The work functions of two metals ($$M_{A}$$ and $$M_{B}$$ ) are in the 1 : 2 ratio. When these metals are exposed to photons of energy 6 eV, the kinetic energy of liberated electrons of $$M_{A}$$ : $$M_{B}$$ is in the ratio of 2.642 : 1. The work functions (in eV) of $$M_{A}$$ and $$M_{A}$$ are respectively.

The photoelectric effect equation states that the kinetic energy $$K$$ of an emitted electron is given by $$K = h\nu - \phi$$, where $$h\nu$$ is the photon energy and $$\phi$$ is the work function of the metal.

Given that the work functions of metals $$M_A$$ and $$M_B$$ are in the ratio 1:2, let $$\phi_A$$ be the work function of $$M_A$$ and $$\phi_B$$ be the work function of $$M_B$$. Therefore, $$\phi_B = 2\phi_A$$.

The photon energy is 6 eV for both metals. Thus:

For $$M_A$$: $$K_A = 6 - \phi_A$$

For $$M_B$$: $$K_B = 6 - \phi_B = 6 - 2\phi_A$$

The ratio of kinetic energies is given as $$K_A : K_B = 2.642 : 1$$, so:

$$\frac{K_A}{K_B} = \frac{6 - \phi_A}{6 - 2\phi_A} = 2.642$$

Solving for $$\phi_A$$:

$$6 - \phi_A = 2.642 \times (6 - 2\phi_A)$$

$$6 - \phi_A = 2.642 \times 6 - 2.642 \times 2\phi_A$$

$$6 - \phi_A = 15.852 - 5.284\phi_A$$

Bringing like terms together:

$$-\phi_A + 5.284\phi_A = 15.852 - 6$$

$$4.284\phi_A = 9.852$$

$$\phi_A = \frac{9.852}{4.284} \approx 2.3 \text{ eV}$$

Then, $$\phi_B = 2 \times \phi_A = 2 \times 2.3 = 4.6 \text{ eV}$$.

Verifying the kinetic energies:

$$K_A = 6 - 2.3 = 3.7 \text{ eV}$$

$$K_B = 6 - 4.6 = 1.4 \text{ eV}$$

Ratio: $$\frac{K_A}{K_B} = \frac{3.7}{1.4} = 2.642857... \approx 2.642$$

This matches the given ratio.

Thus, the work functions of $$M_A$$ and $$M_B$$ are 2.3 eV and 4.6 eV, respectively.

Comparing with the options:

A. 3.1, 6.2

B. 1.4, 2.8

C. 1.5, 3.0

D. 2.3, 4.6

The correct answer is option D.