Question

carbon- carbon bonds form the “backbone” of nearly every organic and biological molecule. The average bond energy of the C-C bond is 347kJ/mol. Calculate the frequency and wavelength of the least energetic photon that can break an average C-C bond. In what region of the electromagnetic spectrum is this radiation?

Short answer

$\mathbf{8}\mathbf{.}\mathbf{696}\mathbf{\times }{\mathbf{10}}^{14}\mathbf{Hz}$frequency and 344.98nm wavelength of the least energetic photon can break an average C-C bond.

This radiation is in the UV region of the electromagnetic spectrum.

Step-by-step solution

energy of the single bond

Firstly, we have to calculate the least energy that is required to break the single bond in C-C. average bond energy divided by the Avogadro number is the least energy required to break a single bond.

$\begin{array}{l}{\mathbf{E}}_{\mathbf{single}\mathbf{bond}}\mathbf{=}\frac{{\mathbf{E}}_{\mathrm{average}}}{{\mathbf{N}}_A}\\ \mathbf{=}\frac{\mathbf{347}\mathbf{\times }{\mathbf{10}}^3\mathbf{J}\mathbf{/}\mathbf{mol}}{\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times }{\mathbf{10}}^{23}{\mathbf{mol}}^{-1}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{763}\mathbf{\times }{\mathbf{10}}^{-19}\mathbf{J}\end{array}$

So,$\mathbf{5}\mathbf{.}\mathbf{763}\mathbf{\times }{\mathbf{10}}^{-19}\mathbf{J}$is theleast energy that is required to break the single bond in C-C.

Frequency calculation

Frequency is the number of waves that pass a certain point in a specified amount of time.

Now, the frequency of the least energetic photon can be calculated by using the Einstein- plank equation.

$\begin{array}{l}E=\mathrm{h\upsilon }\\ \mathbf{\upsilon }\mathbf{=}\frac{E}{h}\mathbf{=}\frac{\mathbf{5}\mathbf{.}\mathbf{762}\mathbf{\times }{\mathbf{10}}^{-19}\mathbf{J}}{\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{\times }{\mathbf{10}}^{-34}\mathbf{J}\mathbf{.}\mathbf{s}}\\ \mathbf{=}\mathbf{8}\mathbf{.}\mathbf{696}\mathbf{\times }{\mathbf{10}}^{14}\mathbf{Hz}\end{array}$

calculation of wavelength

Wavelength can be calculated by using the formula

$\begin{array}{l}\mathbf{\upsilon }\mathbf{=}\frac{c}{\lambda }\\ \mathbf{\lambda }\mathbf{=}\frac{c}{\upsilon }\end{array}$

Where c is the speed of light. Value of c is $\mathbf{3}\mathbf{\times }{\mathbf{10}}^8\mathbf{m}\mathbf{/}\mathbf{s}$.

$\begin{array}{l}\mathbf{\lambda }\mathbf{=}\frac{c}{\upsilon }\mathbf{=}\frac{\mathbf{3}\mathbf{\times }{\mathbf{10}}^8\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{8}\mathbf{.}\mathbf{696}\mathbf{\times }{\mathbf{10}}^{14}\mathbf{Hz}}\\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{34498}\mathbf{\times }{\mathbf{10}}^{-6}\mathbf{m}\\ \mathbf{=}\mathbf{344}\mathbf{.}\mathbf{98}\mathbf{nm}\end{array}$

So, 344.98nm wavelength is required to break an average C-C bond.

calculation of wavenumber

$\begin{array}{l}\mathbf{wavenumber}\mathbf{(}\mathbf{c}{\mathbf{m}}^{-1}\mathbf{)}\mathbf{=}\frac{{\mathbf{10}}^7}{\mathrm{wavelength}\left(\mathrm{nm}\right)}\\ \mathbf{=}\frac{{\mathbf{10}}^7}{\mathbf{344}\mathbf{.}\mathbf{98}\mathbf{nm}}\mathbf{=}\mathbf{28987}\mathbf{.}\mathbf{18}{\mathbf{cm}}^{-1}\end{array}$

Since, the range of the UV region of electromagnetic radiation is 200-400nm or $\mathbf{25000}\mathbf{-}\mathbf{50000}{\mathbf{cm}}^{-1}$, from the calculation value of wavelength and wavenumber it is clear that the radiation is in the UV region.