The heat of atomization of methane and ethane are 360 kJ mol$$^{-1}$$ and 620 kJ mol$$^{-1}$$, respectively. The longest wavelength of light capable of breaking the C - C bond is (Avogadro's number = $$6.023 \times 10^{23}$$, h = $$6.62 \times 10^{-34}$$ J s)

The heat of atomization is the energy required to break all bonds in one mole of a compound to form individual atoms. For methane (CH4), the heat of atomization is 360 kJ mol-1. Methane has four C-H bonds, so the bond energy for one C-H bond is calculated as follows:

$$ E(\text{C-H}) = \frac{360 \text{ kJ mol}^{-1}}{4} = 90 \text{ kJ mol}^{-1} $$

For ethane (C2H6), the heat of atomization is 620 kJ mol-1. Ethane has one C-C bond and six C-H bonds. The total bond energy is the sum of the energy for the C-C bond and six C-H bonds:

$$ E(\text{C-C}) + 6 \times E(\text{C-H}) = 620 \text{ kJ mol}^{-1} $$

Substituting the value of $$ E(\text{C-H}) = 90 \text{ kJ mol}^{-1} $$:

$$ E(\text{C-C}) + 6 \times 90 = 620 $$

$$ E(\text{C-C}) + 540 = 620 $$

$$ E(\text{C-C}) = 620 - 540 = 80 \text{ kJ mol}^{-1} $$

The bond energy for the C-C bond is 80 kJ mol-1. To find the longest wavelength of light capable of breaking one C-C bond, we need the energy required for a single bond. First, convert the bond energy from kJ mol-1 to J per bond. Since 1 kJ = 1000 J and 1 mole contains Avogadro's number of bonds ($$ N_A = 6.023 \times 10^{23} $$):

$$ E_{\text{bond}} = \frac{80 \times 1000}{6.023 \times 10^{23}} \text{ J} = \frac{80000}{6.023 \times 10^{23}} \text{ J} $$

Simplifying:

$$ E_{\text{bond}} = \frac{8 \times 10^4}{6.023 \times 10^{23}} = \frac{8}{6.023} \times 10^{-19} \text{ J} \approx 1.3282 \times 10^{-19} \text{ J} $$

The energy of a photon is given by $$ E = \frac{hc}{\lambda} $$, where $$ h = 6.62 \times 10^{-34} \text{ J s} $$ is Planck's constant, $$ c = 3 \times 10^8 \text{ m s}^{-1} $$ is the speed of light, and $$ \lambda $$ is the wavelength. The longest wavelength corresponds to the minimum energy required to break the bond, so set $$ E = E_{\text{bond}} $$:

$$ \frac{hc}{\lambda} = E_{\text{bond}} $$

$$ \lambda = \frac{hc}{E_{\text{bond}}} $$

First, compute $$ hc $$:

$$ hc = (6.62 \times 10^{-34}) \times (3 \times 10^8) = 1.986 \times 10^{-25} \text{ J m} $$

Now substitute into the wavelength formula:

$$ \lambda = \frac{1.986 \times 10^{-25}}{1.3282 \times 10^{-19}} = \frac{1.986}{1.3282} \times 10^{-6} \text{ m} \approx 1.4952 \times 10^{-6} \text{ m} $$

Convert meters to nanometers (1 m = 109 nm):

$$ \lambda = 1.4952 \times 10^{-6} \times 10^9 \text{ nm} = 1.4952 \times 10^3 \text{ nm} $$

Rounding to three significant figures, $$ \lambda \approx 1490 \text{ nm} = 1.49 \times 10^3 \text{ nm} $$. Comparing with the options, this matches option D.

Hence, the correct answer is Option D.