Monochromatic photons of energy 3.3 eV are incident on a photo sensitive surface of work function 2.4 eV. , maximum velocity of photoelectron is

Options

A. `5.65xx 10^(5)m//s`
B. `6.65xx 10^(5)m//s`
C. `10.65xx 10^(5)m//s`
D. `2.65xx 10^(5)m//s`

Answer

To solve the problem of finding the maximum velocity of photoelectrons when monochromatic photons of energy 3.3 eV are incident on a photosensitive surface with a work function of 2.4 eV, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Energy of the Incident Photon and Work Function**:
- The energy of the incident photon (E_photon) is given as 3.3 eV.
- The work function (Φ) of the surface is given as 2.4 eV.

2. **Calculate the Maximum Kinetic Energy (K.E.) of the Photoelectron**:
- The maximum kinetic energy of the emitted photoelectron can be calculated using the formula:
\[ K.E. = E_{photon} - \Phi \]
- Substituting the values:
\[ K.E. = 3.3 \, \text{eV} - 2.4 \, \text{eV} = 0.9 \, \text{eV} \]

3. **Convert the Kinetic Energy from eV to Joules**:
- To find the maximum velocity, we need to convert the kinetic energy from electron volts to joules. The conversion factor is:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
- Thus, the kinetic energy in joules is:
\[ K.E. = 0.9 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.44 \times 10^{-19} \, \text{J} \]

4. **Use the Kinetic Energy Formula to Find Maximum Velocity**:
- The kinetic energy of the photoelectron can also be expressed as:
\[ K.E. = \frac{1}{2} m v_{max}^2 \]
- Rearranging this formula to solve for \(v_{max}\):
\[ v_{max} = \sqrt{\frac{2 \times K.E.}{m}} \]
- Where \(m\) is the mass of the electron, approximately \(9.1 \times 10^{-31} \, \text{kg}\).

5. **Substituting the Values**:
- Now substituting the values into the equation:
\[ v_{max} = \sqrt{\frac{2 \times 1.44 \times 10^{-19} \, \text{J}}{9.1 \times 10^{-31} \, \text{kg}}} \]

6. **Calculating the Maximum Velocity**:
- Performing the calculations:
\[ v_{max} = \sqrt{\frac{2.88 \times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{3.16 \times 10^{11}} \approx 5.65 \times 10^5 \, \text{m/s} \]

7. **Final Result**:
- Therefore, the maximum velocity of the photoelectron is approximately:
\[ v_{max} \approx 5.65 \times 10^5 \, \text{m/s} \]