Answer:
λ = 150 nm
Explanation:
For C-O bond rupture:
The required energy to rupture C-O bond = bond energy of C-O bond
= 799 kJ/mol
[tex]\mathsf{= 799 \ kJ/mol \times ( \dfrac{1 \ mol }{6.023 \times 10^{23} \ C-O \ bonds })}[/tex]
[tex]\mathsf{= 1.3265 \times 10^{-21} \ kJ/ C-O \ bond}[/tex]
[tex]\mathsf{= 1.33 \times 10^{-18} \ J/C-O \ bond}[/tex]
Recall that the wavelength associated with energy and frequency is expressed as:
[tex]E = \dfrac{hc}{\lambda}[/tex]
[tex]\lambda = \dfrac{hc}{E}[/tex]
[tex]\lambda = \dfrac{(6.626 \times 10^{-34} \ J.s^{-1}) \times (3.0 \times 10^8 \ ms^{-1})}{ 1.33 \times 10^{-18} \ J/C-O \ bond}}[/tex]
[tex]\mathsf{\lambda = 1.50 \times 10^{-7} \ m}[/tex]
λ = 150 nm
