Answer
$h = 6.59\times 10^{-34}~J~s$
Work Step by Step
We can write an expression for the photoelectric effect:
$hf = K_{max}+\Phi$
$\frac{hc}{\lambda} = K_{max}+\Phi$
Then:
$\frac{hc}{\lambda_1} = K_{max,1}+\Phi$
$\frac{hc}{\lambda_2} = K_{max,2}+\Phi$
We can subtract the first equation from the second equation:
$\frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} = K_{max,2} - K_{max,1}$
$hc~(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}) = K_{max,2} - K_{max,1}$
$h = \frac{K_{max,2} - K_{max,1}}{c~(\frac{1}{\lambda_2} - \frac{1}{\lambda_1})}$
$h = \frac{(0.820~eV-1.85~eV)(1.6\times 10^{-19}~J/eV)}{(3.0\times 10^8~m/s)~(\frac{1}{400\times 10^{-9}~m} - \frac{1}{300\times 10^{-9}~m})}$
$h = 6.59\times 10^{-34}~J~s$
