Chapter 7, Problem 7.3PR

(a)

Interpretation Introduction

Interpretation:

The vibrational frequency of the molecule has to be calculated.

Concept Introduction:

The energies are quantized in harmonic oscillator and are generally expressed as follows,

$\begin{array}{c}{\text{E}}_{\text{v}}=\left(\upsilon +\frac{1}{2}\right)h\nu \\ \upsilon =0,1,2,\mathrm{...}\\ \nu =\frac{1}{2\pi }{\left(\frac{k_f}{m}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\to 1\\ here,k_f=\text{force}\text{constant}\\ \upsilon =\text{vibrational quantum number}\\ m=mass\\ \nu =vibrationalfrequency\\ \end{array}$

The force constant, ${\text{k}}_{\text{f}}$ is used to characterize the particle that under harmonic motion subjected to Hooke’s law restoring force.

The vibrational quantum number is used to denote the energy levels of harmonic oscillator.

Explanation of Solution

Given:

The m is replaced by the effective mass $\text{μ = }\frac{{\text{m}}_{\text{A}}{\text{m}}_{\text{B}}}{\left({\text{m}}_{\text{A}}{\text{+m}}_{\text{B}}\right)}$.

The vibrational frequency, of the molecule is calculated as follows by the given m value into equation 1 as follows,

$\begin{array}{c}\nu =\frac{1}{2\pi }{\left(\frac{k_f}{m}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\to 1\\ \\ here,\mu =\frac{m_A{m}_B}{m_A+m_B}=\frac{M_A{M}_B}{\left[N_A\left(M_A+M_B\right)\right]}\\ \\ \nu =\frac{1}{2\pi }{\left(\frac{k_f}{\mu }\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ \\ {\mu }_{{}^{12}{C}^{16}O}=\frac{12\times 16\times {10}^{-3}kg}{6.0221\times {10}^{23}\times 28}\\ \\ =1.139\times {10}^{-26}kg\\ \\ {\nu }_{{}^{12}{C}^{16}O}=\frac{1}{2\pi }{\left(\frac{1860N{m}^{-1}}{1.139\times {10}^{-26}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ \\ =6.432\times {10}^{13}{s}^{-1}\\ \end{array}$

(b)

Interpretation Introduction

Interpretation:

The vibrational wavenumber of the given molecule has to be calculated.

Concept Introduction:

The energies are quantized in harmonic oscillator and are generally expressed as follows,

$\begin{array}{c}{\text{E}}_{\text{v}}=\left(\upsilon +\frac{1}{2}\right)h\nu \\ \upsilon =0,1,2,\mathrm{...}\\ \nu =\frac{1}{2\pi }{\left(\frac{k_f}{m}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\to 1\\ here,k_f=\text{force}\text{constant}\\ \upsilon =\text{vibrational quantum number}\\ m=mass\\ \nu =vibrationalfrequency\\ \end{array}$

The force constant, ${\text{k}}_{\text{f}}$ is used to characterize the particle that under harmonic motion subjected to Hooke’s law restoring force.

The vibrational quantum number is used to denote the energy levels of harmonic oscillator.

The vibrational wavenumber is determined by taking reciprocal of wavelength.

Explanation of Solution

The vibrational wavenumber for given molecule is determined as shown below,

$\begin{array}{c}{\tilde{\nu }}_{{}^{12}{C}^{16}O}=\frac{1}{{\lambda }_{{}^{12}{C}^{16}O}}\\ \\ =\frac{{\nu }_{{}^{12}{C}^{16}O}}{c}\\ \\ {\nu }_{{}^{12}{C}^{16}O}=6.432\times {10}^{13}{s}^{-1}\\ \\ =\frac{6.432\times {10}^{13}{s}^{-1}}{3\times {10}^{8}m{s}^{-1}}\\ \\ =2.144\times {10}^5{m}^{-1}\\ \\ =2144c{m}^{-1}\\ \end{array}$

(c)

Interpretation Introduction

Interpretation:

The vibrational wavenumber of the given molecules has to be calculated.

Concept Introduction:

The energies are quantized in harmonic oscillator and are generally expressed as follows,

$\begin{array}{c}{\text{E}}_{\text{v}}=\left(\upsilon +\frac{1}{2}\right)h\nu \\ \upsilon =0,1,2,\mathrm{...}\\ \nu =\frac{1}{2\pi }{\left(\frac{k_f}{m}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\to 1\\ here,k_f=\text{force}\text{constant}\\ \upsilon =\text{vibrational quantum number}\\ m=mass\\ \nu =vibrationalfrequency\\ \end{array}$

The force constant, ${\text{k}}_{\text{f}}$ is used to characterize the particle that under harmonic motion subjected to Hooke’s law restoring force.

The vibrational quantum number is used to denote the energy levels of harmonic oscillator.

The vibrational wavenumber is determined by taking reciprocal of wavelength.

