Question

$O_2$ undergoes photochemical dissociation into $1$ normal oxygen atom ($O$) and more energetic oxygen $O^{\ast }$. If ($O^{\ast }$) has $1.967eV$ more energy than ($O$) and normal dissociation energy of $O_2$ is $498kJ$ ${mol}^{-1}$, what is the maximum wavelength effective for the photochemical dissociation of $O_2$?

Answer 1

(1) we have one molecule of $O_2$ dissociated to give two normal, oxygen atoms:

$O_2$$\to O$(normal)+$O$(normal) $E_{dis}=E_1=498KJ/mol$

(2) Then, we also have reaction in which instead of normal, one Energetic Oxygen is released.

$O_2$$\to O$(normal)+$O$(enegetic) $E_{dis}=E_2$

(3)According to the question, one oxygen is 1.967ev more energetic than usual oxygen.

$E_2-E_1=1.967eV$

$E_2-498KJ/mol=1.967eV$

$E_2-\frac{498\times {10}^3}{6.022\times {10}^{23}}=1.967\times 1.6\times {10}^{-19}$

$E_2-8.27\times {10}^{-19}=3.15\times {10}^{-19}$

$E_2=11.42\times {10}^{-19}J/molecule$

Since, we have to find for one molecule of $O_2$,

$E_2=11.42\times {10}^{-19}J$

$E_2=\frac{hc}{\lambda }$

$11.42\times {10}^{-19}=\frac{6.626\times {10}^{-34}\times 3\times 10^8}{\lambda }$

$\lambda =\frac{6.626\times 3\times {10}^{-7}}{11.42}=1.740\times {10}^{-7}m$

$\lambda =1740\times {10}^{-10}m=174nm$

Therefore, required wavelength of photon is $174$nm