`N_(2)` undergoes photochemical dissociation into one normal `N-`atom and one `N-`atom having `0.1eV`more energy than normal `N-`atom. The dissociation of `N_(2)` into two normal atom of `N` requires `289.5 KJ//mol` energy. The maximum wavelength effective for photochemical dissociation of `N_(2)` in `nm` is `P nm` (`1 eV = 96.5` KJ/mol). Find the value of `(P)/(100)`.

Answer

4

Explanation

Energy required for `N_(2)` bond `(289.5)/(96.5) = 3`
Total energy given by `= 3 + 0.1 = 3.1 eV`
`lambda(nm) = (1240)/(3.1) = 400`