Step 1: Einstein’s Photoelectric Equation
\[ hf = \phi + K_{\max} \]
Photon energy:
\[ E = \frac{hc}{\lambda}, \quad hc = 12400 \, \text{eV·\AA} \]
Step 2: Calculate photon energies
For \( \lambda_1 = 3100 \, \text{\AA} \):
\[ E_1 = \frac{12400}{3100} = 4 \, \text{eV} \]
For \( \lambda_2 = 1550 \, \text{\AA} \):
\[ E_2 = \frac{12400}{1550} = 8 \, \text{eV} \]
Step 3: Apply given condition
\[ E_1 = \phi + K_1 \] \[ E_2 = \phi + 3K_1 \]
Subtracting:
\[ E_2 - E_1 = 2K_1 \]
\[ 8 - 4 = 2K_1 \Rightarrow K_1 = 2 \, \text{eV} \]
Step 4: Find work function
\[ \phi = E_1 - K_1 = 4 - 2 = 2 \, \text{eV} \]
Final Answer: \( 2 \, \text{eV} \)