Question

The ${\mathbf{Na}}_2$molecule(${}^{23}\mathbf{Na}$isotope) shows a very weak infrared line in its vibrational spectrum at a wavelengthof$\mathbf{6}.\mathbf{28}\times {\mathbf{10}}^{-\mathbf{5}}\mathbf{m}$.Calculate the force constant for the${\mathbf{Na}}_2$ molecule and compare your result with that ofproblem $\mathbf{17}$Givea reason for any difference.

Short answer

The force constant for ${\text{N}}_{\text{2}}$molecule is $\text{k}=17.38\text{kg}/{\text{s}}^2$

Due to sodium molecule will be apart as it is bigger and harder to collide.

Step-by-step solution

Definition of vibration spectrum 

The part of a molecular spectrum in which the bands arise from quantized changes in the energy of mutual atomic vibrations within the molecule.

Step 2:Given information

Calculate the force constant for${\text{N}}_{\text{2}}$molecule. The given data is as follows,

Wavelength$\lambda =6.28\times {10}^{-5}m$

Calculationof force constant ofN2 molecule

The formula of harmonic oscillation is as follows,

$\text{ν}=\frac{1}{2\text{π}}\sqrt{\frac{\text{k}}{\text{μ}}}$

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

Now, use the Table $19.1$for reduced mass of the molecule and mass of sodium atom.

data-custom-editor="chemistry" $\text{m}{[}_{11}^{23}\text{Na}]=22.9897697\text{u}$

Calculate the reduced mass

data-custom-editor="chemistry" $\begin{array}{c}\text{μ}=\frac{\text{m}{{[}_{11}^{23}\text{Na}]}^2}{2\cdot \text{m}{[}_{11}^{23}\text{Na}]}\\ =\frac{{\left(22.9897697\text{u}\right)}^2}{2\cdot 22.9897697\text{u}}\\ =11.49488485\text{u}\\ =11.49488485\text{u}\cdot 1.6605387\cdot {10}^{-27}{\text{kgu}}^{-1}\\ =1.909\cdot {10}^{-26}\text{kg}\end{array}$

The reduced mass of${\text{N}}_{\text{2}}$ molecule is $\text{μ}=1.909\cdot {10}^{-26}\text{kg}$

Add all value in formula of frequency

Lastly, add value to formula of frequency

$\begin{array}{c}\nu =\frac{1}{2\pi }\sqrt{\frac{\text{k}}{\mu }}\\ 4.8\cdot {10}^{12}{\text{s}}^{-1}=0.1591\cdot \sqrt{\frac{\text{k}}{1.909\cdot {10}^{-26}\text{kg}}}\\ \frac{4.8\cdot {10}^{12}{\text{s}}^{-1}}{0.159}=\sqrt{\frac{\text{k}}{1.909\cdot {10}^{-26}\text{kg}}}\\ {\left(\frac{4.8\cdot {10}^{12}{\text{s}}^{-1}}{0.159}\right)}^2=\frac{\text{k}}{1.909\cdot {10}^{-26}\text{kg}}\\ {\left(\frac{4.8\cdot {10}^{12}{\text{s}}^{-1}}{0.159}\right)}^2\cdot 1.909\cdot {10}^{-26}\text{kg}=\text{k}\\ \text{k}=17.38\text{kg}/{\text{s}}^2\end{array}$

The force constant for ${\text{N}}_{\text{2}}$moleculeis $\text{k}=17.38\text{kg}/{\text{s}}^2$

${\text{Li}}_{\text{2}}$

Now, compare with result from problem , $17$where the force of constant for molecule is ,$\text{k}=25.42\text{kg}/{\text{s}}^2$

Due towhich sodium molecule will be apart as it is bigger and harder to collide