Chapter 11, Problem 11C.7P

(a)

Interpretation Introduction

Interpretation:

The fundamental vibrational wavenumber, force constant, moment of inertia and rotational constant has to be stated based on harmonic oscillator and rigid-rotor approximation for ${\text{He}}_{\text{2}}$ complex.

Concept introduction:

Rotational spectroscopy measures the electronic energy between the rotational states.  The rotational constant is inversely proportional to the moment of inertia.  The rotational constant is given by an expression shown below.

  $B=\frac{h}{8{\pi }^2{I}_{B}c}$

Answer to Problem 11C.7P

The value of the fundamental vibrational wavenumber is $\underset{\_}{1.01\times 10^{-1}{m}^{-1}}$.

The value of the force constant is $\underset{\_}{1.20\times 10^{-10}kg{s}^{-1}}$.

The moment of inertia is $\underset{\_}{2.93\times 10^{-46}kg{m}^2}$.

The value of the rotational constant is $\underset{\_}{95.6m^{-1}}$.

Explanation of Solution

The fundamental vibrational wavenumber is given by the expression as shown below.

  $\tilde{v}={\tilde{D}}_o$

Where, $\tilde{v}$ is the fundamental vibrational wavenumber and ${\tilde{D}}_o$ is the depth of the potential energy minimum.

The given value for $hc{\tilde{D}}_o$ is shown below.

  $hc{\tilde{D}}_o=2\times {10}^{-26}\text{J}$        (1)

Where,

• $h$ is the Planck’s constant
• $c$ is the velocity of light.

The value of $h$ is $6.626\times {10}^{-34}\text{J}\text{s}$ and the value of $c$ is $2.998\times {10}^8\text{m}{\text{s}}^{-1}$

Substitute the values of corresponding variables in equation (1) to calculate the fundamental vibrational wavenumber as shown below.

  $\begin{array}{c}hc{\tilde{D}}_o=2\times {10}^{-26}\text{J}\\ {\tilde{D}}_o=\frac{2\times {10}^{-26}\text{J}}{\left(6.626\times {10}^{-34}\text{J}\text{s}\right)\times \left(2.998\times {10}^8\text{m}{\text{s}}^{-1}\right)}\\ \tilde{v}=\underset{\_}{1.01\times 10^{-1}{m}^{-1}}\end{array}$

Therefore, the value of the fundamental vibrational wavenumber is $\underset{\_}{1.01\times 10^{-1}{m}^{-1}}$.

The fundamental vibrational wavenumber in terms of force constant is expressed as shown below.

  $\begin{array}{c}\tilde{v}=\frac{1}{2\pi c}{\left(\frac{k_{\text{f}}}{u}\right)}^{1/2}\\ k_{\text{f}}={\left(2\pi c\tilde{v}\right)}^{2}u\end{array}$        (2)

Where, $k_{\text{f}}$ is the force constant and $u$ is the reduced mass.

The mass of one $\text{He}$ atom is denoted by $m_{\text{He}}$.  The value of $m_{\text{He}}$ is $4.0026m_{\text{u}}$ where $m_u$ is the atomic mass unit.

  $1m_u=1.66054\times {10}^{-27}\text{kg}$

Therefore, the mass of $\text{He}$ in $\text{kg}$ is given below.

  $\begin{array}{c}{m}_{\text{He}}=4.0026m_{\text{u}}\\ =4.0026\times 1.66054\times {10}^{-27}\text{kg}\\ =6.646\times {10}^{-27}\text{kg}\end{array}$

The reduced mass of the ${\text{He}}_{\text{2}}$ is calculated as shown below.

  $\begin{array}{c}\mu =\frac{m_{\text{He}}\times m_{\text{He}}}{m_{\text{He}}+m_{\text{He}}}\\ =\frac{6.646\times {10}^{-27}\text{kg}\times 6.646\times {10}^{-27}\text{kg}}{6.646\times {10}^{-27}\text{kg}+6.646\times {10}^{-27}\text{kg}}\\ =3.323\times {10}^{-27}\text{kg}\end{array}$

Substitute the value of the reduced mass and other corresponding variables in equation (2) as shown below.

  $\begin{array}{c}{k}_{\text{f}}={\left(2\pi c\tilde{v}\right)}^{2}u\\ ={\left(2\times 3.14\times 2.998\times {10}^8\text{m}{\text{s}}^{-1}\times 1.01\times {10}^{-1}{\text{m}}^{-1}\right)}^2\times 3.323\times {10}^{-27}\text{kg}\\ =\underset{\_}{1.20\times 10^{-10}kg{s}^{-1}}\end{array}$

Therefore, the value of the force constant is $\underset{\_}{1.20\times 10^{-10}kg{s}^{-1}}$.

The moment of inertia is calculated by the formula shown below.

  $I=u{R}_e{}^2$        (3)

Where

• $I$ is the moment of inertia.
• $u$ is the reduced mass.
• $R_e$ is the internuclear distance.

