Answer
(a) The work function is $~~4.14~eV$
(b) $h = 4.0\times 10^{-15}~eV~s$
Work Step by Step
(a) From the graph, we can see that the threshold frequency is $1.0\times 10^{15}~Hz$
Thus, the energy of each photon of this frequency is equal to the work function.
We can find the energy of each photon:
$E = h~f$
$E = (6.626\times 10^{-34}~J~s)(1.0\times 10^{15}~Hz)$
$E = 6.626\times 10^{-19}~J$
$E = (6.626\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$E = 4.14~eV$
The work function is $~~4.14~eV$
(b) We can write an equation for the stopping potential:
$K_{max} = hf- E_0$
$\Delta V~e = hf- E_0$
$\Delta V = \frac{h}{e}~f- \frac{E_0}{e}$
This equation is a linear equation where $\frac{h}{e}$ is the slope of the $\Delta V$ versus $f$ graph.
We can find $h$:
$\frac{h}{e} = \frac{8.0~V - 0~V}{(3.0\times 10^{15}~Hz)-(1.0\times 10^{15}~Hz)}$
$h = (e)~(4.0\times 10^{-15}~V~s)$
$h = 4.0\times 10^{-15}~eV~s$