A radiation of wavelength `lamda` illuminates a metal and ejects photoelectrons of maximum kinetic energy of 1eV. Aother radiation of wavelength `(lamda)/(3)`, ejects photoelectrons of maximum kinetic energy of 4eV. What will be the work function of metal?

Options

A. 1eV
B. 2eV
C. 0.5eV
D. 3eV

Explanation

To solve the problem, we will use the photoelectric equation, which relates the energy of the incident photons to the work function of the metal and the kinetic energy of the emitted photoelectrons. The equation is given by:

\[ E = \phi + KE \]

Where:
- \( E \) is the energy of the incident photon,
- \( \phi \) is the work function of the metal,
- \( KE \) is the maximum kinetic energy of the emitted photoelectrons.

### Step 1: Write the equations for both cases

For the first case, where the wavelength is \( \lambda \) and the maximum kinetic energy is 1 eV:

\[ E_1 = \frac{hc}{\lambda} = \phi + 1 \text{ eV} \quad (1) \]

For the second case, where the wavelength is \( \frac{\lambda}{3} \) and the maximum kinetic energy is 4 eV:

\[ E_2 = \frac{hc}{\frac{\lambda}{3}} = \phi + 4 \text{ eV} \quad (2) \]

### Step 2: Simplify the equations

From equation (1):

\[ \frac{hc}{\lambda} = \phi + 1 \text{ eV} \]

From equation (2):

\[ \frac{3hc}{\lambda} = \phi + 4 \text{ eV} \]

### Step 3: Rearranging the equations

Rearranging both equations gives us:

1. \( \phi = \frac{hc}{\lambda} - 1 \text{ eV} \)
2. \( \phi = \frac{3hc}{\lambda} - 4 \text{ eV} \)

### Step 4: Set the equations equal to each other

Since both expressions equal \( \phi \), we can set them equal to each other:

\[ \frac{hc}{\lambda} - 1 = \frac{3hc}{\lambda} - 4 \]

### Step 5: Solve for \( \phi \)

Rearranging gives:

\[ -1 + 4 = \frac{3hc}{\lambda} - \frac{hc}{\lambda} \]

This simplifies to:

\[ 3 = \frac{2hc}{\lambda} \]

Now, we can express \( \phi \):

\[ \phi = \frac{hc}{\lambda} - 1 \]

Substituting \( \frac{hc}{\lambda} = 3 + 1 = 4 \):

\[ \phi = 4 - 1 = 3 \text{ eV} \]

### Step 6: Final Work Function Calculation

Thus, the work function \( \phi \) of the metal is:

\[ \phi = 3 \text{ eV} \]

### Conclusion

The work function of the metal is **3 eV**.

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