Photons of energy 1.5 eV and 2.5 eV are incident on a metal surface of work function 0.5 eV. What is the ratio of the maximum kinetic energy of the photoelectrons ?

Options

A. `1:2`
B. 2
C. 3
D. `1:4`

Explanation

To solve the problem, we need to calculate the maximum kinetic energy of the photoelectrons emitted when photons of energy 1.5 eV and 2.5 eV strike a metal surface with a work function of 0.5 eV.

### Step-by-Step Solution:

1. **Identify the given values:**
- Energy of the first photon (E1) = 1.5 eV
- Energy of the second photon (E2) = 2.5 eV
- Work function of the metal (W0) = 0.5 eV

2. **Use the photoelectric effect equation:**
The maximum kinetic energy (KE) of the emitted photoelectrons can be calculated using the formula:
\[ KE = E - W_0 \]
where \(E\) is the energy of the incident photon and \(W_0\) is the work function.

3. **Calculate the maximum kinetic energy for the first photon:**
\[ KE_1 = E_1 - W_0 = 1.5 \, \text{eV} - 0.5 \, \text{eV} = 1.0 \, \text{eV} \]

4. **Calculate the maximum kinetic energy for the second photon:**
\[ KE_2 = E_2 - W_0 = 2.5 \, \text{eV} - 0.5 \, \text{eV} = 2.0 \, \text{eV} \]

5. **Find the ratio of the maximum kinetic energies:**
\[ \text{Ratio} = \frac{KE_1}{KE_2} = \frac{1.0 \, \text{eV}}{2.0 \, \text{eV}} = \frac{1}{2} \]

### Final Answer:
The ratio of the maximum kinetic energy of the photoelectrons is \( \frac{1}{2} \).