Two differenct photons of energies, 1 eV and 2.5 eV, fall on two identical metal plates having work function 0.5 eV, Then the ratio of maximum KE of the electrons emitted from the two surface is -

Options

A. `1 : 2 `
B. ` 1 :4 `
C. `2:1 `
D. `4 : 1 `

Answer

To solve the problem, we need to calculate the maximum kinetic energy (KE) of the electrons emitted from two identical metal plates when photons of different energies strike them. The work function of the metal plates is given, and we will use the photoelectric effect equation to find the required ratio.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Energy of photon 1, \( E_1 = 1 \, \text{eV} \)
- Energy of photon 2, \( E_2 = 2.5 \, \text{eV} \)
- Work function of the metal, \( \phi = 0.5 \, \text{eV} \)

2. **Use the Photoelectric Effect Equation:**
The maximum kinetic energy of the emitted electrons can be calculated using the equation:
\[ KE_{\text{max}} = E - \phi \]
where \( E \) is the energy of the incident photon and \( \phi \) is the work function.

3. **Calculate Maximum Kinetic Energy for Photon 1:**
\[ KE_{\text{max1}} = E_1 - \phi = 1 \, \text{eV} - 0.5 \, \text{eV} = 0.5 \, \text{eV} \]

4. **Calculate Maximum Kinetic Energy for Photon 2:**
\[ KE_{\text{max2}} = E_2 - \phi = 2.5 \, \text{eV} - 0.5 \, \text{eV} = 2 \, \text{eV} \]

5. **Find the Ratio of Maximum Kinetic Energies:**
To find the ratio of the maximum kinetic energies of the electrons emitted from the two surfaces, we calculate:
\[ \text{Ratio} = \frac{KE_{\text{max1}}}{KE_{\text{max2}}} = \frac{0.5 \, \text{eV}}{2 \, \text{eV}} = \frac{1}{4} \]

6. **Final Answer:**
The ratio of the maximum kinetic energies of the electrons emitted from the two surfaces is:
\[ \text{Ratio} = 1:4 \]