91Ó°ÊÓ

Elastic Potential energy: The elastic potential energy of a spring-like body of spring constant$k$ that has been stretched or compressed a distance x from the undistorted position is:$U_s=\frac{1}{2}k{x}^2$

The frequency of an oscillation in simple harmonic motion is given by

$f=\frac{1}{2\mathrm{Ï€}}\sqrt{\frac{k}{m}}$

Figure P15.63 displays as an approximate SHM for the potential energy of the HCl molecule

The hydrogen atom in Hcl oscillates back and forth while the chlorine atom remains in the experiment

The mass of the hydrogen atom is

$m=1.67×{10}^{−27}\mathrm{kg}$

The objective is to calculate the vibrational frequency of the Hcl molecule

As we can see in Fig p 15.63 , the equilibrium length of the bond$x_0$is $0.13mm$

The elastic energy stored in Hcl molecule is

$U=\frac{1}{2}k(\mathrm{Δ}x{)}^2$

$=\frac{1}{2}k{\left(x−x_0\right)}^2$

Solve

$k=\frac{2U}{{\left(x−x_0\right)}^2}$Equation (3)

Thus, the average value of $k$is

$k_{\mathrm{avg}}=\frac{(400+422+412.5)\mathrm{N}/\mathrm{m}}{3}$

$=411.5\mathrm{N}/\mathrm{m}$

The vibrational frequency of Hcl molecule can be obtained from equation $\left(2\right)$

$f=\frac{1}{2\mathrm{Ï€}}\sqrt{\frac{k}{m}}$

$f=\frac{1}{2\mathrm{π}}\sqrt{\frac{411.5\mathrm{N}/\mathrm{m}}{1.67×{10}^{−27}\mathrm{kg}}}$

$=7.9×{10}^{13}\mathrm{Hz}$