Chapter 11, Problem 11.5P

(a)

Interpretation Introduction

Interpretation:

The moment of inertia of a linear molecule of the form ABC has to be derived.

Concept Introduction:

Moment of inertia:

The moment of inertia of a molecule is the mass of each atom multiplied by the square of its perpendicular distance from the axis of rotation.

$\text{I}={\displaystyle \sum _i{m}_i{r}_i^2}$

Explanation of Solution

Consider a triatomic linear molecule of the form ABC.  Let, $R_A,R_{B}and{R}_C$ are the distances from the center of gravity of atom A, B and C respectively.  Let $m_A,m_{B}and{m}_C$ are the masses of atom A, B and C respectively.

Figure.1

Now,

Moment around the center of gravity,

$I=m_A{R}_A^2+m_B{R}_B^2+m_C{R}_C^2\mathrm{......}\left(1\right)$

Then,

$m_A{R}_A+m_B{R}_B=m_C{R}_C\mathrm{.......}\left(2\right)$

$R_A=R_{BA}+R_B,R_C=R_{BC}-R_B\mathrm{........}\left(3\right)$

Now substitute the value of equation 3 in equation 2.

$\begin{array}{l}{m}_A{R}_A+m_B{R}_B=m_C{R}_C\\ \\ m_A\left(R_{BA}+R_B\right)+m_B{R}_B=m_C\left(R_{BC}-R_B\right)\\ \\ \left(m_A+m_B+m_C\right)R_B=m_C{R}_{BC}-m_C{R}_B\\ \\ M{R}_B=m_C{R}_{BC}-m_C{R}_B\mathrm{.....}\left(4\right)\end{array}$

Now substituting the equation 4 in equation 1,

$\begin{array}{l}I=m_A{\left(R_{AB}+R_B\right)}^2+m_B{R}_B^2+m_C{\left(R_{BC}-R_B\right)}^2\\ \\ =m_A{R}_{AB}^2+m_A{R}_B^2+2m_A{R}_{AB}{R}_B+m_B{R}_B^2+m_C{R}_{BC}^2+m_C{R}_B^2-2m_C{R}_{BC}{R}_B\\ \\ =\left(m_A+m_B+m_C\right)R_B^2+2\left(m_A{R}_{AB}-m_C{R}_{BC}\right)R_B+m_A{R}_{AB}^2+m_C{R}_{BC}^2\\ \\ =M{R}_B^2+2\left(m_A{R}_{AB}-m_C{R}_{BC}\right)R_B+m_A{R}_{AB}^2+m_C{R}_{BC}^2\mathrm{.........}\left(5\right)\end{array}$

Now substituting the value of $r_B$ from equation 4 into equation 5,

$I=M\frac{{\left(m_C{R}_{BC}-m_C{R}_B\right)}^2}{M^2}+2\left(m_A{R}_{AB}-m_C{R}_{BC}\right)\frac{\left(m_C{R}_{BC}-m_C{R}_B\right)}{M}+m_A{R}_{AB}^2+m_C{R}_{BC}^2$

It can be rearranged as

$\boxed{I=m_A{R}_{AB}^2+m_C{R}_{BC}^2-\frac{{\left(m_C{R}_{BC}-m_C{R}_B\right)}^2}{M}}$

This is the expression for the moment of inertia of a triatomic linear molecule of the form ABC.

(b)

Interpretation Introduction

Interpretation:

The rotational constants and moment of inertia of ${}^{16}{O}^{12}{C}^{32}Sand{}^{16}{O}^{12}{C}^{34}S$ has to be calculated.

Concept Introduction:

The gap between two adjacent spectral lines is $2B$, where $B=RotationalCons\tan t$.  Rotational constant can be mathematically represented as

$B=\frac{\hslash }{4\pi I}$

Answer to Problem 11.5P

The rotational constants of ${}^{16}{O}^{12}{C}^{32}Sand{}^{16}{O}^{12}{C}^{34}S$ has been calculated to be $6.08145GHz$ and $11.865035GHz$ respectively.  The moment of inertia of ${}^{16}{O}^{12}{C}^{32}Sand{}^{16}{O}^{12}{C}^{34}S$ has been calculated to be $137.9934\times 10^{-47}kg{m}^2$ and  $70.7288\times 10^{-47}kg{m}^2$ respectively.

Explanation of Solution

Given data:

Figure.2

Calculation of rotational constants:

For ${}^{16}{O}^{12}{C}^{32}S$:

The gap between the two spectral lines, $2B=\left(36.48882-24.32592\right)GHz=12.1629GHz$

Now

$B=\frac{12.1629GHz}{2}=6.08145GHz.$

For ${}^{16}{O}^{12}{C}^{34}S$:

The gap between the two spectral lines, $2B=\left(47.46240-23.73233\right)GHz=23.73007GHz$

Now

$B=\frac{23.73007GHz}{2}=11.865035GHz.$

Therefore, the rotational constants for ${}^{16}{O}^{12}{C}^{32}Sand{}^{16}{O}^{12}{C}^{34}S$ are $6.08145GHz$ and $11.865035GHz$ respectively.

