Q 12-11-035JEE MainJEE Main 2026 (8 Apr, Shift 2)Medium
$K_1$ and $K_2$ be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength $\lambda_1$ and $\lambda_2$, respectively. If $\lambda_1 = 2\lambda_2$ then the work function of material is given by :
Answer: (D) $K_2 - 2K_1$
Let $E = \dfrac{hc}{\lambda_1}$. Since $\lambda_2 = \dfrac{\lambda_1}{2}$, the second photon has energy $2E$.
$K_1 = E - \phi$ and $K_2 = 2E - \phi$.
From the first, $E = K_1 + \phi$. Substituting: $K_2 = 2K_1 + 2\phi - \phi = 2K_1 + \phi$, so $\phi = K_2 - 2K_1$.
Solution by Sreeraj P, M.Sc Physics