The work function of a metal is $$3 eV$$. The color of the visible light that is required to cause emission of photoelectrons is

We are given that the work function of a metal is $$\phi = 3 \, eV$$. We need to find which color of visible light can cause emission of photoelectrons.

For photoelectron emission to occur, the energy of the incident photon must be at least equal to the work function of the metal:
$$E \geq \phi$$

The energy of a photon is related to its wavelength by:
$$E = \frac{hc}{\lambda}$$

where $$hc = 1240 \, eV \cdot nm$$.

For the minimum photon energy of $$3 \, eV$$, the maximum wavelength is:
$$\lambda_{max} = \frac{hc}{\phi} = \frac{1240}{3} \approx 413 \, nm$$

So, only light with wavelength $$\lambda \leq 413 \, nm$$ can cause photoemission.

Now let us check which color of visible light satisfies this condition:

- Red light: $$\lambda \approx 620 - 750 \, nm$$ (energy $$\approx 1.65 - 2.0 \, eV$$) — Not sufficient
- Yellow light: $$\lambda \approx 570 - 590 \, nm$$ (energy $$\approx 2.1 - 2.2 \, eV$$) — Not sufficient
- Green light: $$\lambda \approx 495 - 570 \, nm$$ (energy $$\approx 2.2 - 2.5 \, eV$$) — Not sufficient
- Blue light: $$\lambda \approx 400 - 495 \, nm$$ (energy $$\approx 2.5 - 3.1 \, eV$$) — Sufficient at shorter wavelengths

Blue light at its shorter wavelength end (around $$400 - 413 \, nm$$) has energy of about $$3.0 - 3.1 \, eV$$, which is equal to or greater than the work function of $$3 \, eV$$. Therefore, blue light can cause photoelectron emission.

Red, yellow, and green light all have energies less than $$3 \, eV$$ and cannot cause photoemission from this metal.

Hence, the correct answer is Option B.