Light of wavelength $3500 \AA$ is incident on two metals A and B whose work functions are 4.2 eV and 1.9 eV respectively. Which metal will emit photoelectrons? 

$\mathrm{\lambda }=3500 \mathrm{\AA }=3500\times {10}^{-10} \mathrm{m}=3.5\times {10}^{-7} \mathrm{m}$

The energy of incident light is given by $\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$ 

$\Rightarrow \mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda e}}$ eV

where $\mathrm{h}=6.6\times {10}^{-34}\mathrm{Js}$ is the Planck's constant, $\mathrm{c}=3\times {10}^8 \mathrm{m}/\mathrm{s}$ is the speed of light in air and $\mathrm{e}=1.6\times {10}^{-19}\mathrm{C}$ is the charge of electron. Substituting the values of h, c and e and solving we get $\mathrm{E}\simeq 3.55 \mathrm{eV}$.

Photoelectrons will be emitted from a metal if the energy of the incident light is greater than the work function. Since the work function of metal B is less than E, metal B will emit photoelectrons.