Answer
a) Kinetic energy will be larger for the beam having the smaller wavelength (B).
b)
$KE_{max}=1.6\times 10^{-20}J$ and $KE_{max}=1.79\times 10^{-19}J$
Work Step by Step
(a) We know that energy is inversely proportional to the wavelength. Thus, kinetic energy will be larger for the beam having the smaller wavelength (B).
(b) We know that
$Energy=\frac{hc}{\lambda}$
We plug in the known values to obtain:
$Energy=\frac{6.63\times 10^{-34}\times 3\times 10^8}{620\times 10^{-9}}=3.2\times 10^{-19}J$
As given that $W_{\circ}=1.9eV=1.9\times 10^6\times 10^{-19}=3.2\times 10^{-19}J$
Now $KE_{max}=3.2\times 10^{-19}-3.04\times 10^{-19}=1.6\times 10^{-20}J$
We know that
$Energy=\frac{hc}{\lambda}$
We plug in the known values to obtain:
$Energy=\frac{6.63\times 10^{-34}\times 3\times 10^8}{410\times 10^{-9}}=4.85\times 10^{-19}J$
As given that $W_{\circ}=1.9eV=1.9\times 10^6\times 10^{-19}=3.2\times 10^{-19}J$
Now $KE_{max}=4.85\times 10^{-19}-3.04\times 10^{-19}=1.79\times 10^{-19}J$
