Answer
$\Phi = 2.27~eV$
Work Step by Step
In part (a), we found that $~~h = 6.59\times 10^{-34}~J~s$
We can find the work function:
$hf = K_{max}+\Phi$
$\frac{hc}{\lambda} = K_{max}+\Phi$
$\Phi = \frac{hc}{\lambda} - K_{max}$
$\Phi = \frac{(6.59\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{400\times 10^{-9}~m} - 0.820~eV$
$\Phi = 4.9425\times 10^{-19}~J-0.820~eV$
$\Phi = (4.9425\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})-0.820~eV$
$\Phi = 2.27~eV$
