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 emitted from a photo-sensitive surface when monochromatic photons of energy 3.3 eV are incident on it, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Energy of the incident photon (E) = 3.3 eV
- Work function of the surface (φ) = 2.4 eV

2. **Calculate the Kinetic Energy of the Emitted Photoelectron:**
The kinetic energy (KE) of the emitted photoelectron can be calculated using the equation:
\[ KE = E - \phi \]
Substituting the given values:
\[ KE = 3.3 \, \text{eV} - 2.4 \, \text{eV} = 0.9 \, \text{eV} \]

3. **Convert Kinetic Energy from eV to Joules:**
To convert the kinetic energy from electron volts to joules, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ KE = 0.9 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.44 \times 10^{-19} \, \text{J} \]

4. **Use the Kinetic Energy to Find Maximum Velocity:**
The kinetic energy of the photoelectron can also be expressed in terms of its mass (m) and velocity (v) using the formula:
\[ KE = \frac{1}{2} mv^2 \]
Rearranging for velocity (v):
\[ v = \sqrt{\frac{2 \times KE}{m}} \]
The mass of the electron (m) is approximately \(9.1 \times 10^{-31} \, \text{kg}\).

5. **Substituting Values to Calculate Velocity:**
\[ v = \sqrt{\frac{2 \times 1.44 \times 10^{-19} \, \text{J}}{9.1 \times 10^{-31} \, \text{kg}}} \]
\[ v = \sqrt{\frac{2.88 \times 10^{-19}}{9.1 \times 10^{-31}}} \]
\[ v = \sqrt{3.16 \times 10^{11}} \approx 5.65 \times 10^{5} \, \text{m/s} \]

6. **Final Result:**
The maximum velocity of the emitted photoelectron is approximately:
\[ v \approx 5.65 \times 10^{5} \, \text{m/s} \]