Answer
$2.36\;eV$
Work Step by Step
From the plot, we can write
$V_{stop}=\Big(\frac{h}{e}\Big)f-\frac{\phi}{e}$
or, $\phi=hf-eV_{stop}$
From the previous section of this problem we have calculated the value of $h$
$h=6.72\times 10^{-34}\;J.s$
Thus, at $f=9.6\times 10^{14}\;Hz$ and $V_{stop}=1.67\;V$,
$\phi=6.72\times 10^{-34}\times 9.6\times 10^{14}-1.6\times 10^{-19}\times1.67\;J$
or, $\phi\approx3.78\times 10^{-19}\;J$
or, $\phi\approx 2.36\;eV$
Therefore, the work function for lithium is $2.36\;eV$
