Question

A molecular bond can be modeled as a spring between two atoms that vibrate with simple harmonic motion. The figure shows an SHM approximation for the potential energy of an HCl molecule. For E < 4*10^-19 J it is a good approximation to the more accurate HCl potential-energy curve that was shown in the image attached. Because the chlorine atom is so much more massive than the hydrogen atom, it is reasonable to assume that the hydrogen atom (m = 1.67 x 10^-27)vibrates back and forth while the chlorine atom remains at rest.

Answer 1

The given HCl molecule is vibrating in the simple harmonic motion. In HCl molecule, chlorine

atom is so much more massive than the hydrogen atom, it is reasonable to assume that the

hydrogen atom vibrates back and forth while the chlorine atom remains at rest. Let \(k\) be the

spring constant of the bond between hydrogen atom and chlorine atoms.

From the graph, bond length corresponding to the potential energy \(3.0 \times 10^{-19} \mathrm{~J}\) is \(0.09 \mathrm{nm}\)

and the equilibrium separation between Hydrogen and chlorine atoms is \(0.13 \mathrm{nm} .\)

Calculate the displacement covered by the Hydrogen atom from its mean position as follows:

\(\Delta x=0.13 \mathrm{nm}-0.09 \mathrm{nm}\)

$$ =0.04 \mathrm{nm}\left(\frac{10^{-9} \mathrm{~m}}{1 \mathrm{nm}}\right) $$

\(=0.04 \times 10^{-9} \mathrm{~m}\)

The amount of potential energy stored in the spring is expressed as follows:

$$ U=\frac{1}{2} k(\Delta x)^{2} $$

Rearrange the above formula for \(k\).

$$ k=\frac{2 U}{(\Delta x)^{2}} $$

Substitute \(3.0 \times 10^{-19} \mathrm{~J}\) for \(U\) and \(0.04 \times 10^{-9} \mathrm{~m}\) for \(\Delta x\)

$$ \begin{aligned} k &=\frac{2\left(3.0 \times 10^{-19} \mathrm{~J}\right)}{\left(0.04 \times 10^{-9} \mathrm{~m}\right)^{2}} \\ &=375 \mathrm{~N} / \mathrm{m} \end{aligned} $$

Reduced mass \((\mu)\) of the \(\mathrm{HCl}\) molecule can be calculated by using the following equation.

$$ \mu=\frac{m_{\mathrm{H}} m_{\mathrm{Cl}}}{m_{\mathrm{H}}+m_{\mathrm{Cl}}} $$

Here, \(m_{\mathrm{H}}\) is the mass of the Hydrogen atom and \(m_{\mathrm{C}}\) is the mass of the Chlorine atom.

Since the mass of Chlorine is equal to 17 times the mass of the Hydrogen atom, the above equation becomes:

$$ \begin{aligned} \mu &=\frac{\left(m_{\mathrm{H}}\right)\left(17 m_{\mathrm{H}}\right)}{m_{\mathrm{H}}+17 m_{\mathrm{H}}} \\ \mu &=\frac{17 m_{\mathrm{H}}}{18} \end{aligned} $$

Substitute \(1.67 \times 10^{-27} \mathrm{~kg}\) for \(m_{\mathrm{H}}\).

$$ \mu=\frac{17\left(1.67 \times 10^{-27} \mathrm{~kg}\right)}{18} $$

\(=1.577 \times 10^{-27} \mathrm{~kg}\)

Now, use the following relation to find the vibrational frequency of the \(\mathrm{HCl}\) molecule.

\(f=\frac{1}{2 \pi} \sqrt{\frac{k}{\mu}}\)

Substitute \(375 \mathrm{~N} / \mathrm{m}\) for \(k\) and \(1.577 \times 10^{-27} \mathrm{~kg}\) for \(\mu .\)

$$ \begin{aligned} f &=\frac{1}{2 \pi} \sqrt{\frac{375 \mathrm{~N} / \mathrm{m}}{1.577 \times 10^{-27} \mathrm{~kg}}} \\ &=7.76 \times 10^{13} \mathrm{~Hz} \end{aligned} $$

Thus, the vibrational frequency of the \(\mathrm{HCl}\) molecule is approximately \(7.8 \times 10^{13} \mathrm{~Hz}\)