A light source of wavelength `lambda` illuminates a metal and ejects photo-electrons with `(K.E.)_(max) =1.eV` Another light source of wavelength `(lambda)/(3)`, ejects photo-electrons from same metal with `(K.E.)_(max) = 4eV` Find the value of work function?

Options

A. `1eV`
B. `2eV`
C. `0.5eV`
D. `1.5eV`

Explanation

To find the work function of the metal, we can use the photoelectric effect equation, which relates the energy of the incident photons to the work function and the kinetic energy of the emitted photoelectrons.

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect Equation**:
The equation for the photoelectric effect is given by:
\[ E = \phi + (K.E.)_{max} \]
where \(E\) is the energy of the incident photon, \(\phi\) is the work function, and \((K.E.)_{max}\) is the maximum kinetic energy of the emitted photoelectrons.

2. **Calculate the Energy of the First Light Source**:
For the first light source with wavelength \(\lambda\):
\[ E_1 = \frac{hc}{\lambda} \]
The maximum kinetic energy is given as \(1 \, \text{eV}\). Therefore, we can write:
\[ \frac{hc}{\lambda} = \phi + 1 \]
This is our **Equation 1**.

3. **Calculate the Energy of the Second Light Source**:
For the second light source with wavelength \(\frac{\lambda}{3}\):
\[ E_2 = \frac{hc}{\frac{\lambda}{3}} = \frac{3hc}{\lambda} \]
The maximum kinetic energy here is \(4 \, \text{eV}\). Thus, we can write:
\[ \frac{3hc}{\lambda} = \phi + 4 \]
This is our **Equation 2**.

4. **Set Up the Equations**:
We now have two equations:
- Equation 1: \(\frac{hc}{\lambda} = \phi + 1\)
- Equation 2: \(\frac{3hc}{\lambda} = \phi + 4\)

5. **Multiply Equation 1 by 3**:
To eliminate \(\phi\), we can multiply Equation 1 by 3:
\[ 3 \left(\frac{hc}{\lambda}\right) = 3\phi + 3 \]
This gives us:
\[ \frac{3hc}{\lambda} = 3\phi + 3 \]

6. **Subtract Equation 2 from the Modified Equation 1**:
Now, we can subtract Equation 2 from this modified equation:
\[ (3\phi + 3) - (\phi + 4) = 0 \]
Simplifying this gives:
\[ 2\phi - 1 = 0 \]
Therefore:
\[ 2\phi = 1 \quad \Rightarrow \quad \phi = \frac{1}{2} \, \text{eV} \]

7. **Conclusion**:
The work function \(\phi\) of the metal is:
\[ \phi = 0.5 \, \text{eV} \]