Question

O2 undergoes photochemical dissociation into one normal oxygen atom and one oxygen atom, 1.967 eV more energetic than normal. The dissociation of O2 into two normal atoms of oxygen requires 498 KJmol−1.what is the maximum wavelength effective for photochemical dissociation of O2 ?

Answer 1

Final Answer: 1740Â = 174 nm (approx.) 

Steps:
1) We have one molecule of $O_{2}$ dissociated to give two normal Oxygen atoms:
$O_{2} ----\ \textgreater \ O (normal ) + O (normal) \:\:\:\: E_{diss} = E_{1} = 498KJ/mol$ 

2) Then, we also have reaction in which instead of normal ,one Energetic Oxygen is released .
$O_{2} ----\ \textgreater \ O (normal ) + O (Energetic) \:\:\:\: E_{diss} = E_{2}$

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

[tex]E_{2}-E_{1} = 1.967eV\\=\ \textgreater \ E_{2} - 498KJ/mol =1.967eV \\=\ \textgreater \ E_{2} - \frac{498 * 10^{3} }{6.022* 10^{23} } = 1.967 * 1.6 * 10^{-19} \\ \\ =\ \textgreater \ E_{2} - 8.27* 10^{-19} = 3.15 * 10^{-19} \\=\ \textgreater \ E_{2} = 11.42 * 10^{-19 } J/ molecule [/tex]
 
Since , we have to find for one molecule of $O_{2}$ ,
Therfore,
$E_{2} = 11.42 *10^{-19} J$ 

4) Since light/ photon is used for dissociation, where $\lambda$
 is maximum wavelength of effective for photochemical dissociation reaction .
=>  
$E_{2} = \frac{hc}{\lambda} \\=\ \textgreater \ 11.42 * 10^{-19} = \frac{6.626*10^{-34}*3* 10^{8} }{\lambda} \\=\ \textgreater \ \lambda = \frac{6.626*3*10^{-7}}{11.42} = 1.740 * 10^{-7} m \\=\ \textgreater \ \lambda = 1740 * 10^{-10} m=174 nm$ 

Therefore, Required wavelength of photon is  $\boxed{174 nm (approx)}$

Hope You Enjoyed : D 

Answer 2

Answer:

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