A metal surface having a work function `phi = 2.2 xx 10^(-19) J`, is illuminated by the light of wavelengh `1320Å`. What is the maximum kinetic energy ( in eV ) of the emitted photoelectron ? [Take `h = 6.6 xx 10^(-34) Js `]

Answer

To solve the problem, we need to find the maximum kinetic energy of the emitted photoelectron when a metal surface is illuminated by light of a specific wavelength. We will use the photoelectric effect equation to find the answer.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Work function, \( \phi = 2.2 \times 10^{-19} \, \text{J} \)
- Wavelength of light, \( \lambda = 1320 \, \text{Å} = 1320 \times 10^{-10} \, \text{m} \)
- Planck's constant, \( h = 6.6 \times 10^{-34} \, \text{Js} \)
- Speed of light, \( c = 3 \times 10^{8} \, \text{m/s} \)

2. **Calculate the Energy of the Incident Photon:**
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Substituting the values:
\[ E = \frac{(6.6 \times 10^{-34} \, \text{Js}) \times (3 \times 10^{8} \, \text{m/s})}{1320 \times 10^{-10} \, \text{m}} \]

3. **Perform the Calculation:**
First, calculate \( hc \):
\[ hc = 6.6 \times 3 = 19.8 \times 10^{-26} \, \text{Jm} \]
Now, calculate \( E \):
\[ E = \frac{19.8 \times 10^{-26}}{1320 \times 10^{-10}} = \frac{19.8}{1320} \times 10^{-16} \, \text{J} \]
\[ E \approx 1.5 \times 10^{-19} \, \text{J} \]

4. **Calculate the Maximum Kinetic Energy:**
The maximum kinetic energy of the emitted photoelectron is given by:
\[ K.E_{\text{max}} = E - \phi \]
Substituting the values:
\[ K.E_{\text{max}} = (1.5 \times 10^{-19} \, \text{J}) - (2.2 \times 10^{-19} \, \text{J}) = -0.7 \times 10^{-19} \, \text{J} \]
Since the energy of the photon is less than the work function, no photoelectron will be emitted. However, if we were to calculate the kinetic energy assuming the photon energy was greater than the work function, we would proceed as follows:

5. **Convert Kinetic Energy to Electron Volts:**
To convert joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ K.E_{\text{max}} \text{ (in eV)} = \frac{K.E_{\text{max}} \text{ (in J)}}{1.6 \times 10^{-19}} \]
If we had a positive kinetic energy value, we would substitute it here.

### Final Answer:
Since the calculated photon energy is less than the work function, the maximum kinetic energy of the emitted photoelectron is zero, meaning no photoelectron is emitted.