The vibration frequencies of atoms in solids at normal temperatures are of the order of${\mathbf{10}}^{\mathbf{13}}\mathbf{}\mathbf{Hz}$. Imagine the atoms to be connected to one another by springs. Suppose that a single silver atom in a solid vibrates with this frequency and that all the other atoms are at rest. Compute the effective spring constant. One mole of silver ($\mathbf{6}\mathbf{.}\mathbf{021023}$atoms) has a mass of 108 g.

• Mass of one mole of silver is, m=0.108kg.
• Frequency (f)=${10}^{13}\mathrm{Hz}$.

We can find the total mass from the mass of one mole of silver and the number of atoms in silver. Then, using the given frequency, we can find the angular frequency. From mass and angular frequency we can find the spring constant.

Formulae:

Angular speed in terms of mass and spring constant,$\omega =\sqrt{\frac{k}{m}}or\omega =2\pi \mathrm{f}$ (i)

Mass of single silver atom:

$$\begin{gathered} \mathrm{m}=\frac{\mathrm{mass}\mathrm{of}\mathrm{one}\mathrm{mole}\mathrm{of}\mathrm{silver}}{\mathrm{no}.\mathrm{of}\mathrm{atoms}\mathrm{in}\mathrm{one}\mathrm{mole}\mathrm{of}\mathrm{silver}\mathrm{atom}} \\ =\frac{0.108\mathrm{kg}}{6.023\times {10}^{23}} \\ =1.79\times {10}^{-25}\mathrm{kg} \end{gathered}$$

Using equation (i), we can calculate the angular frequency of the oscillation as:

$$\begin{gathered} \omega =2\mathrm{\pi }\times {10}^{13} \\ =6.28\times {10}^{13}\mathrm{rad}/\mathrm{s} \end{gathered}$$

Using equation (i) and the value of angular frequency, we get the spring constant as:

$$\begin{gathered} k=m{\omega }^2 \\ =1.79\times {10}^{-25}\times (6.28\times {10}^{13}) \\ =706\mathrm{N}/\mathrm{m} \end{gathered}$$

By rounding off to appropriate significant figures:

$\mathrm{k}=7.0\times {10}^2\mathrm{N}/\mathrm{m}$

Hence, the value of effective spring constant is$7.0\times {10}^2\mathrm{N}/\mathrm{m}$.