Question

The “signature” infrared absorption that indicates the presence ofa $\mathbf{C}\mathbf{-}\mathbf{H}$stretchingmotion in a molecule occurs at wavelengthsnear.$\mathbf{3}.\mathbf{4}\times {\mathbf{10}}^{-\mathbf{6}}\mathbf{m}$Use this information to estimate the force constant of the $\mathbf{C}\mathbf{-}\mathbf{H}$stretch. Take the reduced mass in this motion to be approximately equal to the mass of the hydrogen atom (a good approximation whenthe $\mathit{H}$atomis attached to a heavy atom).

Short answer

The force constant is $k=510.16\text{kg}/{\text{s}}^2$

Step-by-step solution

 Step 1: Definition of infrared absorption

The interaction between the IR electric field vector and the molecular dipole transition moments connected to molecular vibrations causes IR absorption. When the electric field vector and the dipole transition moment are parallel to one other, absorption is greatest.

Step 2:Given information

The wavelength of $C-H$stretching molecule is near to.$3.4\times {10}^{-6}m$ Find the force constant of $C-H$stretching molecule.

Calculationof force constant ofC−H moiety

The formula of harmonic oscillation is as follows,

$\nu =\frac{1}{2\pi }\sqrt{\frac{k}{\mu }}$

where$k$, is the force constant and $\mu$ is the mass,$\nu$ is frequency

The reduced mass of hydrogen atom is $\mu =1.66\cdot {10}^{-27}\text{kg}$

The wave length is$\lambda =3.4\cdot {10}^{-6}\text{m}$ . Use wave length in formula of frequency is as follows,

$\begin{array}{c}\nu =\frac{c}{\lambda }\\ =\frac{3.0\cdot {10}^8\text{m}/\text{s}}{3.4\cdot {10}^{-6}\text{m}}\\ =8.82\cdot {10}^{13}{\text{s}}^{-1}\end{array}$

The frequency is .$\nu =8.82\cdot {10}^{13}{\text{s}}^{-1}$

Adding all value in formula

Lastly, add all value in formula of frequency,

$\begin{array}{l}\nu =\frac{1}{2\pi }\sqrt{\frac{k}{\mu }}\\ 8.82\cdot {10}^{13}{\text{s}}^{-1}=0.1591\cdot \sqrt{\frac{k}{1.66\cdot {10}^{-27}\text{kg}}}\\ \frac{8.82\cdot {10}^{13}{\text{s}}^{-1}}{0.1591}=\sqrt{\frac{k}{1.66\cdot {10}^{-27}\text{kg}}}\\ {\left(\frac{8.82\cdot {10}^{13}{\text{s}}^{-1}}{0.1591}\right)}^2=\frac{k}{1.66\cdot {10}^{-27}\text{kg}}\\ {\left(\frac{8.82\cdot {10}^{13}{\text{s}}^{-1}}{0.1591}\right)}^2\cdot 1.66\cdot {10}^{-27}\text{kg}=k\\ k=510.16\text{kg}/{\text{s}}^2\end{array}$

The force constant is $k=510.16\text{kg}/{\text{s}}^2$