times 10^{-15} \, \text{eV·s} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)
- \( \lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \)

Substituting the values:
\[ E = \frac{(4.1357 \times 10^{-15} \, \text{eV·s})(3 \times 10^8 \, \text{m/s})}{400 \times 10^{-9} \, \text{m}} \approx 3.1 \, \text{eV} \]

2. **Calculate the Energy after the First Collision:**
The electron loses 10% of its energy in each collision. Therefore, after the first collision:
\[ E_1 = 0.9 \times E = 0.9 \times 3.1 \, \text{eV} = 2.79 \, \text{eV} \]

3. **Check if the Electron Can Escape After the First Collision:**
The work function \( W \) is given as 2.2 eV. Since \( E_1 = 2.79 \, \text{eV} > 2.2 \, \text{eV} \), the electron can escape after the first collision.

4. **Calculate the Energy after the Second Collision:**
After the second collision, the energy will be:
\[ E_2 = 0.9 \times E_1 = 0.9 \times 2.79 \, \text{eV} = 2.511 \, \text{eV} \]

5. **Calculate the Final Kinetic Energy:**
The kinetic energy of the electron as it comes out of the metal is given by:
\[ KE = E_2 - W = 2.511 \, \text{eV} - 2.2 \, \text{eV} = 0.311 \, \text{eV} \]

### Final Answer for Part (a):
The final kinetic energy of the electron as it comes out of the metal is approximately **0.31 eV**.

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### Part (b): Finding the Maximum Number of Collisions

1. **Continue Calculating Energy After Each Collision:**
We will continue calculating the energy after each collision until it drops below the work function.

- After the third collision:
\[ E_3 = 0.9 \times E_2 = 0.9 \times 2.511 \, \text{eV} \approx 2.2599 \, \text{eV} \]

- After the fourth collision:
\[ E_4 = 0.9 \times E_3 = 0.9 \times 2.2599 \, \text{eV} \approx 2.0339 \, \text{eV} \]

- After the fifth collision:
\[ E_5 = 0.9 \times E_4 = 0.9 \times 2.0339 \, \text{eV} \approx 1.8305 \, \text{eV} \]

2. **Check if the Electron Can Escape After Each Collision:**
- After the third collision: \( E_3 = 2.2599 \, \text{eV} > 2.2 \, \text{eV} \) (can escape)
- After the fourth collision: \( E_4 = 2.0339 \, \text{eV} > 2.2 \, \text{eV} \) (can escape)
- After the fifth collision: \( E_5 = 1.8305 \, \text{eV} < 2.2 \, \text{eV} \) (cannot escape)

### Final Answer for Part (b):
The maximum number of collisions the electron can suffer before it becomes unable to come out of the metal is **4**.

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