JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Modern Physics

Einstein's photoelectric relation gives the maximum kinetic energy as $K=\frac{hc}{\lambda }-\varphi$. At ${\lambda }_1=500$ nm we have $K_1=\frac{hc}{{\lambda }_1}-\varphi$, and at ${\lambda }_2=200$ nm the energy triples, $K_2=\frac{hc}{{\lambda }_2}-\varphi =3K_1$. Substituting and clearing, $\frac{hc}{{\lambda }_2}-\varphi =3\left(\frac{hc}{{\lambda }_1}-\varphi \right)$, so $2\varphi =hc\left(\frac{3}{{\lambda }_1}-\frac{1}{{\lambda }_2}\right)$. With $hc=1240$ eV nm, $\frac{3}{500}-\frac{1}{200}=0.001$ nm${}^{-1}$, giving $\varphi =\frac{1240\times 0.001}{2}=0.62$ eV, which is close to $0.61$ eV.