JEE Main 2025 Apr 3 Shift 1, Physics Q15: Introduction and Photoelectric Effect

For photoemission, the photon energy must satisfy $\frac{hc}{\lambda }\ge \varphi$, equivalently $\lambda \le \frac{hc}{\varphi }.$

Using $hc\approx 1242\text{eV-nm}$ and $\varphi =3$ eV:

${\lambda }_{\max }=\frac{1242}{3}=414\text{nm}.$

$\therefore$ Blue is the correct option.