$(a)\;1645.3A^{\large\circ}\qquad(b)\;1740.2A^{\large\circ}\qquad(c)\;1545.2A^{\large\circ}\qquad(d)\;1942.3A^{\large\circ}$

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$O_2 \rightarrow O_{Normal}+O_{Excited}$

$O_2 \rightarrow O_{Normal}+O_{Normal}$

Energy required for simple dissociation of $O_2$ into two normal atoms

$=498\times 10^3 J/mol$

$=\large\frac{498\times10^3}{6.023\times10^{23}}$ J/Molecule

If one atom is excited state has more energy i.e 1.967 eV

$=1.967\times 1.602\times 10^{-19} J$

Energy required for photochemical dissociation of $O_2$

$=\large\frac{498\times 10^3}{6.023\times 10^{23}} + 1.967\times1.602\times 10^{-19}$

$= 82.68\times 10^{-20} + 31.51\times 10^{-20}$

$= 114.19\times 10^{-20} J$

$\therefore E = \large\frac{hc}{\lambda}$

$114.19\times 10^{-20} = \large\frac{6.625\times10^{-34}\times3.0\times 10^8}{\lambda}$

$\lambda=1740.2\times 10^{-10} m$

$\lambda = 1740.2 A^{\large\circ}$

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