Answer
(a) Cadmium
(b) $Cad:0.30eV;Zn:0.19eV$
Work Step by Step
(a) We know that $K_{max}=\frac{hc}{\lambda}-W_{\circ}$. This equation shows that maximum kinetic energy of the photo electrons depend on the work function of the metal. Since the work function of the cadmium is less than the work function of zinc, therefore, the cadmium metal surface ejects the electrons with maximum kinetic energy.
(b) We know that
$K_{max,Zn}=\frac{hc}{\lambda}-(W_{\circ}Cd)$
We plug in the known values to obtain:
$K_{max,Zn}=\frac{(6.63\times 10^{-34}J.s)(3\times 10^8m/s)}{(275nm)(\frac{10^{-9}m}{1nm})-4.33eV}$
$\implies K_{max,Zn}=0.19eV$
Now for Cadmium:
$K_{max,Cd}=\frac{hc}{\lambda}-(W_{\circ})Cd$
We plug in the known values to obtain:
$K_{max, Cd}=\frac{(6.63\times 10^{-34}J.s)(3\times 10^8m/s)}{(275nm)(\frac{10^{-9}m}{1nm})}-4.22eV=0.30eV$