A metal has threshold (cut-off) wavelength $\lambda_0 = 500\,\text{nm}$ for photoelectric emission. The work function of the metal is closest to (Take $hc = 1240\,\text{eV\,nm}$.)

1. At the threshold, the photon has just enough energy to free an electron with zero kinetic energy: $h\nu_0 = \phi_0$. 2. In wavelength form, this becomes $\phi_0 = \dfrac{hc}{\lambda_0}$. 3. Plug in: $\phi_0 = \dfrac{1240\,\text{eV\,nm}}{500\,\text{nm}} = 2.48\,\text{eV}$. 4. Sanity check: $2.48\,\text{eV}$ is in the alkali-metal range (Cs is $2.14$, Na is $2.28$, K is $2.30$, Rb is $2.16$). Consistent with a green/blue-cutoff metal. 5. Option B ($3.10\,\text{eV}$) corresponds to $\lambda_0 = 400\,\text{nm}$. Options C and D are work functions of heavier metals (e.g. zinc $4.3\,\text{eV}$, tungsten $4.5\,\text{eV}$, platinum $\sim 5.6\,\text{eV}$) — these don't match the given threshold. _Source: NCERT Class 12 Physics Part 2, Ch 11, §11.6 (relation $\nu_0 = \phi_0/h$, with $\lambda_0 = c/\nu_0$), p. 6–7._