Answer
(a) $\phi = 4.31~eV$
(b) $K_{max} = 4.56~eV$
Work Step by Step
(a) We can find the work function $\phi$:
$\phi = \frac{hc}{\lambda_0}$
$\phi = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{288\times 10^{-9}~m}$
$\phi = 6.9\times 10^{-19}~J$
$\phi = (6.9\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$\phi = 4.31~eV$
(b) We can find the maximum kinetic energy:
$K_{max} = \frac{hc}{\lambda}-\phi$
$K_{max} = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{140\times 10^{-9}~m}-4.31~eV$
$K_{max} = (1.42\times 10^{-18}~J)-(4.31~eV)$
$K_{max} = (1.42\times 10^{-18}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})-(4.31~eV)$
$K_{max} = 4.56~eV$
