Chapter 7, Problem 7D.2P

(i)

Interpretation Introduction

Interpretation:

The value of $n$ for the nitrogen molecule that has energy of $\frac{3}{2}kT$ at $300\text{K}$ has to be calculated.

Concept introduction:

The model in which the particle is free to move in a small space which is enclosed by the impenetrable walls is called the particle in a box model.  The particle in a box is significant application of the Schrödinger equation and it provides many basic concepts of quantum mechanics.

Answer to Problem 7D.2P

The value of $n$ for the molecule that has energy of $\frac{3}{2}kT$ at $300\text{K}$ is $\underset{\_}{7.2\times 10^{10}}$.

Explanation of Solution

The expression for translational kinetic energy is given below.

    $E_{\text{k}}=\frac{3kT}{2}$                                                                                                      (1)

Where,

• $k$ is the Boltzmann constant.
• $T$ is the temperature.

The expression for the energy for the cubic box is calculated as shown below.

    $E_n=\frac{\left(n_x^2+n_y^2+n_z^2\right)h^2}{8m{L}^2}$                                                                                   (2)

Where,

• $n_x^2$, $n_y^2$ and $n_z^2$ are the quantum numbers of energy levels.
• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-1}\right)$.
• $m$ is the mass of the nitrogen molecule $\left(0.028\text{kg}{\text{mol}}^{-1}\right)$.
• $L$ is the length of the box.

Equate equation (1) and equation (2).

    $\frac{3kT}{2}=\frac{\left(n_x^2+n_y^2+n_z^2\right)h^2}{8m{L}^2}$                                                                               (3)

The value of $L^3$ is $\text{1}{\text{m}}^3$, therefore, the value of $L^2$ is $\text{1}{\text{m}}^2$.

The value of $k$ is $1.38\times {10}^{-23}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-2}{\text{K}}^{-1}$.

The value of $T$ is $300\text{K}$.

Substitute the values of $L^2$, $k$, $T$, $h$ and $m$ in equation (3).

    $\begin{array}{c}\frac{3\times 1.38\times {10}^{-23}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-2}{\text{K}}^{-1}\times 300\text{K}}{2}=\frac{\left(n_x^2+n_y^2+n_z^2\right)\times {\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-1}\right)}^2}{8\times \left(\frac{0.028\text{kg}{\text{mol}}^{-1}}{6.022\times {10}^{23}{\text{mol}}^{-1}}\right)\times \text{1}{\text{m}}^2}\\ \left(n_x^2+n_y^2+n_z^2\right)=\frac{6.21\times {10}^{-21}\times 3.72\times {10}^{-25}}{4.39\times {10}^{-67}}\\ =0.52\times {10}^{22}\end{array}$

It is given that,

    $n={\left(n_x^2+n_y^2+n_z^2\right)}^{1/2}$

Substitute the value of $n_x^2+n_y^2+n_z^2$ in the above expression.

    $\begin{array}{c}n={\left(0.52\times {10}^{20}\right)}^{1/2}\\ =\underset{\_}{7.2\times 10^{10}}\end{array}$

Hence, the value of $n$ for the molecule that has energy of $\frac{3}{2}kT$ at $300\text{K}$ is $\underset{\_}{7.2\times 10^{10}}$.

(ii)

Interpretation Introduction

Interpretation:

The energy separation between the energy levels $n$ and $n+1$ has to be calculated.

Concept introduction:

The separation between the adjacent energy levels with quantum numbers $n$ and $n+1$ is given by the expression,

    $E_{n+1}-E_n=\left(2n+1\right)\frac{h^2}{8m{L}^2}$

Answer to Problem 7D.2P

The energy separation between the energy levels $n$ and $n+1$ is $\underset{\_}{1.7\times 10^{-31}kg{m}^2{s}^{-1}}$.

Explanation of Solution

The expression for the separation between the energy levels is given below.

    $E_{n+1}-E_n=\left(2n+1\right)\frac{h^2}{8m{L}^2}$                                                                              (1)

Where,

• $n\text{and}n+1$ are the quantum numbers of the energy levels.
• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-1}\right)$.
• $m$ is the mass of the molecule $\left(0.028\text{kg}{\text{mol}}^{-1}\right)$.
• $L$ is the length of the box $\left(1\text{m}\right)$.

