The photoelectric work function of a metal is 3 eV . The maximum kinetic energy of the emitted electrons when light of wavelength 3000 Å falls on it is

Options

A. 1.14 eV
B. 11.4 eV
C. 4.14 eV
D. 1.44 eV

Answer

To find the maximum kinetic energy of the emitted electrons when light of wavelength 3000 Å falls on a metal with a work function of 3 eV, we can follow these steps:

### Step-by-Step Solution

1. **Identify the given values:**
- Work function (\( \phi_0 \)) = 3 eV
- Wavelength (\( \lambda \)) = 3000 Å = \( 3000 \times 10^{-10} \) m

2. **Convert the wavelength to meters:**
\[ \lambda = 3000 \, \text{Å} = 3000 \times 10^{-10} \, \text{m} = 3 \times 10^{-7} \, \text{m} \]

3. **Calculate the energy of the incident photons using the formula:**
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) (Planck's constant) = \( 6.626 \times 10^{-34} \, \text{J s} \)
- \( c \) (speed of light) = \( 3 \times 10^8 \, \text{m/s} \)

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3 \times 10^8 \, \text{m/s})}{3 \times 10^{-7} \, \text{m}} \]

4. **Calculate \( E \):**
\[ E = \frac{(6.626 \times 3) \times 10^{-26}}{3} = \frac{19.878 \times 10^{-26}}{3} = 6.626 \times 10^{-26} \, \text{J} \]

5. **Convert the energy from Joules to electron volts:**
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
\[ E \text{ (in eV)} = \frac{6.626 \times 10^{-26} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 4.14 \, \text{eV} \]

6. **Calculate the maximum kinetic energy (\( KE_{\text{max}} \)):**
\[ KE_{\text{max}} = E - \phi_0 \]
\[ KE_{\text{max}} = 4.14 \, \text{eV} - 3 \, \text{eV} = 1.14 \, \text{eV} \]

### Final Answer
The maximum kinetic energy of the emitted electrons is approximately **1.14 eV**.