Ans : (a) $6.592 \times 10^{-34} Js$
The slope of the cut-off voltage(V) versus frequency(v)
of an incident light is given as:
$V/v = 4.12 \times 10^{-15} Vs$
$V$ is related to frequency by the equation: $hv=eV$
So, $h=e \times V/v = 1.6 \times 10^{-19} \times 4.12 \times 10^{-15} = 6.592 \times 10^{-34} Js$