The value of $R_e$ is $297\text{pm}$ and the value of $u$ is $3.323\times {10}^{-27}\text{kg}$.  Substitute the values of corresponding variables in equation (3) to calculate the moment of inertia as shown below.

  $\begin{array}{c}I=u{R}_e{}^2\\ =\left(3.323\times {10}^{-27}\text{kg}\right)\times {\left(297\text{pm}\times \frac{{10}^{-12}\text{m}}{1\text{pm}}\right)}^2\\ =\left(3.323\times {10}^{-27}\text{kg}\right)\times {\left(2.97\times {10}^{-10}\text{m}\right)}^2\\ =\underset{\_}{2.93\times 10^{-46}kg{m}^2}\end{array}$

Therefore, the moment of inertia is $\underset{\_}{2.93\times 10^{-46}kg{m}^2}$.

The rotational constant is given by an expression shown below.

  $B=\frac{h}{8{\pi }^{2}Ic}$        (4)

Where,

• $I$ is the moment of inertia.
• $h$ is the planks constant.
• $\text{B}$ is the rotational constant.
• $c$ is the velocity of light.

Substitute the values of corresponding variables in equation (4) to calculate the moment of inertia as shown below.

$\begin{array}{c}B=\frac{h}{8{\pi }^{2}Ic}\\ =\frac{6.626\times {10}^{-34} {\text{m}}^2{\text{kg s}}^{-1}}{8\times {\left(3.14\right)}^2\times \left(2.93\times {10}^{-46}\text{kg}{\text{m}}^{\text{2}}\right)\times \left(2.998\times {10}^8{\text{ms}}^{\text{-1}}\right)}\\ =\underset{\_}{95.6m^{-1}}\end{array}$

Therefore, the value of the rotational constant is $\underset{\_}{95.6m^{-1}}$.

(b)

Interpretation Introduction

Interpretation:

The vibrational wavenumber and anharmonicity constant has to be calculated on the basis of Morse potential energy.

Concept introduction:

Rotational spectroscopy measures the electronic energy between the rotational states.  The rotational constant is inversely proportional to the moment of inertia.  The Morse potential energy function is shown below.

  $V=hc{\tilde{D}}_{\text{e}}{\left\{1-{\text{e}}^{-a\left(R-R_e\right)}\right\}}^2$

Answer to Problem 11C.7P

The value the vibrational wavenumber is $\underset{\_}{76.0m^{-1}}$.

The anharmonicity constant $\left(x_{\text{e}}\right)$ is $\underset{\_}{0.25}$.

Explanation of Solution

The Morse potential energy is given by the expression as shown below.

  $V=hc{\tilde{D}}_{\text{e}}{\left\{1-{\text{e}}^{-a\left(R-R_e\right)}\right\}}^2$

Where, $h$ is the Planck’s constant, $c$ is the velocity of light, $a$ is a constant, ${\tilde{D}}_{\text{e}}$ is the depth of the potential minimum and $R$ represents internuclear distance.

The vibrational wavenumber $\left(\tilde{v}\right)$ is given by the expression as shown below.

  $\tilde{v}={\tilde{D}}_{\text{e}}$

The given value for $hc{\tilde{D}}_{\text{e}}$ is shown below.

  $hc{\tilde{D}}_{\text{e}}=1.51\times {10}^{-23}\text{J}$        (1)

Substitute the values of corresponding variables in equation (1) to calculate the fundamental vibrational wavenumber as shown below.

  $\begin{array}{c}hc{\tilde{D}}_{\text{e}}=1.51\times {10}^{-23}\text{J}\\ {\tilde{D}}_{\text{e}}=\frac{1.51\times {10}^{-23}\text{J}}{\left(6.626\times {10}^{-34}\text{J}\text{s}\right)\times \left(2.998\times {10}^8\text{m}{\text{s}}^{-1}\right)}\\ \tilde{v}=\underset{\_}{76.0m^{-1}}\end{array}$

Therefore, the value the vibrational wavenumber is $\underset{\_}{76.0m^{-1}}$.

The anharmonicity constant $\left(x_{\text{e}}\right)$ is given by the expression as shown below.

  $x_{\text{e}}=\frac{\tilde{v}}{4{\tilde{D}}_{\text{e}}}$        (2)

Substitute the values of corresponding variables in equation (2) to calculate the anharmonicity constant $\left(x_{\text{e}}\right)$ as shown below.

  $\begin{array}{c}{x}_{\text{e}}=\frac{\tilde{v}}{4{\tilde{D}}_{\text{e}}}\\ =\frac{\tilde{v}}{4\tilde{v}}\\ =\frac{1}{4}\\ =\underset{\_}{0.25}\end{array}$

Therefore, the anharmonicity constant $\left(x_{\text{e}}\right)$ is $\underset{\_}{0.25}$.