Chapter 11, Problem 11.2P

(a)

Interpretation Introduction

Interpretation:

The bond length of the molecule ${}^{\text{1}}{\text{H}}^{\text{127}}\text{I}$ has to be determined using the given data.

Concept Introduction:

Rotational constant of a molecule can be determined using the given formula.

  $\begin{array}{c}\text{Rotational constant, B = }\frac{\hslash }{\text{4πI}}\\ \\ \text{where,}\\ \text{I = moment of inertia }\end{array}$

Moment of inertia of diatomic molecules is given by,

  $\begin{array}{l}\text{Moment of inertia, I = }\mu {\text{R}}^2\\ \\ where,\mu =\frac{m_A{m}_B}{\left(m_A+m_B\right)}\end{array}$

Explanation of Solution

Given information is shown below,

  $\begin{array}{c}{\text{2B = 384 GHz = 384×10}}^{\text{9}}\text{ Hz}\\ {\text{= 384×10}}^{\text{9}}{\text{ s}}^{\text{-1}}\end{array}$

Therefore, $\text{B}\text{=}{\text{384×10}}^{\text{9}}{\text{ s}}^{\text{-1}}/\text{2}\text{=}\text{1}{\text{.92×10}}^{\text{11}}{\text{s}}^{\text{-1}}$

Effective mass of ${}^{\text{1}}{\text{H}}^{\text{127}}\text{I}$ is given by,

  $\begin{array}{c}\mu =\frac{m_A{m}_B}{\left(m_A+m_B\right)}\\ \\ =\frac{\left(126.90m_u\right)\left(1.008m_u\right)}{\left(126.90m_u\right)+\left(1.008m_u\right)}\\ \\ =\text{ 1}{\text{.0 m}}_u\\ \\ =\text{ 1}\text{.0}\times 1.66054\times {10}^{-27}kg\\ \\ =1.66054\times {10}^{-27}kg\end{array}$

Moment of inertia in ${}^{\text{1}}{\text{H}}^{\text{127}}\text{I}$ is determined as given,

  $\begin{array}{c}\text{Rotational constant, B = }\frac{\hslash }{\text{4πI}}\\ \\ \text{1}{\text{.92×10}}^{\text{11}}{\text{s}}^{\text{-1}}\text{ = }\frac{1.05457\times {10}^{-34}Js}{\text{4}\times \text{3}\text{.14}\times I}\\ \\ I=\frac{1.05457\times {10}^{-34}Js}{\text{4}\times \text{3}\text{.14}\times \left(\text{1}{\text{.92×10}}^{\text{11}}{\text{s}}^{\text{-1}}\right)}\\ \\ =\text{ 4}\text{.37}\times {\text{ 10}}^{-47}\text{ kg}{\text{.m}}^2\end{array}$

Bond length in ${}^{\text{1}}{\text{H}}^{\text{127}}\text{I}$ is calculated using the equation for moment of inertia.

  $\begin{array}{c}I\text{ = }\mu R^2\\ \\ \left(\text{4}\text{.37}\times {\text{ 10}}^{-47}\text{ kg}{\text{.m}}^2\right)=\left(1.66054\times {10}^{-27}kg\right)R^2\\ \\ R^2=\frac{\left(\text{4}\text{.37}\times {\text{ 10}}^{-47}\text{ kg}{\text{.m}}^2\right)}{\left(1.66054\times {10}^{-27}kg\right)}\\ \\ =\text{ 2}\text{.63}\times {\text{10}}^{-20}{\text{ m}}^2\\ \\ R\text{ = }\sqrt{\left(\text{5}\text{.39}\times {\text{10}}^{-20}{\text{ m}}^2\right)}\\ \\ =\text{ 1}\text{.62}\times {\text{10}}^{-10}m\end{array}$

Bond length in ${}^{\text{1}}{\text{H}}^{\text{127}}\text{I}$ is $\text{1}\text{.62}\times {\text{10}}^{-10}m$

(b)

Interpretation Introduction

Interpretation:

The separation of the lines in ${}^2{\text{H}}^{\text{127}}\text{I}$ has to be determined.

