The threshold wavelength of tungsten is 2300 Å. If ultra voilet light of wavelength 1800 Å is incident on it, then the maximum kinetic energy of photoelectrons would be

Options

A. `1.5 eV `
B. `2.2 eV `
C. `3.0 eV `
D. `5.0 eV `

Answer

To solve the problem, we need to find the maximum kinetic energy of photoelectrons emitted from tungsten when ultraviolet light of wavelength 1800 Å is incident on it. The threshold wavelength for tungsten is given as 2300 Å.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Threshold wavelength of tungsten, \( \lambda_0 = 2300 \, \text{Å} = 2300 \times 10^{-10} \, \text{m} = 2.3 \times 10^{-7} \, \text{m} \)
- Incident wavelength, \( \lambda_i = 1800 \, \text{Å} = 1800 \times 10^{-10} \, \text{m} = 1.8 \times 10^{-7} \, \text{m} \)

2. **Use the Photoelectric Effect Equation:**
The maximum kinetic energy (\( KE_{max} \)) of the emitted photoelectrons can be calculated using the equation:
\[ KE_{max} = h c \left( \frac{1}{\lambda_i} - \frac{1}{\lambda_0} \right) \]
where:
- \( h \) is Planck's constant, \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- \( c \) is the speed of light, \( c = 3 \times 10^8 \, \text{m/s} \)

3. **Calculate the Terms:**
- First, calculate \( \frac{1}{\lambda_i} \) and \( \frac{1}{\lambda_0} \):
\[ \frac{1}{\lambda_i} = \frac{1}{1.8 \times 10^{-7}} \approx 5.56 \times 10^6 \, \text{m}^{-1} \]
\[ \frac{1}{\lambda_0} = \frac{1}{2.3 \times 10^{-7}} \approx 4.35 \times 10^6 \, \text{m}^{-1} \]

4. **Calculate the Difference:**
\[ \frac{1}{\lambda_i} - \frac{1}{\lambda_0} = 5.56 \times 10^6 - 4.35 \times 10^6 = 1.21 \times 10^6 \, \text{m}^{-1} \]

5. **Substitute into the Kinetic Energy Equation:**
\[ KE_{max} = h c \left( 1.21 \times 10^6 \right) \]
\[ KE_{max} = (6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s}) \times (1.21 \times 10^6 \, \text{m}^{-1}) \]

6. **Calculate \( KE_{max} \):**
\[ KE_{max} = 6.626 \times 3 \times 1.21 \times 10^{-34} \times 10^8 \times 10^6 \]
\[ KE_{max} \approx 2.41 \times 10^{-19} \, \text{J} \]

7. **Convert Joules to Electron Volts:**
To convert Joules to electron volts, use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ KE_{max} \approx \frac{2.41 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.506 \, \text{eV} \approx 1.5 \, \text{eV} \]

### Final Answer:
The maximum kinetic energy of the photoelectrons is approximately **1.5 eV**.