Use Einstein’s photoelectric condition

For photoelectric emission to occur, $h\nu \ge \phi$ where $\phi$ is the work function.

Given: $\phi = 3\,\text{eV}$

Find the threshold wavelength

Using $E = \frac{hc}{\lambda}$ and in convenient units, $E(\text{eV}) = \frac{1240}{\lambda(\text{nm})}$

At threshold, $\lambda_0 = \frac{1240}{3} \approx 413\,\text{nm}$

Interpret this wavelength in the visible spectrum

Typical visible wavelength ranges are:

• Red: $\sim 620$–$750\,\text{nm}$
• Yellow: $\sim 570$–$590\,\text{nm}$
• Green: $\sim 495$–$570\,\text{nm}$
• Blue: $\sim 450$–$495\,\text{nm}$
• Violet: $\sim 380$–$450\,\text{nm}$

The threshold wavelength $413\,\text{nm}$ lies in the violet region.

Check the given options

To emit photoelectrons, the incident light must have wavelength $\le 413\,\text{nm}$, i.e. frequency at least the threshold frequency.

Among the options:

• Red: too low energy
• Yellow: too low energy
• Green: too low energy
• Blue: wavelengths around $450$–$495\,\text{nm}$, still longer than $413\,\text{nm}$, so also insufficient

Therefore, none of the listed visible colors is correct. The correct visible color should be violet, which is not given.

Conclusion

The physically correct answer is that violet light is required. Since violet is not among the options, the stored answer $\text{C (Blue)}$ is not correct.