Atmospheric Nitrogen

We have to find the energy required to break the bond in just one molecule, so,
$\displaystyle E_{bond} = 945\frac{KJ}{mol} = \frac{945\cdot 1000}{6\times10^{23}} = 1.575\times 10^{-18}J$

Thus,
$\displaystyle\frac{hc}{\lambda} = E_{bond}$

$\displaystyle\lambda = \frac{6.6\times 10^{-34}\times 3\times 10^{8}}{1.575\times 10^{-18}} \approx\boxed{127nm}$

I don't know much about plants, but I think the plants use this process to absorb the required Nitrogen...They cannot directly consume the atmospheric nitrogen, so they rely on the sun to break this bond, and then they can use it.

For one mole, i.e. $6.02\times10^{23}$ molecules, the bond enthalpy is 945kJ. Thus for one molecule the bond enthalpy is E= $\frac{945}{6.02\times10^{23}}$ kJ. The energy of a light wave is given by $E=\frac{hc}{\lambda}$ , where h=planck's constant= $6.63\times10^{-34} Js^{-1}$ , c= speed of light= $3\times10^8 ms^{-1}$ and $\lambda$ =wavelength of light. Thus $\lambda = \frac{hc}{E}$ . So using the values of h,c and E, we get $\lambda$ =127nm. Due to such low value of wavelenth, the dinitrogen bond does not dissociate under visible light. Thus, the dinitrogen gas does not react with the oxygen gas in the atmosphere which maintains the amount of oxygen in the atmosphere and keeps it safe for respiration.