Photoelectric Effect

Problems from IIT JEE

Problem (IIT JEE 2014): A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_1$ and $u_2$, respectively. If the ratio $u_1:u_2=2:1$ and $hc={1240}\;\mathrm{eV\text{-}nm}$, the work function of the metal is nearly

  1. 3.7 eV
  2. 3.2 eV
  3. 2.8 eV
  4. 2.5 eV

Solution: The maximum kinetic energy of ejected photoelectrons is given by, \begin{align} \label{txb:eqn:1} K_\text{max}=\frac{1}{2}mu^2=\frac{hc}{\lambda}-\phi. \end{align} Substitute the values in above equation to get, \begin{align} \label{txb:eqn:2} &\frac{1}{2}mu_1^2=\frac{1240}{248}-\phi, \\ \label{txb:eqn:3} &\frac{1}{2}mu_2^2=\frac{1240}{310}-\phi. \end{align} Divide above equations and use $u_1/u_2=2$ to get $\phi={3.7}\;\mathrm{eV}$.