Chapter 7, Problem 7E.1P

(a)

Interpretation Introduction

Interpretation:

It has to be shown that in a diatomic molecule $\text{A}-\text{B}$, when an isotopic substitution is made for atom $\text{A}$, the vibrational frequency of $\text{A'}-\text{B}$ can be expressed in term of vibrational frequency of $\text{A}-\text{B}$.

Concept introduction:

 Vibrational frequency is an important measurement that describes the information to identify phase, observe the bonding between atoms in a molecule and to study the relationships between vibrational frequency and phase transitions.

Answer to Problem 7E.1P

It has been proved that when an isotopic substitution is made for atom $\text{A}$, the vibrational frequency of $\text{A'}-\text{B}$ can be expressed in term of vibrational frequency of $\text{A}-\text{B}$.

Explanation of Solution

For a diatomic molecule $\text{A}-\text{B}$, the vibrational frequency is expressed as,

$w_{\text{A}-\text{B}}={\left(\frac{K_{\text{f}}}{{\mu }_{\text{A}-\text{B}}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ (1)

Where,

• $w_{\text{A}-\text{B}}$ is the vibational frequency of molecule $\text{A}-\text{B}$.
• $K_{\text{f}}$ is the force constant.
• ${\mu }_{\text{A}-\text{B}}$ is the effective mass for the molecule $\text{A}-\text{B}$.

The effective mass for the molecule $\text{A}-\text{B}$ is calculated by the formula,

  ${\mu }_{\text{A}-\text{B}}=\frac{m_{\text{A}}{m}_{\text{B}}}{m_{\text{A}}+m_{\text{B}}}$                                                                                            (2)

Where,

• $m_{\text{A}}$ is the mass of atom $\text{A}$.
• $m_{\text{B}}$ is the mass of atom $\text{B}$.

When an isotopic substitution is made for atom $\text{A}$, its mass $m_{\text{A}}$ changes to $m_{\text{A'}}$.  Hence, the effective mass ${\mu }_{\text{A}-\text{B}}$ changes to ${\mu }_{\text{A'}-\text{B}}$ and the vibrational frequency $w_{\text{A}-\text{B}}$ changes to $w_{\text{A'}-\text{B}}$.

Therefore,

  $w_{\text{A'}-\text{B}}={\left(\frac{K_{\text{f}}}{{\mu }_{\text{A'}-\text{B}}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$                                                                                          (3)

Divide the equation (3) by equation (1).

  $\begin{array}{c}\frac{w_{\text{A'}-\text{B}}}{w_{\text{A}-\text{B}}}={\left(\frac{K_{\text{f}}}{{\mu }_{\text{A'}-\text{B}}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times {\left(\frac{{\mu }_{\text{A}-\text{B}}}{K_{\text{f}}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ ={\left(\frac{{\mu }_{\text{A}-\text{B}}}{{\mu }_{\text{A'}-\text{B}}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}$

Or,

  $w_{\text{A'}-\text{B}}=w_{\text{A}-\text{B}}{\left(\frac{{\mu }_{\text{A}-\text{B}}}{{\mu }_{\text{A'}-\text{B}}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Therefore, it has been proved that when an isotopic substitution is made for atom $\text{A}$, the vibrational frequency of $\text{A'}-\text{B}$ can be expressed in term of vibrational frequency of $\text{A}-\text{B}$.

(b)

Interpretation Introduction

Interpretation:

Under the given conditions, the vibrational frequencies have to be calculated.

Concept introduction:

Vibrational frequency is an important measurement that describes the information to identify phase, observe the bonding between atoms in a molecule and to study the relationships between vibrational frequency and phase transitions.

Answer to Problem 7E.1P

The vibrational frequencies for the diatomic molecules, ${\text{H}}^{\text{2}}{\text{Cl}}^{\text{35}}$ and ${\text{H}}^1{\text{Cl}}^{\text{37}}$ are $\underset{\_}{4.036\times 10^{14}{s}^{-1}}$ and $\underset{\_}{5.626\times 10^{14}{s}^{-1}}$, respectively.

Explanation of Solution

(i)

For a diatomic molecule ${\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}$, when an isotopic substitution is made for atom ${\text{H}}^{\text{1}}$, its mass $m_{{\text{H}}^{\text{1}}}$ changes to $m_{{\text{H}}^2}$.  Hence, the effective mass ${\mu }_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}$ changes to ${\mu }_{{\text{H}}^2{\text{Cl}}^{\text{35}}}$ and the vibrational frequency $w_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}$ changes to $w_{{\text{H}}^2{\text{Cl}}^{\text{35}}}$.

Therefore, the vibrational frequency of $w_{{\text{H}}^2{\text{Cl}}^{\text{35}}}$ can be expressed in term of vibrational frequency of $w_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}$.

  $w_{{\text{H}}^2{\text{Cl}}^{\text{35}}}=w_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}{\left(\frac{{\mu }_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}}{{\mu }_{{\text{H}}^2{\text{Cl}}^{\text{35}}}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$                                                                             (1)

Where,

• $w_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}$ is the vibrational frequency for molecule ${\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}$.
• $w_{{\text{H}}^2{\text{Cl}}^{\text{35}}}$ is the vibrational frequency for molecule ${\text{H}}^{\text{2}}{\text{Cl}}^{\text{35}}$.
• ${\mu }_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}$ is the effective mass for molecule ${\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}$.
• ${\mu }_{{\text{H}}^2{\text{Cl}}^{\text{35}}}$ is the effective mass for molecule ${\text{H}}^{\text{2}}{\text{Cl}}^{\text{35}}$.

