Chapter 7, Problem 7E.5P

(a)

Interpretation Introduction

Interpretation:

The wavenumber of the radiation corresponding to photons, $\tilde{\nu }$ is given by the relation $\tilde{\nu }=\omega /2\pi c$ has to be shown.

Concept Introduction:

A harmonic oscillator consists of a particle of mass $m$ that undergoes a restoring force which is proportional to its amplitude $x$.  The restoring force is represented by $k_{\text{f}}$ which is known as the force constant.  The vibrational frequency is $\omega$.  It is directly proportional to the force constant.

Answer to Problem 7E.5P

The wavenumber of the radiation corresponding to photons, $\tilde{\nu }$ is given by the relation $\tilde{\nu }=\omega /2\pi c$.

Explanation of Solution

The energy of the photons is given by the relation given below.

    $E=\hslash \omega$        (1)

Where,

$\hslash$ is a constant.

$\omega$ is the vibrational frequency.

The value of $\hslash$ is given below.

    $\hslash =\frac{h}{2\pi }$        (2)

Where,

$h$ is Planck’s constant.

The energy of photons is also given by the relation below.

    $E=\frac{hc}{\lambda }$        (3)

Where,

 $h$ is Planck’s constant.

 $c$ is the speed of light.

 $\lambda$ is the wavelength.

Substitute the equations (3) and (2) in (1).

    $\begin{array}{c}\frac{hc}{\lambda }=\frac{h}{2\pi }\omega \\ c=\frac{\omega \lambda }{2\pi }\\ \lambda =\frac{2\pi c}{\omega }\end{array}$        (4)

The relation wavelength $\left(\lambda \right)$ and wavenumber $\left(\tilde{\nu }\right)$ is given below.

    $\frac{1}{\lambda }=\tilde{\nu }$        (5)

Substitute equation (5) in (4).

    $\begin{array}{c}\frac{1}{\tilde{\nu }}=\frac{2\pi c}{\omega }\\ \tilde{\nu }=\frac{\omega }{2\pi c}\end{array}$

Therefore, the wavenumber of the radiation corresponding to photons, $\tilde{\nu }$ is given by the relation $\tilde{\nu }=\omega /2\pi c$.

(b)

Interpretation Introduction

Interpretation:

The $\tilde{\nu }$ of the molecule, ${}^1{\text{H}}^{35}\text{Cl}$ with $\omega =5.63\times {10}^{14}{\text{s}}^{-1}$ has to be calculated.

Concept Introduction:

Refer part (a).

Answer to Problem 7E.5P

The $\tilde{\nu }$ of the molecule, ${}^1{\text{H}}^{35}\text{Cl}$ with $\omega =5.63\times {10}^{14}{\text{s}}^{-1}$ is $\underset{\_}{0.299\times 10^6{m}^{-1}}$.

Explanation of Solution

The formula to calculate $\tilde{\nu }$ is given by the expression below.

    $\tilde{\nu }=\frac{\omega }{2\pi c}$        (6)

Where,

 $c$ is the speed of light.

 $\omega$ is the vibrational frequency.

The speed of light ($c$) is $3\times {10}^8\text{m}{\text{s}}^{-1}$.

The vibrational frequency ($\omega$) is $5.63\times {10}^{14}{\text{s}}^{-1}$

Substitute the value of $c$ and $\omega$ in equation (6).

    $\begin{array}{c}\tilde{\nu }=\frac{5.63\times {10}^{14}{\text{s}}^{-1}}{2\times 3.14\times 3\times {10}^8\text{m}{\text{s}}^{-1}}\\ =\underset{\_}{0.299\times 10^6{m}^{-1}}\end{array}$

Therefore, the $\tilde{\nu }$ of the molecule, ${}^1{\text{H}}^{35}\text{Cl}$ with $\omega =5.63\times {10}^{14}{\text{s}}^{-1}$ is $\underset{\_}{0.299\times 10^6{m}^{-1}}$.

(c)

Interpretation Introduction

Interpretation:

An expression for the force constant, $k_{\text{f}}$ has to derived in terms of $\tilde{\nu }$.

Concept introduction:

Refer part (a).

Answer to Problem 7E.5P

The expression for the force constant, $k_{\text{f}}$ is derived in terms of $\tilde{\nu }$ as shown below.

