Question

Repeat the calculation of the problem$\mathbf{19}$for the$\mathbf{N}-\mathbf{H}$stretch, where absorption occurs near$\mathbf{2}.\mathbf{9}\times {\mathbf{10}}^{-\mathbf{6}}\mathbf{m}$. Which bond is stiffer:$\mathbf{N}-\mathbf{H}$or$\mathit{C}\mathbf{-}\mathbf{H}$?

Short answer

The force constant for$\text{N-H}$stretch is$k=695.73\text{kg}/{\text{s}}^2$.

In the Questiondata-custom-editor="chemistry" $19$. Calculate the $C-H$stretch was$510.16\text{kg}/{\text{s}}^2$. Calculated the force constant for$\text{N}-\text{H}$stretch is bigger, so the stiffer bond is$\text{N}-\text{H}$

Step-by-step solution

Definition of radiation adsorption

Absorption of electromagnetic radiation is the process by which matter absorbs a photon's energy and converts it into the absorber's internal energy.

Given information

Calculate the force constant forstretch. The given data are follows,

Wavelength$\text{λ}=2.85\times {10}^{-5}m$

 Step 3: Calculationof force constant of N−Hstretch

The formula for frequency is as follows,

$\begin{array}{c}\nu =\frac{c}{\lambda }\\ =\frac{3.0\cdot {10}^8\text{m}/\text{s}}{2.9\cdot {10}^{-6}\text{m}}\\ \nu =1.03\cdot {10}^{14}{\text{s}}^{-1}\end{array}$

The frequency is$\nu =1.03\cdot {10}^{14}{\text{s}}^{-1}$

 Step 4: 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 }}\\ 1.03\cdot {10}^{14}{\text{s}}^{-1}=0.1591\cdot \sqrt{\frac{k}{1.66\cdot {10}^{-27}\text{kg}}}\\ \frac{1.03\cdot {10}^{14}{\text{s}}^{-1}}{0.1591}=\sqrt{\frac{k}{1.66\cdot {10}^{-27}\text{kg}}}\\ {\left(\frac{1.03\cdot {10}^{14}{\text{s}}^{-1}}{0.1591}\right)}^2=\frac{k}{1.66\cdot {10}^{-27}\text{kg}}\\ {\left(\frac{1.03\cdot {10}^{14}{\text{s}}^{-1}}{0.1591}\right)}^2\cdot 1.66\cdot {10}^{-27}\text{kg}=k\\ k=695.73\text{kg}/{\text{s}}^2\end{array}$$\begin{array}{c}\nu =\frac{1}{2\pi }\sqrt{\frac{k}{\mu }}\\ 1.03\cdot {10}^{14}{\text{s}}^{-1}=0.1591\cdot \sqrt{\frac{k}{1.66\cdot {10}^{-27}\text{kg}}}\\ \frac{1.03\cdot {10}^{14}{\text{s}}^{-1}}{0.1591}=\sqrt{\frac{k}{1.66\cdot {10}^{-27}\text{kg}}}\\ {\left(\frac{1.03\cdot {10}^{14}{\text{s}}^{-1}}{0.1591}\right)}^2=\frac{k}{1.66\cdot {10}^{-27}\text{kg}}\\ {\left(\frac{1.03\cdot {10}^{14}{\text{s}}^{-1}}{0.1591}\right)}^2\cdot 1.66\cdot {10}^{-27}\text{kg}=k\\ k=695.73\text{kg}/{\text{s}}^2\end{array}$

The force constant for$\text{N-H}$stretch is$k=695.73\text{kg}/{\text{s}}^2$

In the Question.data-custom-editor="chemistry" $19$ Calculate the $\text{C-H}$stretch was.$510.16\text{kg}/{\text{s}}^2$ Calculated the force constant for $\text{N}-\text{H}$stretch is bigger, so the stiffer bond is$\text{N}-\text{H}$