Answer
a) $ 6.56\times10^{-34 }\;\rm J.s$
b) $5.52\times10^{14}\;\rm Hz$
c) $2.29\;\rm eV$
Work Step by Step
First of all, we need to plot the graphs where the frequency is in the $x$-direction while the energy is in the $y$-direction, as you see below.
a) We know that the energy is given by
$$ E=hf$$
where $h$ is Planck’s constant and $f$ is the frequency.
Thus,
$$h=\dfrac{E}{f} =\rm slope $$
Thus, the slope in the graph below is given by
$$h={\rm slope}=\dfrac{0.41\rm \;eV}{10^{14} \;\rm Hz} $$
$$h =\rm\dfrac{0.41\rm \;eV}{10^{14} \;\rm Hz} \times \dfrac{1.6\times10^{-19}\;\rm J}{1\;eV}=\color{red}{\bf 6.56\times10^{-34}}\;\rm J.s $$
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b) We know that the the cutoff frequency of sodium is given by the $x$-intercept. So, as we see in the figure below, the cutoff frequency is
$$f_{\rm cutoff}=\color{red}{\bf 5.52\times10^{14}}\;\rm Hz$$
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c) We know that the work function is given by the opposite of the $y$-intercept.
Thus, as we see below,
$$W_0=-W_{y-{\rm intercept}}=-(-2.29)=\color{red}{\bf2.29}\;\rm eV$$
