The ozone molecule ${\mathrm{O}}_3$ has a permanent dipole moment of 1$1.8\times {10}^{-30}Cm$. Although the molecule is very slightly bent— which is why it has a dipole moment—it can be modeled as a uniform rod of length $2.5\times {10}^{-10}m$ with the dipole moment perpendicular to the axis of the rod. Suppose an ozone molecule is in a $5000N/C$ uniform electric field. In equilibrium, the dipole moment is aligned with the electric field. But if the molecule is rotated by a small angle and released, it will oscillate back and forth in simple harmonic motion. What is the frequency $f$ of oscillation?

Given :

The ozone molecule has a dipole moment : $1.8\times {10}^{-30}Cm$.

It can be modeled as a uniform rod of length : $2.5\times {10}^{-10}m$

Ozone molecule is in an electric field : $5000N/C$

In equilibrium, the dipole moment is aligned with the electric field.

Theory used :

The motion of the molecule data-custom-editor="chemistry" ${\mathrm{O}}_3$is periodic, recurring in a sinusoidal pattern with a constant amplitude $A$.

A basic harmonic oscillator's motion is defined by its period $T=2\pi /w$, which is the time it takes for a single oscillation, or its frequency $f=1/T$.

The phase, which controls the starting point on the sine wave, also influences the position at a given time $t$. The size of the mass $m$and the force constant $k$determine the period and frequency, whereas the starting location and velocity dictate the amplitude and phase.

A dipole is torqued by the electric field$\tau =pE\sin \theta$. The dipoles are aligned by the torque.

$$\begin{gathered} F_e=F_c \\ \Rightarrow \frac{1}{4\pi {\epsilon }_0}\times \frac{e^2}{d^2}=m{w}^2\left(\frac{d}{2}\right) \\ \Rightarrow w^2=\frac{9\times {10}^{-9}\times (1.8\times {10}^{-30})²\times 2}{(9.11x{10}^{-31})(2\times {10}^{-10})³} \\ \Rightarrow w=\sqrt{4.09712\times {10}^{-9}} \\ =\underline{6.40088x{10}^{-8}rad/s} \end{gathered}$$

Period is :

$$\begin{gathered} T=\frac{2\pi }{w}=\frac{2\times 3.14}{6.40088x{10}^{-8}} \\ =\underline{9.8115x{10}^7} \end{gathered}$$

Oscillation frequency :

$f=\frac{1}{T}=\frac{1}{9.8115x{10}^7}=\boxed{\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}\mathbf{}}\mathit{H}\mathit{z}}$