Light of two different frequencies whose photons have energies 1eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum kinetic energy of emitted electrons will be:

Options

A. `1:2`
B. `1:1`
C. `1:5`
D. `1:4`

Answer

To solve the problem, we will use the photoelectric equation, which relates the energy of the incident photons, the work function of the material, and the maximum kinetic energy of the emitted electrons.

### Step-by-Step Solution:

1. **Identify the energies of the photons and the work function**:
- Energy of photon 1 (\(E_1\)) = 1 eV
- Energy of photon 2 (\(E_2\)) = 2.5 eV
- Work function (\(\phi\)) = 0.5 eV

2. **Use the photoelectric equation**:
The photoelectric equation is given by:
\[ K_{\text{max}} = E - \phi \]
where \(K_{\text{max}}\) is the maximum kinetic energy of the emitted electrons, \(E\) is the energy of the incident photon, and \(\phi\) is the work function.

3. **Calculate the maximum kinetic energy for each photon**:
- For photon 1:
\[ K_{\text{max1}} = E_1 - \phi = 1 \text{ eV} - 0.5 \text{ eV} = 0.5 \text{ eV} \]
- For photon 2:
\[ K_{\text{max2}} = E_2 - \phi = 2.5 \text{ eV} - 0.5 \text{ eV} = 2.0 \text{ eV} \]

4. **Find the ratio of the maximum kinetic energies**:
We need to find the ratio \( \frac{K_{\text{max1}}}{K_{\text{max2}}} \):
\[ \frac{K_{\text{max1}}}{K_{\text{max2}}} = \frac{0.5 \text{ eV}}{2.0 \text{ eV}} = \frac{1}{4} \]

5. **Express the ratio in simplest form**:
The ratio of maximum kinetic energies of emitted electrons is:
\[ 1 : 4 \]

### Final Answer:
The ratio of maximum kinetic energy of emitted electrons will be \(1 : 4\).

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