Q 12-11-142JEE MainJEE Main 2020 (5 Sep, Shift 2)Easy
The surface of a metal is illuminated alternately with photons of energies $E_1 = 4$ eV and $E_2 = 2.5$ eV respectively. The ratio of the maximum speeds of the photoelectrons emitted in the two cases is $2$. The work function of the metal (in eV) is ______.
Numerical value type. Enter your answer.
Answer: 2
$\tfrac12mv^2 = E - \phi$. A speed ratio of $2$ means a kinetic energy ratio of $4$:
$$4 - \phi = 4(2.5 - \phi) \Rightarrow 3\phi = 6 \Rightarrow \phi = 2\ \text{eV}$$
Solution by Sreeraj P, M.Sc Physics