Light of wavelength 3500 Å is incident on two metals A and B, A of work function 4.2 eV and B of work function 1.19 eV respectively. The photoelectrons will be emitted by:

metal B

Energy of the incident light, $E=\frac{hc}{λ}=\frac{6.63×10^{34}×3×10^8}{3500×10^{-10}×1.6×10^{-19}}eV=3.55eV$

So, only B will emit photoelectrons as E > work function of B.