Given that

Determining the value of Planck’s constant h
Step 1: Finding the frequencies of the different lines of the wavelengths given
Relation between the wavelength and frequency
\upsilon = \frac{c}{\lambda}
Where, speed of light c = 3*10^{8} m/s
Frequency \upsilon_1 = \frac{3*10^{8}}{3650 * 10^{-10}} = 8.22* 10^{14} Hz
\upsilon_2=\frac{3*10^8}{4047*10^{-10}}=7.412*10^{14}\ Hz
\upsilon_3 = \frac{3*10^{8}}{4358 * 10^{-10}} = 6.88 * 10^{14} Hz
\upsilon_4 = \frac{3*10^{8}}{5461 * 10^{-10}} = 5.493 * 10^{14} Hz
\upsilon_5 = \frac{3*10^{8}}{6901 * 10^{-10}} = 4.343 * 10^{14} Hz
Step 2: Make a graph with the frequency and voltage relationship.

Graph:

Step 3: Calculating the Planck’s constant
Energy of proton E=h\upsilon\ -\ \phi
eV=h\upsilon-\phi E=h\upsilon\ =\ \frac{hc}{\lambda}
Where, h - Planck's constant, speed of light c = 3*10^{8} m/s , \phi - work function, and \upsilon - frequency.
V = \frac{h}{e} \upsilon - \phi
Comparing the above equation with equation of straight line y = mx + c
Slope of the line m = \frac{h}{e} ......................(1)
Slope of line from the graph
The slope of the line connecting the two points A (5.493 * 10^{14}, 0.16) and B (8.219 * 10^{14}, 1.28)
Slope of the line = \frac{1.28 - 0.16}{8.219 * 10^{14} - 5.493 * 10^{14}}
Slope of the line = 4.1 * 10^{-15} ............................(2)
From equation (1) and (2)
\frac{h}{e} = 4.1 * 10^{-15}
h = 4.1 * 10^{-15} * 1.6 * 10^{-19}
h = 6.4 * 10^{-34} Js
Hence, Planck’s constant h = 6.4 * 10^{-34} Js
Step 4: Finding the work function
Observation from the graph:
The obtained curve is a straight line, as can be shown. It touches the frequency-axis at 5.4 * 10^{14} Hz, which is the
material's threshold frequency ( \upsilon_0 ). A frequency less than the threshold frequency corresponds to point H. As a result,
the \lambda_5 line produces no photoelectric emission, and thus no stopping voltage is needed to stop the current.
Relation between work function and frequency
\phi = h \upsilon_0
\phi = 6.4 * 10^{-34} * 5.4 * 10^{14}
\phi = 3.38 * 10^{-19} J
\phi = 2 ev
Therefore, work function \phi=2eV