The heat of atomisation of methane and ethane are 'x' kJ $$mol^{-1}$$ and 'y' kJ $$mol^{-1}$$ respectively. The longest wavelength ($$\lambda$$) of light capable of breaking the C-C bond
can be expressed in SI unit as:

We first express the C-C bond energy in terms of x and y. For methane (CH$$_4$$), the heat of atomisation is x kJ/mol. This breaks 4 C-H bonds, giving $$E_{C-H} = x/4$$.

For ethane (C$$_2$$H$$_6$$), the heat of atomisation is y kJ/mol. This breaks 6 C-H bonds and 1 C-C bond, so $$E_{C-C} = y - 6 \times \frac{x}{4} = \frac{4y - 6x}{4} \text{ kJ/mol}$$.

Next, we determine the wavelength associated with this bond energy. The energy per molecule is $$E = \frac{(4y-6x) \times 1000}{4N_A}$$ J.

Using the relation $$E = hc/\lambda$$, we find $$\lambda = \frac{hc}{E} = \frac{hc \times 4N_A}{(4y - 6x) \times 1000} = \frac{N_A hc}{250(4y - 6x)}$$.

The correct answer is Option 3: $$\frac{N_A hc}{250(4y - 6x)}$$.