Answer
Please see the work below.
Work Step by Step
(a) We know that according to the photoelectric effect, the maximum kinetic energy and photon wavelength are related as
$K_{max}=\frac{hc}{\lambda}-W_{\circ}$
But $K_{max}=\frac{1}{2}mv_{max}^2$
$\implies \frac{1}{2}mv_{max}^2=\frac{hc}{\lambda}-W_{\circ}$
This simplifies to:
$W_{\circ}=\frac{hc}{\lambda}-\frac{1}{2}mv_{max}^2$
(b) We know that
$W_{\circ}=\frac{hc}{\lambda}-\frac{1}{2}mv_{max}^2$
We plug in the known values to obtain:
$W_{\circ}=\frac{(6.63\times 10^{-34}J.s)(3\times 10^8m/s)}{545\times 10^{-9}m}-\frac{1}{2}(9.11\times 10^{-31}Kg)(3.10\times 10^5m/s)^2=3.2123\times 10^{-19}J=2.01eV$
Now, we can determine the cut-off frequency as
$f_{\circ}=\frac{W_{\circ}}{h}$
We plug in the known values to obtain:
$f_{\circ}=\frac{2.01eV}{(6.63\times 10^{-34}J.s)(\frac{1eV}{1.6\times 10^{-19}J})}$
$f_{\circ}=4.85\times 10^{14}Hz$