It is given that,

The vibrational frequency for ${\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}$ is $5.63\times {10}^{14}{\text{s}}^{-1}$.

The effective mass of ${\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}$ is calculated as,

  $\begin{array}{c}{\mu }_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}=\frac{m_{{\text{H}}^{\text{1}}}{m}_{{\text{Cl}}^{\text{35}}}}{m_{{\text{H}}^{\text{1}}}+m_{{\text{Cl}}^{\text{35}}}}\\ =\frac{1\times 35}{1+35}\text{amu}\\ \text{=}\frac{35}{36}\text{amu}\\ =\text{0}\text{.9723}\text{amu}\end{array}$

The effective mass of ${\text{H}}^{\text{2}}{\text{Cl}}^{\text{35}}$ is calculated as,

  $\begin{array}{c}{\mu }_{{\text{H}}^2{\text{Cl}}^{\text{35}}}=\frac{m_{{\text{H}}^2}{m}_{{\text{Cl}}^{\text{35}}}}{m_{{\text{H}}^2}+m_{{\text{Cl}}^{\text{35}}}}\\ =\frac{2\times 35}{2+35}\text{amu}\\ \text{=}\frac{70}{37}\text{amu}\\ =\text{1}\text{.892}\text{amu}\end{array}$

Substitute the values of $w_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}$, ${\mu }_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}$ and ${\mu }_{{\text{H}}^2{\text{Cl}}^{\text{35}}}$ in equation (1).

  $\begin{array}{c}{w}_{{\text{H}}^2{\text{Cl}}^{\text{35}}}=5.63\times {10}^{14}{\text{s}}^{-1}{\left(\frac{\text{0}\text{.9723}\text{amu}}{\text{1}\text{.892}\text{amu}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =5.63\times {10}^{14}\times 0.7168{\text{s}}^{-1}\\ =\underset{\_}{4.036\times 10^{14}{s}^{-1}}\end{array}$

(ii)

For a diatomic molecule ${\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}$, when an isotopic substitution is made for atom ${\text{Cl}}^{\text{35}}$, its mass $m_{{\text{Cl}}^{\text{35}}}$ changes to $m_{{\text{Cl}}^{\text{37}}}$.  Hence, the effective mass ${\mu }_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}$ changes to ${\mu }_{{\text{H}}^1{\text{Cl}}^{\text{37}}}$ and the vibrational frequency $w_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}$ changes to $w_{{\text{H}}^1{\text{Cl}}^{\text{37}}}$.

Therefore, the vibrational frequency of $w_{{\text{H}}^2{\text{Cl}}^{\text{35}}}$ can be expressed in term of vibrational frequency of $w_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}$.

  $w_{{\text{H}}^1{\text{Cl}}^{\text{37}}}=w_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}{\left(\frac{{\mu }_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}}{{\mu }_{{\text{H}}^1{\text{Cl}}^{\text{37}}}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$                                                                             (1)

Where,

• $w_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}$ is the vibrational frequency for molecule ${\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}$.
• $w_{{\text{H}}^1{\text{Cl}}^{\text{37}}}$ is the vibrational frequency for molecule ${\text{H}}^1{\text{Cl}}^{\text{37}}$.
• ${\mu }_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}$ is the effective mass for molecule ${\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}$.
• ${\mu }_{{\text{H}}^1{\text{Cl}}^{\text{37}}}$ is the effective mass for molecule ${\text{H}}^1{\text{Cl}}^{\text{37}}$.

It is given that,

The vibrational frequency for ${\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}$ is $5.63\times {10}^{14}{\text{s}}^{-1}$.

The effective mass of ${\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}$ is calculated as,

  $\begin{array}{c}{\mu }_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}=\frac{m_{{\text{H}}^{\text{1}}}{m}_{{\text{Cl}}^{\text{35}}}}{m_{{\text{H}}^{\text{1}}}+m_{{\text{Cl}}^{\text{35}}}}\\ =\frac{1\times 35}{1+35}\text{amu}\\ \text{=}\frac{35}{36}\text{amu}\\ =\text{0}\text{.9723}\text{amu}\end{array}$

The effective mass of ${\text{H}}^1{\text{Cl}}^{\text{37}}$ is calculated as,

  $\begin{array}{c}{\mu }_{{\text{H}}^1{\text{Cl}}^{\text{37}}}=\frac{m_{{\text{H}}^1}{m}_{{\text{Cl}}^{\text{37}}}}{m_{{\text{H}}^1}+m_{{\text{Cl}}^{\text{37}}}}\\ =\frac{1\times 37}{1+37}\text{amu}\\ \text{=}\frac{37}{38}\text{amu}\\ =\text{0}\text{.9736}\text{amu}\end{array}$

Substitute the values of $w_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}$, ${\mu }_{{\text{H}}^{\text{1}}{\text{Cl}}^{\text{35}}}$ and ${\mu }_{{\text{H}}^1{\text{Cl}}^{\text{37}}}$ in equation (1).

  $\begin{array}{c}{w}_{{\text{H}}^1{\text{Cl}}^{\text{37}}}=5.63\times {10}^{14}{\text{s}}^{-1}{\left(\frac{\text{0}\text{.9723}\text{amu}}{\text{0}\text{.9736}\text{amu}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =5.63\times {10}^{14}\times 0.9993{\text{s}}^{-1}\\ =\underset{\_}{5.626\times 10^{14}{s}^{-1}}\end{array}$