Question

The infrared spectrum of ${}^{\mathbf{1}}{\mathbf{H}}^{\mathbf{79}}\mathit{B}\mathit{r}$shows the role="math" localid="1663823566071" $\mathbf{v}\mathbf{=}\mathbf{0}\mathbf{}\mathit{t}\mathit{o}\mathbf{}\mathbf{v}\mathbf{=}\mathbf{1}$transition at $\mathbf{2650}\mathit{c}{\mathit{m}}^{\mathbf{-}\mathbf{1}}$. Determine the vibrational force constant for the HBr molecule. The atomic mass of ${}^1\mathbf{H}$is 1.0078, and the atomic mass of ${}^{\mathbf{79}}\mathit{B}\mathit{r}$is 78.918u.

Short answer

The vibrational force constant for the HBr molecule is $411.5J{m}^{-2}$

Step-by-step solution

Step 1:The formula used for fundamental vibrational frequency

The formula used to find the fundamental vibrational frequency is given as below:

$v_0=\frac{1}{2\pi }{\left(\frac{k}{\mu }\right)}^{\frac{1}{2}}$

Where k is the vibrational force constant and $\mu$is the reduced mass of the molecule.

To find the value of 𝜇 

Now, we can find the value of 𝜇 by using the following formula

$$\begin{gathered} \mu =\frac{m\left(H\right)\times m\left(Br\right)}{m\left(H\right)+m\left(Br\right)} \\ =\frac{1.0078\times 1.0078}{1.0078+1.0078} \\ =0.9951 \end{gathered}$$

To calculate the frequency

To calculate frequency

$$\begin{gathered} \nu =\frac{1}{\lambda }\times c \\ \nu =\left(2650c{m}^{-1}\right)\left(2.998\times {10}^{8}cm{s}^{-1}\right) \\ =7.944\times {10}^{13}{s}^{-1} \end{gathered}$$

To calculate the vibrational frequency

Using the formula,

$$\begin{gathered} {\nu }_0=\frac{1}{2\pi }{\left(\frac{k}{\mu }\right)}^{\frac{1}{2}} \\ k={\nu }^{2}4{\pi }^2\mu \\ =\left(7.994\times {10}^{13}{s}^{-1}\right)4{\pi }^2\left(0.9951\right)\left(1.66\times {10}^{-27}kg\right) \\ =411.5J{m}^{-2} \end{gathered}$$