The heat of atomization of methane and ethane are $360\mathrm{kJ}/\mathrm{mol}$ and $620\mathrm{kJ}/\mathrm{mol}$, respectively the longest wavelength light capable of breaking the $\mathrm{C}-\mathrm{C}$ bond is (Avodrago constant $=6.02\times {10}^{23}{\mathrm{mol}}^{-1}$). 

Given data as

$\begin{array}{l}{\mathrm{CH}}_4\left(\mathrm{g}\right)\to \mathrm{C}\left(\mathrm{g}\right)+4\mathrm{H}\left(\mathrm{g}\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{\Delta }{H}_1=4{\epsilon }_{\mathrm{C}-\mathrm{H}}\\ {\mathrm{C}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)\to 2\mathrm{C}\left(\mathrm{g}\right)+6\mathrm{H}\left(\mathrm{g}\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{\Delta }{H}_2={\epsilon }_{\mathrm{C}-\mathrm{C}}+6{\epsilon }_{\mathrm{C}-\mathrm{H}}\end{array}$

From  $\mathrm{\Delta }{H}_1$, we get   ${\epsilon }_{\mathrm{C}-\mathrm{H}}=(360/4)\mathrm{kJ}{\mathrm{mol}}^{-1}=90\mathrm{kJ}{\mathrm{mol}}^{-1}$

From  $\mathrm{\Delta }{H}_2$, we get   ${\epsilon }_{\mathrm{C}-\mathrm{H}}=\mathrm{\Delta }{H}_2-6{\epsilon }_{\mathrm{C}-\mathrm{H}}=(620-6\times 90){\mathrm{kJ}}^{-1}=80\mathrm{kJ}{\mathrm{mol}}^{-1}$

For a single $\mathrm{C}-\mathrm{H}$  bond, we have

$E=\frac{{\epsilon }_{\mathrm{C}}-\mathrm{H}}{N_{\mathrm{A}}}=\frac{\left(80\mathrm{kJ}{\mathrm{mol}}^{-1}\right)}{\left(6.02\times {10}^{23}{\mathrm{mol}}^{-1}\right)}=1.33\times {10}^{-22}\mathrm{kJ}=1.33\times {10}^{-14}\mathrm{J}$

The longest wavelength required to break this bond is

$\begin{array}{l}\lambda =\frac{hc}{E}=\frac{\left(6.626\times {10}^{-34}\mathrm{Js}\right)\left(3\times {10}^8{\mathrm{ms}}^{-1}\right)}{\left(1.33\times {10}^{-19}\mathrm{J}\right)}=1.495\times {10}^{-6}\mathrm{m}\\ =1.495\times {10}^3\mathrm{nm}\end{array}$