1. Use Einstein’s photoelectric equation

For incident light of wavelength $\lambda$, $K_{\max} = \frac{hc}{\lambda} - \phi$ where $\phi$ is the work function.

Let the maximum kinetic energy for $800\,\text{nm}$ be $K$. Then for $500\,\text{nm}$, it is given to be $2K$.

So, $\frac{hc}{800} - \phi = K \quad ...(1)$ $\frac{hc}{500} - \phi = 2K \quad ...(2)$

2. Compute photon energies

Given: $hc = 1230\,\text{eV-nm}$

Thus, $\frac{hc}{800} = \frac{1230}{800} = 1.5375\,\text{eV}$ $\frac{hc}{500} = \frac{1230}{500} = 2.46\,\text{eV}$

So equations become: $1.5375 - \phi = K \quad ...(1)$ $2.46 - \phi = 2K \quad ...(2)$

3. Eliminate $K$

From (1), $K = 1.5375 - \phi$

Substitute into (2): $2.46 - \phi = 2(1.5375 - \phi)$ $2.46 - \phi = 3.075 - 2\phi$

Bring terms together: $2.46 + \phi = 3.075$ $\phi = 3.075 - 2.46 = 0.615\,\text{eV}$

4. Match with options

$\phi = 0.615\,\text{eV}$ So the correct option is C.