Light of wavelength `5000 Å` falls on a sensitive plate with photoelectric work function of `1.9 eV`. The kinetic energy of the photoelectron emitted will be

Options

A. `0.58eV`
B. `2.48eV`
C. `1.24eV`
D. `1.16eV`

Answer

To find the kinetic energy of the photoelectron emitted when light of wavelength `5000 Å` falls on a sensitive plate with a work function of `1.9 eV`, we can use the photoelectric effect equation derived from Einstein's theory.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Wavelength of light, \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} \)
- Work function of the metal, \( \phi = 1.9 \, \text{eV} \)

2. **Calculate the Energy of the Incident Radiation**:
The energy \( E \) of the incident radiation can be calculated using the formula:
\[ E = \frac{12400}{\lambda} \quad \text{(where } \lambda \text{ is in Å)} \]
Substituting the value of \( \lambda \):
\[ E = \frac{12400}{5000} = 2.48 \, \text{eV} \]

3. **Apply the Photoelectric Effect Equation**:
According to the photoelectric effect equation:
\[ E = \phi + KE \]
where \( KE \) is the kinetic energy of the emitted photoelectron. Rearranging the equation gives:
\[ KE = E - \phi \]

4. **Substitute the Known Values**:
Now substituting the values we have:
\[ KE = 2.48 \, \text{eV} - 1.9 \, \text{eV} = 0.58 \, \text{eV} \]

5. **Conclusion**:
The kinetic energy of the photoelectron emitted is \( 0.58 \, \text{eV} \).

### Final Answer:
The kinetic energy of the photoelectron emitted will be \( 0.58 \, \text{eV} \).

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