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

$K_{\max }=\frac{hC}{{\lambda }_{\min }}-\varphi$

For A $\Rightarrow K_{\max }=3.1-1.55=1.55eV$ 
For B  $\Rightarrow K_{\max }=3.1-2.48=0.62eV$
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.

$\begin{array}{l}\underset{400}{\overset{700}{\int }}\frac{P}{\left(\frac{hC}{\lambda }\right)}d\lambda ={\text{K\hspace{0.17em}I}}_1\\ \underset{400}{\overset{500}{\int }}\frac{P}{\left(\frac{hC}{\lambda }\right)}d\lambda ={\text{K\hspace{0.17em}I}}_1\\ {\text{I}}_{\text{2}}=\frac{18}{11}\text{mA}\end{array}$