The work function of a metal is 3.4 eV. A light of wavelength `3000Å` is incident on it. The maximum kinetic energy of the ejected electron will be :

Options

A. `1.18 xx 10^(-19)J`
B. `2.36 xx 10^(-19)J`
C. `3.4 xx 10^(-19)J`
D. Electron will not be ejected

Answer

To solve the problem, we need to determine the maximum kinetic energy of the ejected electron when light of wavelength 3000 Å is incident on a metal with a work function of 3.4 eV.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Work function (Φ) = 3.4 eV
- Wavelength (λ) = 3000 Å

2. **Convert Wavelength to Meters:**
- 1 Ångström (Å) = \(10^{-10}\) meters
- Therefore, \(3000 \, \text{Å} = 3000 \times 10^{-10} \, \text{m} = 3 \times 10^{-7} \, \text{m}\)

3. **Calculate the Energy of the Incident Light:**
- The energy (E) of the incident light can be calculated 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. **Perform the Calculation:**
- Calculate \(E\):
\[ E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^{8})}{3 \times 10^{-7}} = 6.626 \times 10^{-19} \, \text{J} \]

5. **Convert Work Function from eV to Joules:**
- Work function in Joules:
\[ \Phi = 3.4 \, \text{eV} \times (1.6 \times 10^{-19} \, \text{J/eV}) = 5.44 \times 10^{-19} \, \text{J} \]

6. **Apply the Photoelectric Equation:**
- The maximum kinetic energy (K.E.) of the ejected electron is given by:
\[ K.E. = E - \Phi \]
- Substituting the values:
\[ K.E. = (6.626 \times 10^{-19} \, \text{J}) - (5.44 \times 10^{-19} \, \text{J}) = 1.186 \times 10^{-19} \, \text{J} \]

7. **Final Answer:**
- The maximum kinetic energy of the ejected electron is:
\[ K.E. \approx 1.186 \times 10^{-19} \, \text{J} \]