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

$eV_0 = h\nu - \phi$

where $\phi$ is the work function.

So,

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

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

2. From the graph, the threshold frequency $\nu_0$ is the frequency where $V_0 = 0$.

At threshold,

$0 = \frac{h}{e}\nu_0 - \frac{\phi}{e}$

Hence,

$\phi = h\nu_0$

3. From the plotted graph, the intercept on frequency axis is approximately

$\nu_0 = 4.0 \times 10^{14}\ \text{Hz}$

4. Therefore,

$\phi = h\nu_0 = 6.63 \times 10^{-34} \times 4.0 \times 10^{14}$

$\phi = 2.652 \times 10^{-19}\ \text{J}$

5. Convert into electron volt:

$\phi = \frac{2.652 \times 10^{-19}}{1.6 \times 10^{-19}}\ \text{eV}$

$\phi = 1.6575\ \text{eV} \approx 1.66\ \text{eV}$

6. Checking options:

• A: $1.95\ \text{eV}$ ❌
• B: $2.12\ \text{eV}$ ❌
• C: $1.82\ \text{eV}$ ❌
• D: $1.66\ \text{eV}$ ✅

Therefore, the work function of sodium is

$\boxed{1.66\ \text{eV}}$