Concept Introduction:

Rotational constant of a molecule can be determined using the given formula.

  $\begin{array}{c}\text{Rotational constant, B = }\frac{\hslash }{\text{4πI}}\\ \\ \text{where,}\\ \text{I = moment of inertia }\end{array}$

Moment of inertia of diatomic molecules is given by,

  $\begin{array}{l}\text{Moment of inertia, I = }\mu {\text{R}}^2\\ \\ where,\mu =\frac{m_A{m}_B}{\left(m_A+m_B\right)}\end{array}$

Explanation of Solution

Given information is shown below,

  $\begin{array}{c}{\text{2B = 384 GHz = 384×10}}^{\text{9}}\text{ Hz}\\ {\text{= 384×10}}^{\text{9}}{\text{ s}}^{\text{-1}}\end{array}$

Therefore, $\text{B}\text{=}{\text{384×10}}^{\text{9}}{\text{ s}}^{\text{-1}}/\text{2}\text{=}\text{1}{\text{.92×10}}^{\text{11}}{\text{s}}^{\text{-1}}$

Effective mass of ${}^2{\text{H}}^{\text{127}}\text{I}$ is given by,

  $\begin{array}{c}\mu =\frac{m_A{m}_B}{\left(m_A+m_B\right)}\\ \\ =\frac{\left(126.90m_u\right)\left(\text{2}\text{.0141}{m}_u\right)}{\left(126.90m_u\right)+\left(\text{2}\text{.0141}{m}_u\right)}\\ \\ =\text{ 1}{\text{.98 m}}_u\\ \\ =\text{ 1}\text{.98}\times 1.66054\times {10}^{-27}kg\\ \\ =\text{ 3}\text{.29}\times {10}^{-27}kg\end{array}$

The relationship between rotational constants and reduced masses are given below:

  $\begin{array}{l}\frac{B}{B^{\text{'}}}=\frac{{\mu }^{\text{'}}}{\mu }\\ \\ B-Rotationalcons\tan tof{}^{\text{1}}{\text{H}}^{\text{127}}\text{I}\\ B^{\text{'}}-Rotationalcons\tan tof{}^2{\text{H}}^{\text{127}}\text{I}\\ \mu -\mathrm{Re}ducedmassof{}^{\text{1}}{\text{H}}^{\text{127}}\text{I}\\ {\mu }^{\text{'}}-\mathrm{Re}ducedmassof{}^2{\text{H}}^{\text{127}}\text{I}\end{array}$

$\begin{array}{c}\frac{\text{B}}{{\text{B}}^{\text{'}}}\text{=}\frac{{\text{μ}}^{\text{'}}}{\text{μ}}\\ \\ \frac{\text{1}{\text{.92×10}}^{\text{11}}{\text{s}}^{\text{-1}}}{{\text{B}}^{\text{'}}}\text{=}\frac{\text{3}{\text{.29×10}}^{\text{-27}}\text{kg}}{\text{1}{\text{.66054×10}}^{\text{-27}}\text{kg}}\\ {\text{B}}^{\text{'}}\text{=}\frac{\text{1}{\text{.66054×10}}^{\text{-27}}\text{kg×}\text{1}{\text{.92×10}}^{\text{11}}{\text{s}}^{\text{-1}}}{\text{3}{\text{.29×10}}^{\text{-27}}\text{kg}}\\ \\ \text{=}\text{9}{\text{.69×10}}^{\text{10}}{\text{s}}^{\text{-1}}\end{array}$

The separation is expressed as $\text{2}{\text{B}}^1$

$\begin{array}{c}\text{2}{\text{B}}^{\text{1}}\text{=}\text{2}\text{×9}{\text{.69×10}}^{\text{10}}{\text{s}}^{\text{-1}}\\ \\ \text{=1}{\text{.94×10}}^{\text{11}}{\text{s}}^{\text{-1}}\end{array}$

The separation of the lines in ${}^2{\text{H}}^{\text{127}}\text{I}$ is $\text{1}{\text{.94×10}}^{\text{11}}{\text{s}}^{\text{-1}}$.