Question

Question: The carbon monoxide molecule CO has an effective spring constant of 1860N/m and a bond length of 0.113nnm . Determine four wavelengths of light that CO might absorb in vibration-rotation transitions.

Short answer

Answer

The wavelengths that might be absorbed in vibration-rotation transition are $4.66\mu m,4.67\mu m,4.69\mu m,4.70\mu m$, etc.

Step-by-step solution

Given Data

The effective spring constant is 1860 N/m

The bond length is0.113 nm

Concept of Bond length

The mean separation between the nuclei of two bonded atoms in a molecule is known as bond length. The following factors affect the bond length between two bonded atoms- hybridization, the number of atoms present in a molecule, the number of bonds present and the size of the atom.

Calculation of energy of photon

The energy of a photon that might be absorbed is given as-

$E=\hslash \sqrt{\frac{\kappa }{\mu }}\pm I\frac{{\hslash }^2}{\mu a^2}·····································\left(1\right)$

Here,$\hslash$is the modified Planck’s constant, k is the spring constant,$\mu$is the effective mass of the atom andis the bond length.

The effective mass is given as-

$\mu =\frac{mm\text{'}}{m+m\text{'}}$

Where and are the masses of the atoms participating in bonding.

The mass number of carbon and oxygen are respectively, so the effective mass of CO molecule is-

$$\begin{gathered} \mu =\frac{12.01\times 16.00}{12.01+16.00}\times 1.66\times {10}^{-27}\text{kg} \\ \text{= 1.14}\times {10}^{-26}\text{kg} \end{gathered}$$

For $\kappa =1860N/ma=0,113nm\text{and}\mu =1.14\times {10}^{-26}\text{kg}$, the energy of photon is-

$$\begin{gathered} E=\left(1.05\times {10}^{-34}\text{Js}\right)\sqrt{\frac{1860N/m}{1.14\times {10}^{-26}\text{kg}}}\pm I\frac{{\left(1.05\times {10}^{-34}\text{Js}\right)}^2}{\left(1.14\times {10}^{-26}\text{kg}\right)\left(0.113\times {10}^{-9}\text{m}\right)} \\ E=\left(0.265\text{eV}\right)\pm I\left(0.0005\text{eV}\right) \\ =0.265\text{eV}\left(1\pm I\frac{0.0005\text{eV}}{0.265\text{eV}}\right) \\ =0.265\text{eV}\left(1\pm I0.002\right) \end{gathered}$$

$$\begin{gathered} E=\left(1.05\times {10}^{-34}\text{Js}\right)\sqrt{\frac{1860N/m}{1.14\times {10}^{-26}\text{kg}}}\pm I\frac{{\left(1.05\times {10}^{-34}\text{Js}\right)}^2}{\left(1.14\times {10}^{-26}\text{kg}\right)\left(0.113\times {10}^{-9}\text{m}\right)} \\ E=\left(0.265\text{eV}\right)\pm I\left(0.0005\text{eV}\right) \\ =0.265\text{eV}\left(1\pm I\frac{0.0005\text{eV}}{0.265\text{eV}}\right) \\ =0.265\text{eV}\left(1\pm I0.002\right) \end{gathered}$$

Calculation of wavelength of photon

It is a common observation that, for the $\Delta n=1$ transition and the photon of energy $0.265\text{eV}$, there are many wavelengths present around the photon. These wavelengths are-

So, the required wavelengths are- $4.66\mu m,4.67\mu m,4.69\mu m,4.70\mu m$.

$$\begin{gathered} \lambda =\frac{hc}{E} \\ =\frac{\left(6.626\times {10}^{-34}\text{Js}\right)\left(6.24\times {10}^{18}\right)\left(3\times {10}^8\right)}{0.265\text{eV}\left(1\pm I0.002\right)} \\ =\frac{4.68\times {10}^{-6}\text{m}}{\left(1\pm I0.002\right)} \end{gathered}$$