The light of two different frequencies whose photons have energies $$3.8$$ eV and $$1.4$$ eV respectively, illuminate a metallic surface whose work function is $$0.6$$ eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :

We are given two photon energies $$E_1 = 3.8$$ eV and $$E_2 = 1.4$$ eV, and the work function $$\phi = 0.6$$ eV. We need to find the ratio of maximum speeds of emitted electrons.

By Einstein's photoelectric equation, the maximum kinetic energy is

$$\frac{1}{2}mv_{max}^2 = E_{photon} - \phi$$

For the first photon, substituting gives

$$KE_1 = E_1 - \phi = 3.8 - 0.6 = 3.2 \text{ eV}$$

Similarly, for the second photon,

$$KE_2 = E_2 - \phi = 1.4 - 0.6 = 0.8 \text{ eV}$$

Since kinetic energy is related to speed by $$KE = \frac{1}{2}mv^2$$, the ratio of squared speeds is

$$\frac{v_1^2}{v_2^2} = \frac{KE_1}{KE_2} = \frac{3.2}{0.8} = 4$$

Taking the square root, we find

$$\frac{v_1}{v_2} = \sqrt{4} = 2$$

From this, the ratio of maximum speeds is $$v_1 : v_2 = 2 : 1$$. Therefore, the correct option is Option A: $$2 : 1$$.