Chapter 11, Problem 11C.1E

(a)

Interpretation Introduction

Interpretation:

Vibrational wavenumber of ${}^{12}C={}^{16}\text{O}$ has to be determined where the force constant of the bond is ${\text{908 Nm}}^{-1}$.

Concept Introduction:

Vibrational frequency of a molecule can be determined using the given formula.

  $\begin{array}{c}Vibrationalfrequency,\nu \text{ = }\frac{1}{2\pi }{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\\ \\ \text{where,}\\ \text{μ = Effective mass}\\ {\text{k}}_{\text{f}}\text{= Force constant}\end{array}$

Effective mass of a diatomic molecule is given by,

  $\mu =\frac{m_A{m}_B}{\left(m_A+m_B\right)}$

Wavenumber is the total number of wavelengths observed at a unit distance. Equation for wavenumber can be given as,

$\begin{array}{c}\text{Wavenumber, }\overline{\text{ν}}\text{ =}\frac{\text{1}}{\text{λ}}\text{ = }\frac{\text{ν}}{\text{c}}\\ \\ \text{where, }\\ \text{λ}\text{=}\text{wavelength}\\ \text{ν}=\text{frequency}\\ \text{c = speed of light}\end{array}$

Explanation of Solution

Given information is shown below,

  ${\text{force constant of the bond, k}}_f\text{ = }908{\text{ Nm}}^{-1}$

Effective mass of ${}^{12}{C}^{16}O$ is given by,

  $\begin{array}{c}\mu =\frac{m_A{m}_B}{\left(m_A+m_B\right)}\\ \\ =\frac{\left(\text{12}\text{.01}{m}_u\right)\left(16.00m_u\right)}{\left(\text{12}\text{.01}{m}_u\right)+\left(16.00m_u\right)}\\ \\ =\text{ 6}{\text{.86 m}}_u\\ \\ =\text{ 6}\text{.86}\times 1.66054\times {10}^{-27}kg\\ \\ =\text{ 1}\text{.14}\times {10}^{-26}\text{ kg}\end{array}$

Force constant of the bond is calculated as shown,

  $\begin{array}{c}\nu \text{ = }\frac{1}{2\pi }{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\\ \\ \text{ = }\frac{1}{2\pi }{\left(\frac{\left(908{\text{ Nm}}^{-1}\right)}{\left(\text{1}\text{.14}\times {10}^{-26}\text{ kg}\right)}\right)}^{\text{1/2}}\\ \\ \text{= }4.49\times {10}^{13}{\text{ s}}^{-1}\end{array}$

Wavenumber at which the force constant of the bond in ${}^{12}{C}^{16}O$ is ${\text{908 Nm}}^{-1}$ can be determined as follows,

  $\begin{array}{c}\overline{\text{ν}}\text{ =}\frac{\text{ν}}{\text{c}}\\ \\ =\frac{\left(4.49\times {10}^{13}{\text{ s}}^{-1}\right)}{\left(2.998\times {10}^8{\text{ ms}}^{-1}\right)}\\ \\ ={\text{ 149766 m}}^{-1}\\ \\ ={\text{ 1498 cm}}^{-1}\end{array}$

Vibrational wavenumber of ${}^{12}C={}^{16}\text{O}$ is ${\text{1498 cm}}^{-1}$.

(b)

Interpretation Introduction

Interpretation:

Vibrational wavenumber of ${}^{13}C={}^{16}\text{O}$ has to be determined where the force constant of the bond is ${\text{908 Nm}}^{-1}$.

Concept Introduction:

Vibrational frequency of a molecule can be determined using the given formula.

  $\begin{array}{c}Vibrationalfrequency,\nu \text{ = }\frac{1}{2\pi }{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\\ \\ \text{where,}\\ \text{μ = Effective mass}\\ {\text{k}}_{\text{f}}\text{= Force constant}\end{array}$

Effective mass of a diatomic molecule is given by,

  $\mu =\frac{m_A{m}_B}{\left(m_A+m_B\right)}$

Wavenumber is the total number of wavelengths observed at a unit distance. Equation for wavenumber can be given as,

$\begin{array}{c}\text{Wavenumber, }\overline{\text{ν}}\text{ =}\frac{\text{1}}{\text{λ}}\text{ = }\frac{\text{ν}}{\text{c}}\\ \\ \text{where, }\\ \text{λ}\text{=}\text{wavelength}\\ \text{ν}=\text{frequency}\\ \text{c = speed of light}\end{array}$

Explanation of Solution

Given information is shown below,

  ${\text{force constant of the bond, k}}_f\text{ = }908{\text{ Nm}}^{-1}$

Effective mass of ${}^{13}C={}^{16}\text{O}$ is given by,

  $\begin{array}{c}\mu =\frac{m_A{m}_B}{\left(m_A+m_B\right)}\\ \\ =\frac{\left(\text{13}\text{.00 }{m}_u\right)\left(16.00m_u\right)}{\left(\text{13}\text{.00}{m}_u\right)+\left(16.00m_u\right)}\\ \\ =\text{ 7}{\text{.17 m}}_u\\ \\ =\text{ 7}\text{.17}\times 1.66054\times {10}^{-27}kg\\ \\ =\text{ 1}\text{.19}\times {10}^{-26}\text{ kg}\end{array}$

Force constant of the bond is calculated as shown,

  $\begin{array}{c}\nu \text{ = }\frac{1}{2\pi }{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\\ \\ \text{ = }\frac{1}{2\pi }{\left(\frac{\left(908{\text{ Nm}}^{-1}\right)}{\left(\text{1}\text{.19}\times {10}^{-26}\text{ kg}\right)}\right)}^{\text{1/2}}\\ \\ \text{= }4.40\times {10}^{13}{\text{ s}}^{-1}\end{array}$

Wavenumber at which the force constant of the bond in ${}^{13}C={}^{16}\text{O}$ is ${\text{908 Nm}}^{-1}$ can be determined as follows,

  $\begin{array}{c}\overline{\text{ν}}\text{ =}\frac{\text{ν}}{\text{c}}\\ \\ =\frac{\left(4.40\times {10}^{13}{\text{ s}}^{-1}\right)}{\left(2.998\times {10}^8{\text{ ms}}^{-1}\right)}\\ \\ ={\text{ 146764 m}}^{-1}\\ \\ ={\text{ 1468 cm}}^{-1}\end{array}$

Vibrational wavenumber of ${}^{13}C={}^{16}\text{O}$ is ${\text{1468 cm}}^{-1}$.