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

Answer

To solve the problem, we need to find the ratio of the maximum kinetic energy of emitted electrons when light of two different frequencies illuminates a metallic surface. The energies of the photons are given as 1 eV and 2.5 eV, and the work function of the metal is 0.5 eV.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The energy of a photon is given by the equation:
\[ E = h \nu \]
where \(E\) is the energy of the photon, \(h\) is Planck's constant, and \(\nu\) is the frequency of the light. The energy of the emitted electrons can be calculated using the equation:
\[ K.E. = E - \phi \]
where \(K.E.\) is the kinetic energy of the emitted electrons and \(\phi\) is the work function of the metal.

2. **Calculating Kinetic Energy for the First Photon (1 eV)**:
For the first photon with energy \(E_1 = 1 \, \text{eV}\):
\[ K.E._1 = E_1 - \phi = 1 \, \text{eV} - 0.5 \, \text{eV} = 0.5 \, \text{eV} \]

3. **Calculating Kinetic Energy for the Second Photon (2.5 eV)**:
For the second photon with energy \(E_2 = 2.5 \, \text{eV}\):
\[ K.E._2 = E_2 - \phi = 2.5 \, \text{eV} - 0.5 \, \text{eV} = 2.0 \, \text{eV} \]

4. **Finding the Ratio of Maximum Kinetic Energies**:
Now, we need to find the ratio of the maximum kinetic energies of the emitted electrons:
\[ \text{Ratio} = \frac{K.E._1}{K.E._2} = \frac{0.5 \, \text{eV}}{2.0 \, \text{eV}} = \frac{1}{4} \]

### Final Answer:
The ratio of the maximum kinetic energy of emitted electrons is:
\[ \frac{1}{4} \]