Question

The bond dissociation energy of $N_2$ is $958.38{\text{kJ mol}}^{-1}$ at certain temperature. Hypothetically two moles of photons are necessary to dissociate one moles of $N_2$ molecule into atoms.Find the wavelength of the incident radiation.
$\text{Given}:\text{Planck's constamt}\left(h\right)=6.63\times {10}^{-34}{\text{J s}}^{},c=3\times {10}^8{\text{m s}}^{-1}$

• A. 355 nm
• B. 215 nm
• C. 295 nm
• D. 250 nm

Answer 1

The correct option is D 250 nm

$\text{Energy of photon}E=nh\nu =\frac{nhc}{\lambda }$
where n = number of photons
$\lambda$ = wavelength of the incident radiation

Given, bond dissociation energy of $N_2$ $=958.38\times {10}^3\text{J /mole}$
Number of photons n = $2\times 6.023\times {10}^{23}$
$\therefore$ Wavelength required to dissociate 1 mol of $N_2$ molecules into its atoms $\lambda =\frac{2\times 6.023\times {10}^{23}\times 6.63\times {10}^{-34}\text{J s}\left)\right(3\times {10}^8{\text{m s}}^{-1})}{(958.38\times {10}^{3}J/mole)}$ = 250 nm