Calculation of moment of inertia:

For ${}^{16}{O}^{12}{C}^{32}S$:

As $B=\frac{\hslash }{4\pi I}$. Moment of inertia can be calculated as

$I=\frac{\hslash }{4\pi B}$

So for ${}^{16}{O}^{12}{C}^{32}S$,

$I=\frac{\hslash }{4\pi B}=\frac{1.05457\times {10}^{-34}Js}{\left(4\pi \right)\left(6.08145GHz\right)}=137.9934\times {10}^{-47}kg{m}^2.$

So for ${}^{16}{O}^{12}{C}^{34}S$,

$I=\frac{\hslash }{4\pi B}=\frac{1.05457\times {10}^{-34}Js}{\left(4\pi \right)\left(11.865035GHz\right)}=70.7288\times {10}^{-47}kg{m}^2.$

Therefore, the moment of inertia of ${}^{16}{O}^{12}{C}^{32}Sand{}^{16}{O}^{12}{C}^{34}S$ are found to be $137.9934\times {10}^{-47}kg{m}^2$ and  $70.7288\times {10}^{-47}kg{m}^2$ respectively.

(c)

Interpretation Introduction

Interpretation:

The $C-OandC-S$ bond lengths of $OCS$ has to be determined.

Concept Introduction:

The bond length of a given molecule can be found out from its moment of inertia as moment of inertia is the product of mass and bond length.

$\begin{array}{l}I=\mu R^2\\ \\ R=\sqrt{\frac{I}{\mu }}\end{array}$

Answer to Problem 11.5P

The $C-OandC-S$ bond lengths of $OCS$ has been determined to be $116.28pm$ and $145.97pm$ respectively.

Explanation of Solution

The moment of inertia of $OCS$

$I=m_O{R}_{CO}^2+m_S{R}_{CS}^2-\frac{{\left(m_O{R}_{CO}-m_S{R}_{CS}\right)}^2}{M}$

It can be arranged further,

$I=\frac{m_O{m}_S}{M}{\left(R_{CO}+R_{CS}\right)}^2+\frac{m_O{m}_C}{M}{R}_{CO}^2+\frac{m_O{m}_S}{M}{R}_{CS}^2$

Now

$\begin{array}{l}I\left({}^{16}{O}^{12}{C}^{32}S\right)=\left(\frac{m\left({}^{16}O\right)m\left({}^{32}S\right)}{m\left({}^{16}{O}^{12}{C}^{32}S\right)}\right)\times {\left(R_{CO}+R_{CS}\right)}^2+\left(\frac{m\left({}^{12}C\right)m\left({}^{16}O\right)R_{CO}^2+m\left({}^{12}C\right)m\left({}^{32}S\right)R_{CS}^2}{m\left({}^{16}{O}^{12}{C}^{32}S\right)}\right)\\ \\ I\left({}^{16}{O}^{12}{C}^{34}S\right)=\left(\frac{m\left({}^{16}O\right)m\left({}^{34}S\right)}{m\left({}^{16}{O}^{12}{C}^{34}S\right)}\right)\times {\left(R_{CO}+R_{CS}\right)}^2+\left(\frac{m\left({}^{12}C\right)m\left({}^{16}O\right)R_{CO}^2+m\left({}^{12}C\right)m\left({}^{34}S\right)R_{CS}^2}{m\left({}^{16}{O}^{12}{C}^{34}S\right)}\right)\end{array}$

Where,

$\begin{array}{l}m\left({}^{16}O\right)=15.9949u\\ \\ m\left({}^{12}C\right)=12.0000u\\ \\ m\left({}^{32}S\right)=31.9721u\\ \\ m\left({}^{34}S\right)=33.9679u\end{array}$

$\begin{array}{l}I\left({}^{16}{O}^{12}{C}^{32}S\right)/u=\left(8.5279\right)\times {\left(R_{CO}+R_{CS}\right)}^2+\left(0.20011\right)\times \left(15.994R_{CO}^2+31.9721R_{CS}^2\right)\\ \\ I\left({}^{16}{O}^{12}{C}^{34}S\right)/u=\left(8.7684\right)\times {\left(R_{CO}+R_{CS}\right)}^2+\left(0.19366\right)\times \left(15.994R_{CO}^2+33.9679R_{CS}^2\right)\end{array}$

After putting the value of moment of inertia into the above equation,

$\begin{array}{l}137.9934\times {10}^{-47}{m}^2=\left(1.4161\times {10}^{-26}\right)\times {\left(R_{CO}+R_{CS}\right)}^2+\left(5.3150\times {10}^{-27}{R}_{CO}^2\right)+\left(1.0624\times {10}^{-26}{R}_{CS}^2\right)\\ \\ 70.7288\times {10}^{-47}{m}^2=\left(1.4560\times {10}^{-26}\right)\times {\left(R_{CO}+R_{CS}\right)}^2+\left(5.1437\times {10}^{-27}{R}_{CO}^2\right)+\left(1.0923\times {10}^{-26}{R}_{CS}^2\right)\end{array}$

After solving this complex equations, the outcome is $R_{CO}=116.28pm$ and $R_{CS}=145.97pm$.