1. Use Einstein’s photoelectric equation

For photoelectric effect, $K_{\max}=h\nu-\phi$ where $\phi$ is the work function (minimum energy needed to eject the electron).

Also, if we plot maximum kinetic energy $K_{\max}$ versus frequency $\nu$, then:

• slope $= h$
• intercept on the energy axis $= -\phi$

2. Extract threshold frequency from the graph

From the given plotted data points, the straight line through the points cuts the frequency axis at approximately $\nu_0 \approx 5.5\times 10^{14}\ \text{Hz}$ This is the threshold frequency.

At threshold, $\phi = h\nu_0$

3. Substitute the values

Given: $h=6.62\times 10^{-34}\ \text{J s}$

So, $\phi=(6.62\times 10^{-34})(5.5\times 10^{14})$ $\phi=3.641\times 10^{-19}\ \text{J}$

4. Convert joule to electron-volt

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

Therefore, $\phi=\frac{3.641\times 10^{-19}}{1.6\times 10^{-19}}\ \text{eV}$ $\phi\approx 2.28\ \text{eV}$

This matches closest with: $\boxed{2.27\ \text{eV}}$

5. Check options

• A: $2.10\ \text{eV}$ — not matching
• B: $2.27\ \text{eV}$ — correct
• C: $2.59\ \text{eV}$ — not matching
• D: $1.93\ \text{eV}$ — not matching

Hence the minimum energy for ejection of electron is $\boxed{2.27\ \text{eV}}$