    $\tilde{\nu }=\frac{1}{2\pi c}\sqrt{\frac{k_{\text{f}}}{\mu }}$

Explanation of Solution

The formula to calculate $\tilde{\nu }$ is given by the expression below.

    $\tilde{\nu }=\frac{\omega }{2\pi c}$        (6)

Where,

 $c$ is the speed of light.

 $\omega$ is the vibrational frequency.

The force constant, $k_{\text{f}}$ is given by the expression below.

    $\omega =\sqrt{\frac{k_{\text{f}}}{\mu }}$

Where,

$\mu$ is the reduced mass of the molecule.

Substitute the equation (7) in equation (6).

    $\tilde{\nu }=\frac{1}{2\pi c}\sqrt{\frac{k_{\text{f}}}{\mu }}$

(d)

Interpretation Introduction

Interpretation:

The force constant and the wavenumber corresponding to the transition $v=0\to 1$ in ${}^{\text{12}}{\text{C}}^{\text{16}}\text{O}$ has to be calculated for the molecule ${}^{\text{14}}{\text{C}}^{\text{16}}\text{O}$.

Concept introduction:

Refer part (a).

Answer to Problem 7E.5P

The force constant of the ${}^{\text{14}}{\text{C}}^{\text{16}}\text{O}$ is $\underset{\_}{1902.1N{m}^{-1}}$.  The wavenumber at which the transition takes place from transition from $v=0\to 1$ for ${}^{\text{14}}{\text{C}}^{\text{16}}\text{O}$ is $\underset{\_}{2.07\times 10^5{m}^{-1}}$.

Explanation of Solution

The formula to calculate the wavenumber of the transition from $v=0\to 1$ is given below.

    $\tilde{\nu }=\frac{1}{2\pi c}\sqrt{\frac{k_{\text{f}}}{\mu }}$        (7)

Where,

$k_{\text{f}}$ is force constant.

$\mu$ is the reduced mass of the molecule.

$c$ is the speed of light.

The reduced mass of ${}^{\text{12}}{\text{C}}^{\text{16}}\text{O}$ is calculated by the formula shown below.

    $\mu =\frac{m_{{}^{\text{12}}\text{C}}{m}_{{}^{\text{16}}\text{O}}}{m_{{}^{\text{12}}\text{C}}+m_{{}^{\text{16}}\text{O}}}$        (8)

The mass of ${}^{\text{12}}\text{C}$ ($m_{{}^{\text{12}}\text{C}}$) is $\text{12}\text{amu}$.

The mass of ${}^{\text{16}}\text{O}$ ($m_{{}^{\text{16}}\text{O}}$) is $\text{16}\text{amu}$.

Substitute the value of $m_{{}^{\text{12}}\text{C}}$ and $m_{{}^{\text{16}}\text{O}}$ in equation (8).

    $\begin{array}{c}\mu =\frac{12\text{amu}\times 16\text{amu}}{12\text{amu}+16\text{amu}}\\ =6.857\text{amu}\end{array}$

The conversion of $\text{1}\text{amu}$ to $\text{kg}$ is shown below.

    $\text{1}\text{amu}=\text{1}\text{.66}\times {\text{10}}^{-27}\text{kg}$

Therefore, the conversion of $6.857\text{amu}$ to $\text{kg}$ is shown below.

    $\begin{array}{c}6.857\text{amu}=6.857\times \text{1}\text{.66}\times {\text{10}}^{-27}\text{kg}\\ \text{=1}\text{.138}\times {\text{10}}^{-26}\text{kg}\end{array}$

The value of $\tilde{\nu }$ is $2170{\text{cm}}^{-1}$.

The conversion of $1{\text{cm}}^{-1}$ to ${\text{m}}^{-1}$ is shown below.

    $1{\text{cm}}^{-1}={10}^2{\text{m}}^{-1}$

Therefore, the conversion of $2170{\text{cm}}^{-1}$ to ${\text{m}}^{-1}$ is shown below.

    $2170{\text{cm}}^{-1}=2170\times {10}^2{\text{m}}^{-1}$

The speed of light ($c$) is $3\times {10}^8\text{m}{\text{s}}^{-1}$.

Substitute the value of $\tilde{\nu }$, $c$ and $\mu$ in equation (7).