Explanation of Solution

First, the ${\mu }_{{}^{12}{C}^{16}O}$ is determined and then the ${\nu }_{{}^{12}{C}^{16}O}$ is calculated which finally used to identify the wave number value as follows,

The vibrational frequency, of the molecule is calculated as follows by the given m value into equation 1 as follows,

$\begin{array}{c}\nu =\frac{1}{2\pi }{\left(\frac{k_f}{m}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\to 1\\ \\ here,\mu =\frac{m_A{m}_B}{m_A+m_B}=\frac{M_A{M}_B}{\left[N_A\left(M_A+M_B\right)\right]}\\ \\ \nu =\frac{1}{2\pi }{\left(\frac{k_f}{\mu }\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ \\ {\mu }_{{}^{12}{C}^{16}O}=\frac{12\times 16\times {10}^{-3}kg}{6.0221\times {10}^{23}\times 28}\\ \\ =1.139\times {10}^{-26}kg\\ \\ {\nu }_{{}^{12}{C}^{16}O}=\frac{1}{2\pi }{\left(\frac{1860N{m}^{-1}}{1.139\times {10}^{-26}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ \\ =6.432\times {10}^{13}{s}^{-1}\\ \end{array}$

The vibrational wavenumber for given molecule is determined as shown below,

$\begin{array}{c}{\tilde{\nu }}_{{}^{12}{C}^{16}O}=\frac{1}{{\lambda }_{{}^{12}{C}^{16}O}}\\ \\ =\frac{{\nu }_{{}^{12}{C}^{16}O}}{c}\\ \\ {\nu }_{{}^{12}{C}^{16}O}=6.432\times {10}^{13}{s}^{-1}\\ \\ =\frac{6.432\times {10}^{13}{s}^{-1}}{3\times {10}^{8}m{s}^{-1}}\\ \\ =2.144\times {10}^5{m}^{-1}\\ \\ =2144c{m}^{-1}\\ \end{array}$

Next, the ${\mu }_{{}^{13}{C}^{16}O}$ is determined and then the ${\nu }_{{}^{13}{C}^{16}O}$ is calculated which finally used to identify the wave number value as follows,

The vibrational frequency, of the molecule is calculated as follows by the given m value into equation 1 as follows,

$\begin{array}{c}\nu =\frac{1}{2\pi }{\left(\frac{k_f}{m}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\to 1\\ \\ here,\mu =\frac{m_A{m}_B}{m_A+m_B}=\frac{M_A{M}_B}{\left[N_A\left(M_A+M_B\right)\right]}\\ \\ \nu =\frac{1}{2\pi }{\left(\frac{k_f}{\mu }\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ \\ {\mu }_{{}^{13}{C}^{16}O}=\frac{13\times 16\times {10}^{-3}kg}{6.0221\times {10}^{23}\times 29}\\ \\ =1.191\times {10}^{-26}kg\\ \end{array}$

The vibrational wavenumber is identified by considering that the given isotopic CO molecules are differ only in reduced masses since they have similar identical force constants. The frequency and wavenumber are inversely proportional to ${\text{μ}}^{\text{1/2}}$ the wavenumber calculation is as follows,

$\begin{array}{c}{\tilde{\nu }}_{{}^{13}{C}^{16}O}={\left(\frac{{\mu }_{{}^{12}{C}^{16}O}}{{\mu }_{{}_{{}^{13}{C}^{16}O}}}\right)}^{1/2}{\tilde{\nu }}_{{}^{12}{C}^{16}O}\\ \\ ={\left(\frac{1.139}{1.191}\right)}^{1/2}\times 2146c{m}^{-1}\\ =2099c{m}^{-1}\\ \end{array}$

Similarly, the ${\mu }_{{}^{12}{C}^{18}O}$ is calculated as follows,

$\begin{array}{c}\nu =\frac{1}{2\pi }{\left(\frac{k_f}{m}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\to 1\\ \\ here,\mu =\frac{m_A{m}_B}{m_A+m_B}=\frac{M_A{M}_B}{\left[N_A\left(M_A+M_B\right)\right]}\\ \\ \nu =\frac{1}{2\pi }{\left(\frac{k_f}{\mu }\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ \\ {\mu }_{{}^{12}{C}^{18}O}=\frac{12\times 18\times {10}^{-3}kg}{6.0221\times {10}^{23}\times 30}\\ \\ =1.196\times {10}^{-26}kg\\ \end{array}$

$\begin{array}{c}{\tilde{\nu }}_{{}^{12}{C}^{18}O}={\left(\frac{{\mu }_{{}^{12}{C}^{16}O}}{{\mu }_{{}_{{}^{12}{C}^{18}O}}}\right)}^{1/2}{\tilde{\nu }}_{{}^{12}{C}^{16}O}\\ \\ ={\left(\frac{1.139}{1.196}\right)}^{1/2}\times 2146c{m}^{-1}\\ =2094c{m}^{-1}\\ \end{array}$

Similarly, the ${\mu }_{{}^{13}{C}^{18}O}$ is calculated as follows,

$\begin{array}{c}\nu =\frac{1}{2\pi }{\left(\frac{k_f}{m}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\to 1\\ \\ here,\mu =\frac{m_A{m}_B}{m_A+m_B}=\frac{M_A{M}_B}{\left[N_A\left(M_A+M_B\right)\right]}\\ \\ \nu =\frac{1}{2\pi }{\left(\frac{k_f}{\mu }\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ \\ {\mu }_{{}^{13}{C}^{18}O}=\frac{13\times 18\times {10}^{-3}kg}{6.0221\times {10}^{23}\times 31}\\ \\ =1.253\times {10}^{-26}kg\\ \end{array}$

$\begin{array}{c}{\tilde{\nu }}_{{}^{13}{C}^{18}O}={\left(\frac{{\mu }_{{}^{12}{C}^{16}O}}{{\mu }_{{}_{{}^{13}{C}^{18}O}}}\right)}^{1/2}{\tilde{\nu }}_{{}^{12}{C}^{16}O}\\ \\ ={\left(\frac{1.139}{1.25}\right)}^{1/2}\times 2146c{m}^{-1}\\ =2046c{m}^{-1}\\ \end{array}$