Question

Carbon dioxide is a linear molecule. The carbon–oxygen bonds in this molecule act very much like springs. Figure 11–58 shows one possible way the oxygen atoms in this molecule can oscillate: the oxygen atoms oscillate symmetrically in and out, while the central carbon atom remains at rest. Hence each oxygen atom acts like a simple harmonic oscillator with a mass equal to the mass of an oxygen atom. It is observed that this oscillation occurs at a frequency \(f{\bf{ = 2}}{\bf{.83 \times 1}}{{\bf{0}}^{{\bf{13}}}}\;{\bf{Hz}}\). What is the spring constant of the C-O bond?

Short answer

The spring constant of the C-O bond is \(8.40 \times {10^2}\;{\rm{N/m}}\).

Step-by-step solution

Concept of frequency oscillating atoms in terms of mass of the atoms

The frequency\(f\)of SHM is

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

Here,\(k\)is the spring constant and\(m\)is the mass.

Given data

The mass of the oxygen atom is\(m = 16\;{\rm{u}}\).

The frequency of oscillation is\(f = 2.83 \times {10^{13}}\;{\rm{Hz}}\).

Calculation of spring constant

When oxygen atoms oscillate along the carbon atom, then effective spring constant is taken into consideration as shown in the figure below:

Thus, the effective spring constant is calculated as:

\(\begin{aligned}{c}f &= \frac{1}{{2{\rm{\pi }}}}\sqrt {\frac{k}{m}} \\k &= 4{{\rm{\pi }}^2}{f^2}m\\ &= 4{{\rm{\pi }}^2}{\left( {2.83 \times {{10}^{13}}\;{\rm{Hz}}} \right)^2}\left( {16\;{\rm{u}}} \right)\frac{{\left( {1.66 \times {{10}^{ - 27}}\;{\rm{kg}}} \right)}}{{\left( {1\;{\rm{u}}} \right)}}\\ &= 8.40 \times {10^2}\;{\rm{N/m}}\end{aligned}\)

Hence, effective spring constant is \(8.40 \times {10^2}\;{\rm{N/m}}\).