    $\begin{array}{c}2170\times {10}^2{\text{m}}^{-1}=\frac{1}{2\times 3.14\times 3\times {10}^8\text{m}{\text{s}}^{-1}}\sqrt{\frac{k_{\text{f}}}{\text{1}\text{.138}\times {\text{10}}^{-26}\text{kg}}}\\ k_{\text{f}}=4\times {\left(3.14\right)}^2\times {\left(3\times {10}^8\text{m}{\text{s}}^{-1}\right)}^2\times {\left(2170\times {10}^2{\text{m}}^{-1}\right)}^2\times \text{1}\text{.138}\times {\text{10}}^{-26}\text{kg}\\ =1902.1\text{kg}{\text{m}}^2{\text{s}}^{-2}{\text{m}}^{-2}\times \frac{1\text{N}{\text{m}}^{\text{-1}}}{\text{kg}{\text{m}}^2{\text{s}}^{-2}{\text{m}}^{-2}}\\ =\text{1902}\text{.1}\text{N}{\text{m}}^{\text{-1}}\end{array}$

The force constant of ${}^{\text{12}}{\text{C}}^{\text{16}}\text{O}$ is equal to force constant of ${}^{\text{14}}{\text{C}}^{\text{16}}\text{O}$.

Therefore, the force constant of the ${}^{\text{12}}{\text{C}}^{\text{16}}\text{O}$ is $\underset{\_}{1902.1N{m}^{-1}}$.

The reduced mass of ${}^{\text{14}}{\text{C}}^{\text{16}}\text{O}$ is calculated by the formula shown below.

    $\mu =\frac{m_{{}^{\text{14}}\text{C}}{m}_{{}^{\text{16}}\text{O}}}{m_{{}^{\text{14}}\text{C}}+m_{{}^{\text{16}}\text{O}}}$        (8)

The mass of ${}^{\text{14}}\text{C}$ ($m_{{}^{\text{12}}\text{C}}$) is $\text{14}\text{amu}$.

The mass of ${}^{\text{16}}\text{O}$ ($m_{{}^{\text{16}}\text{O}}$) is $\text{16}\text{amu}$.

Substitute the value of $m_{{}^{\text{14}}\text{C}}$ and $m_{{}^{\text{16}}\text{O}}$ in equation (8).

    $\begin{array}{c}\mu =\frac{14\text{amu}\times 16\text{amu}}{14\text{amu}+16\text{amu}}\\ =7.47\text{amu}\end{array}$

The conversion of $\text{1}\text{amu}$ to $\text{kg}$ is shown below.

    $\text{1}\text{amu}=\text{1}\text{.66}\times {\text{10}}^{-27}\text{kg}$

Therefore, the conversion of $7.47\text{amu}$ to $\text{kg}$ is shown below.

    $\begin{array}{c}7.47\text{amu}=7.47\times \text{1}\text{.66}\times {\text{10}}^{-27}\text{kg}\\ =\text{1}\text{.239}\times {\text{10}}^{-26}\text{kg}\end{array}$

The force constant ($k_{\text{f}}$) is $\text{1902}\text{.1}\text{N}{\text{m}}^{\text{-1}}$.

Substitute the value of $c$, $k_{\text{f}}$ and $\mu$ in equation (1).

    $\begin{array}{c}\tilde{\nu }=\frac{1}{2\times 3.14\times 3\times {10}^8\text{m}{\text{s}}^{-1}}\sqrt{\frac{\text{1902}\text{.1}\text{N}{\text{m}}^{\text{-1}}}{\text{1}\text{.239}\times {\text{10}}^{-26}\text{kg}}}\\ =\frac{1}{6.28\times 3\times {10}^8\text{m}{\text{s}}^{-1}}\sqrt{\frac{\text{1902}\text{.1}\text{N}{\text{m}}^{\text{-1}}}{\text{1}\text{.239}\times {\text{10}}^{-26}\text{kg}}\times \frac{\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-2}}{1\text{N}}}\\ =\frac{1}{6.28\times 3\times {10}^8\text{m}{\text{s}}^{-1}}\times 39.18\times {10}^{13}{\text{s}}^{-1}\\ =\underset{\_}{2.07\times 10^5{m}^{-1}}\end{array}$

Therefore, the wavenumber at which the transition takes place from transition from $v=0\to 1$ is $\underset{\_}{2.07\times 10^5{m}^{-1}}$.