1. For photoelectric effect, the stopping potential and frequency are related by

$eV_0 = h\nu - \phi$

where:

• $\phi$ = work function
• $h$ = Planck's constant
• $e$ = electronic charge

So,

$V_0 = \frac{h}{e}\nu - \frac{\phi}{e}$

This is a straight line in $V_0$ vs $\nu$ graph.

2. The work function is obtained from the threshold frequency $\nu_0$, where stopping potential becomes zero:

$V_0=0 \Rightarrow h\nu_0 = \phi$

Hence,

$\phi = h\nu_0$

3. From the graph, the intercept on the frequency axis (threshold frequency) is approximately

$\nu_0 \approx 5 \times 10^{14}\ \text{Hz}$

4. Now calculate work function:

Using

$h = 6.626 \times 10^{-34}\ \text{J s}$

$\phi = h\nu_0 = 6.626 \times 10^{-34} \times 5 \times 10^{14}$

$\phi = 3.313 \times 10^{-19}\ \text{J}$

5. Convert joule into electron-volt:

$1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$

So,

$\phi = \frac{3.313 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.07\ \text{eV}$

6. Therefore, the work function is

$\boxed{2.07\ \text{eV}}$

7. Checking options:

• A: $1.36\ \text{eV}$ — incorrect
• B: $18.6\ \text{eV}$ — incorrect
• C: $2.98\ \text{eV}$ — incorrect
• D: $2.07\ \text{eV}$ — correct