Answer
$6.72\times 10^{-34}\;J.s$
Work Step by Step
The following is the stopping potential data for lithium where we have converted the wavelength into frequency
Wavelength (nm):$\;\;\;\;\;\;433.9\;\;\;404.7\;\;\;365.0\;\;\;312.5\;\;\;253.5$
Frequency ($10^{14}$ Hz):$\;\;6.914\;\;\;7.413\;\;\;8.219\;\;\;9.600\;\;\; 11.834$
Stopping potential (V):$\;0.55\;\;\;\;0.73\;\;\;\;1.09 \;\;\;\;\;1.67\;\;\;\;\;2.57$
The plot of the measured stopping potential $(V_{stop})$ versus the frequency $(f)$ of the light is a straight line. Further, the slope of that straight line should be $\frac{h}{e}$. Therefore, from the following plot, we can write
$\frac{h}{e}=\frac{ab}{bc}$
or, $h=e\Big(\frac{ab}{bc}\Big)$
or, $h=1.6\times 10^{-19}\times\Big(\frac{1.67-1.09}{9.6\times 10^{14}-8.219\times 10^{14}}\Big)\;J.s$
or $h=6.72\times 10^{-34}\;J.s$
Therefore, the value of Planck constant is $6.72\times 10^{-34}\;J.s$
