Consider white light whose wavelength spread is from 400 nm to 700 nm. Its energy is uniformly distributed in this spectrum. The light is incident on a metal A of work function 1.55 eV. Saturation photo current is 6 mA. Now the same light is incident on metal B, work function 2.48 eV. [Take hc = 1240 eV nm. Assume photo efficiency remains same]

$\begin{array}{l}{\mathrm{K}}_{max}=\frac{\mathrm{hC}}{{\lambda }_{min}}-\varphi \\ Energy of photon having wavelength 400nm=\frac{12400}{4000}ev = 3.1 ev\\ Energy of photon having wavelength 700 nm =\frac{12400}{7000}ev = 1.77 ev\\ Energy of photon having wavelength 500 nm = \frac{12400}{5000}ev =2.48 ev\\ \\ \text{ For }A\Rightarrow {\mathrm{K}}_{max}=3.1-1.55=1.55\mathrm{eV}\\ \text{ For }\mathrm{B}\Rightarrow {\mathrm{K}}_{max}=3.1-2.48=0.62\mathrm{eV}\end{array}$

In experiment with metal A complete spectrum is able to eject photo electrons but for metal B spectrum from 400nm to 500nm, will be able to eject photo electrons. Number of photons in spectrum 400nm to 500nm will be less than 1/3 of total photons in white light.