Answer
(a) $K_{max} = 0.848~eV$
(b) $\lambda = 575~nm$
Work Step by Step
(a) We can find the maximum kinetic energy:
$K_{max} = hf-\phi$
$K_{max} = \frac{hc}{\lambda}-\phi$
$K_{max} = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{413\times 10^{-9}~m}-2.16~eV$
$K_{max} = (4.813\times 10^{-19}~J)-(2.16~eV)$
$K_{max} = (4.813\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})-(2.16~eV)$
$K_{max} = (3.008~eV)-(2.16~eV)$
$K_{max} = 0.848~eV$
(b) We can find the threshold wavelength:
$E = \frac{hc}{\lambda}$
$\lambda = \frac{hc}{E}$
$\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(2.16~eV)(1.6\times 10^{-19}~J/eV)}$
$\lambda = 5.75\times 10^{-7}~m$
$\lambda = 575~nm$
