Q 12-11-109JEE MainJEE Main 2022 (25 Jul, Shift 1)Medium
A metal exposed to light of wavelength $800\ \text{nm}$ emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electrons doubles when light of wavelength $500\ \text{nm}$ is used. The work function of the metal is (Take $hc = 1230\ \text{eV nm}$)
Answer: (C) $0.615\ \text{eV}$
Photon energies: $\dfrac{1230}{800} = 1.5375\ \text{eV}$ and $\dfrac{1230}{500} = 2.46\ \text{eV}$.
$$2.46 - \phi = 2(1.5375 - \phi)\ \Rightarrow\ \phi = 3.075 - 2.46 = 0.615\ \text{eV}$$
Solution by Sreeraj P, M.Sc Physics