The value of $n$ is $7.2\times {10}^{10}$.

Substitute the value of $h$, $m$, $L$, $n$ in equation (1).

    $\begin{array}{c}{E}_{n+1}-E_n=\left(2\times 7.2\times {10}^{10}+1\right)\frac{{\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-1}\right)}^2}{8\times \left(\frac{0.028\text{kg}{\text{mol}}^{-1}}{6.022\times {10}^{23}{\text{mol}}^{-1}}\right)\times {\left(1\text{m}\right)}^2}\\ =\left(14.4\times {10}^{10}\right)\left(\frac{4.39\times {10}^{-67}}{3.719\times {10}^{-25}}\right)\\ =\underset{\_}{1.7\times 10^{-31}kg{m}^2{s}^{-1}}\end{array}$

Hence, the energy separation between the energy levels $n$ and $n+1$ is $\underset{\_}{1.7\times 10^{-31}kg{m}^2{s}^{-1}}$.

(iii)

Interpretation Introduction

Interpretation:

The de Broglie wavelength for the molecule has to be calculated.

Concept introduction:

The small packet of energy is known as the quanta.  Light is emitted in the form of quanta or photons.  The de Broglie wavelength is given by the expression.

    $\lambda =\frac{h}{mv}$

Answer to Problem 7D.2P

The de Broglie wavelength for the molecule is $\underset{\_}{2.75\times 10^{-11}m}$.

Explanation of Solution

The expression for the kinetic energy is given below.

    $E_{\text{k}}=\frac{1}{2}m{v}^2$                                                                                                    (1)

Where,

• $E_{\text{k}}$ is the kinetic energy.
• $m$ is the mass of the nitrogen molecule $\left(0.028\text{kg}{\text{mol}}^{-1}\right)$.
• $v$ is the velocity of the particle.

The expression for the translational kinetic energy is shown below.

        $E_{\text{k}}=\frac{3kT}{2}$                                                                                          (2)

Where,

• $k$ is the Boltzmann constant.
• $T$ is the temperature.

Equate equation (1) and (2).

    $\begin{array}{c}\frac{1}{2}m{v}^2=\frac{3kT}{2}\\ v^2=\frac{3kT}{m}\end{array}$                                                                                                 (3

The value of $m$ is $\left(0.028\text{kg}{\text{mol}}^{-1}\right)$.

The value of $T$ is $300\text{K}$.

The value of $k$ is $1.38\times {10}^{-23}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-2}{\text{K}}^{-1}$.

Substitute the values of $m$, $T$ and $k$ in equation (3).

    $\begin{array}{c}{v}^2=\frac{3\times 1.38\times {10}^{-23}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-2}{\text{K}}^{-1}\times 300\text{K}}{\frac{0.028\text{kg}{\text{mol}}^{-1}}{6.022\times {10}^{23}{\text{mol}}^{-1}}}\\ =\frac{1.242\times {10}^{-20}}{4.65\times {10}^{-26}}\\ v=\sqrt{0.267\times {10}^6}\\ v=516.7\text{m}{\text{s}}^{-1}\end{array}$

The expression for de Broglie wavelength is given below.

    $\lambda =\frac{h}{mv}$                                                                                                         (4)

Where,

• $\lambda$ is the de Broglie wavelength.
• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-1}\right)$.

Substitute the value of $h$ and $v$ in equation (4).

    $\begin{array}{c}\lambda =\frac{\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-1}\right)}{\left(\frac{0.028\text{kg}{\text{mol}}^{-1}}{6.022\times {10}^{23}{\text{mol}}^{-1}}\right)\times 516.7\text{m}{\text{s}}^{-1}}\\ =\frac{6.626\times {10}^{-34}}{4.65\times {10}^{-26}\times 516.7}\\ =\underset{\_}{2.75\times 10^{-11}m}\end{array}$

Hence, the de Broglie wavelength for the molecule is $\underset{\_}{2.75\times 10^{-11}m}$.