The following relation can be derived from photoelectric effect,
$eV_{o}=hv-\phi_{o}$
Work function of the metal, $\Phi_{0}=\mathrm{hv}-\mathrm{eV}_{0}$
$\Phi_{0}=(h c / \lambda)-e V_{0}$
$V_{0}=\frac{h}{e} \nu-\frac{\phi_{0}}{e}—(1)$
Here,
$\mathrm{V}_{0}$ is the Stopping potential
$\mathrm{h}$ is the Planck’s constant
$\mathrm{e}$ is the Charge on an electron
$\mathrm{v}$ is the Frequency of radiation
$\Phi_{0}$ is the Work function of a material
As we know, stopping proportional is directly proportional to the frequency from the relation,
Frequency, $v=$ Speed of light $(c) /$ Wavelength $(\lambda)$
Using this above equation, we can calculate the frequency of various lines
$v_{1}=c/\lambda_{1}=3 \times 10^{8} / 3650 \times 10^{-10}$
$=8.219 \times 10^{14} \mathrm{~Hz}$
$v_{1}=c/\lambda_{2}=3 \times 10^{8} / 4047 \times 10^{-10}$
$=7.412 \times 10^{14} \mathrm{~Hz}$
$v_{1}=c/\lambda_{3}=3 \times 10^{8} / 4358 \times 10^{-10}$
$=6.88 \times 10^{14} \mathrm{~Hz}$
$v_{1}=c/\lambda_{4}=3 \times 10^{8} / 5461 \times 10^{-10}$
$=5.493 \times 10^{14} \mathrm{~Hz}$
$v_{1}=c/\lambda_{5}=3 \times 10^{8} / 6907 \times 10^{-10}$
$=4.343 \times 10^{14} \mathrm{~Hz}$
The above values can be plotted in a graph
A straight line can be seen in the graph. At $5 \times 10^{14} \mathrm{~Hz}$, it crosses the y-axis. This is the frequency at which the threshold is reached. The frequency less than the threshold frequency is represented by point D.
Slope of the straight line will be $=A B / C B=(1.28-0.16) /(8.214-5.493) \times 10^{14}$
Using equation (1), the slope can be written as
$h / e=(1.28-0.16) /(8.214-5.493) \times 10^{14}$
$h=\left(1.12 \times 1.6 \times 10^{-19}\right) /\left(2.726 \times 10^{14}\right)$
$=6.573 \times 10^{-34} \mathrm{~J} \mathrm{~s}$
The work function of the metal is expressed as,
$\Phi_{0}=\mathrm{hv}_{0}$
$=\left(6.573 \times 10^{-34} \times 5 \times 10^{14}\right)$
$=3.286 \times 10^{-19} \mathrm{~J}$
$=3.286 \times 10^{-19} / 1.6 \times 10^{-19}$
$\Phi_{0}=2.054 \mathrm{eV}$