Answer
Please see the work below.
Work Step by Step
(a) We know that the metal surface with less work function ($W_{\circ}$) ejects electrons with greater maximum kinetic energy. It is given that the work function of the platinum metal surface is greater than the work function of the iron metal surface. Thus, the electrons ejected from the iron surface have greater maximum kinetic energy.
(b) We can find the maximum kinetic energy for the platinum surface as follows:
$K_{max}=hf-W_{pt}$
We plug in the known values to obtain:
$K_{max}=(6.63\times 10^{-34}J.s)(1.88\times 10^{15}Hz)-(6.35eV)$
$\implies K_{max}=1.44eV$
Similarly, for the iron surface
$K_{max}=hf-W_{Fe}$
We plug in the known values to obtain:
$K_{max}=(6.63\times 10^{-34}J.s)(1.88\times 10^5Hz)-4.50eV$
$K_{max}=(12.4644\times 10^{-19}J)(\frac{1eV}{1.6\times 10^{-19}J})-4.50eV$
$\implies K_{max}=3.29eV$
