Chapter 7, Problem 7D.3BE

(i)

Interpretation Introduction

Interpretation:

The energy separation in joules, kilojoules per mole, electronvolts and reciprocal centimetres between the levels $n=3$ and $n=2$ of an electron in a box of length $1.50\text{nm}$ 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.3BE

The energy separation in joules, kilojoules per mole, electronvolts and reciprocal centimetres between the levels  $n=3$ and $n=2$ of an electron in a box of length $1.50\text{nm}$ is $\underset{\_}{1.33\times 10^{-19}J}$, $\underset{\_}{800.926kJmo{l}^{-1}}$, $\underset{\_}{0.8246eV}$ and $\underset{\_}{6.695\times 10^{3}c{m}^{-1}}$, respectively.

Explanation of Solution

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

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

Where,

• $n\text{and}n+1$ are the quantum numbers of the energy levels.
• $h$ is the Planck’s constant.
• $m$ is the mass of the electron $\left(9.109\times {10}^{-31}\text{kg}\right)$.
• $L$ is the length of the box $\left(1.50\text{nm}\right)$ .

The value of $n$ is $2$.

The value of $n+1$ is $3$.

The value of $h$ is $6.626\times {10}^{-34}\text{kg}{\text{m}}^2{\text{s}}^{-1}$.

The conversion of $\text{nm}$ to $\text{m}$ is done as,

  $\text{1}\text{nm}={10}^{-9}\text{m}$

The conversion of $1.50\text{nm}$ to $\text{m}$ is done as,

  $1.50\text{nm}=1.50\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}$

The value of $L$ is $1.5\times {10}^{-9}\text{m}$

Substitute the values of $n$, $n+1$, $h$ and $L$ in equation (1).

    $\begin{array}{c}{E}_2-E_1=\left(2\times 2+1\right)\frac{{\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^2{\text{s}}^{-1}\right)}^2}{8\times 9.109\times {10}^{-31}\text{kg}{\left(1.5\times {10}^{-9}\text{m}\right)}^2}\\ =\frac{2.195\times {10}^{-66}}{1.64\times {10}^{-47}}\\ =1.33\times {10}^{-19}\text{kg}{\text{m}}^2{\text{s}}^{-1}\end{array}$

The unit conversion of $\text{kg}{\text{m}}^2{\text{s}}^{-1}$ into $\text{J}$ is,

    $\text{1}\text{kg}{\text{m}}^2{\text{s}}^{-1}=1\text{J}$

The conversion of $1.33\times {10}^{-19}\text{kg}{\text{m}}^2{\text{s}}^{-1}$ into $\text{J}$ is,

    $\begin{array}{c}1.33\times {10}^{-19}\text{kg}{\text{m}}^2{\text{s}}^{-1}=1.33\times {10}^{-19}\text{kg}{\text{m}}^2{\text{s}}^{-1}\times \frac{1\text{J}}{\text{1}\text{kg}{\text{m}}^2{\text{s}}^{-1}}\\ =\underset{\_}{1.33\times 10^{-19}J}\end{array}$

Hence, the energy separation in joules between the levels $n=2$ and $n=3$ of an electron in a box of length $1.50\text{nm}$ is $\underset{\_}{1.33\times 10^{-19}J}$.

The conversion of $\text{J}$ into $\text{kJ}$ is,

    $1\text{kJ}={10}^3\text{J}$

The value of Avogadro number is $6.022\times {10}^{23}{\text{mol}}^{-1}$

The conversion of $1.33\times {10}^{-19}\text{J}$ into $\text{kJ}{\text{mol}}^{-1}$ is,

    $\begin{array}{c}1.33\times {10}^{-19}\text{J}=1.33\times {10}^{-19}\text{J}\times \frac{{10}^{-3}\text{kJ}}{1\text{J}}\times 6.022\times {10}^{23}{\text{mol}}^{-1}\\ =\underset{\_}{800.926kJmo{l}^{-1}}\end{array}$

Hence, the energy separation in kilojoules per mole between the levels $n=2$ and $n=3$ of an electron in a box of length $1.50\text{nm}$ is $\underset{\_}{800.926kJmo{l}^{-1}}$.

The conversion of $\text{J}$ into $\text{eV}$ is,

    $1\text{J}=6.2\times {10}^{18}\text{eV}$

The conversion of $1.33\times {10}^{-19}\text{J}$ into $\text{eV}$ is,

    $\begin{array}{c}1.33\times {10}^{-19}\text{J}=1.33\times {10}^{-19}\text{J}\times \frac{6.2\times {10}^{18}\text{eV}}{1\text{J}}\\ =\underset{\_}{0.8246eV}\end{array}$

Hence, the energy separation in electronvolts between the levels $n=2$ and $n=3$ of an electron in a box of length $1.50\text{nm}$ is $\underset{\_}{0.8246eV}$.

The conversion of $\text{J}$ to ${\text{cm}}^{-1}$ is,

    $1\text{J}=5.034\times {10}^{22}{\text{cm}}^{-1}$

The conversion of $1.33\times {10}^{-19}\text{J}$ into ${\text{cm}}^{-1}$ is,

    $\begin{array}{c}1.33\times {10}^{-19}\text{J}=1.33\times {10}^{-19}\text{J}\times \frac{5.034\times {10}^{22}{\text{cm}}^{-1}}{1\text{J}}\\ =\underset{\_}{6.695\times 10^{3}c{m}^{-1}}\end{array}$

Hence, the energy separation in reciprocal centimetres between the levels $n=2$ and $n=3$ of an electron in a box of length $1.50\text{nm}$ is $\underset{\_}{6.695\times 10^{3}c{m}^{-1}}$.

(ii)

Interpretation Introduction

Interpretation:

The energy separation in joules, kilojoules per mole, electronvolts and reciprocal centimetres between the levels $n=7$ and $n=6$ of an electron in a box of length $1.50\text{nm}$ 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.3BE

The energy separation in joules, kilojoules per mole, electronvolts and reciprocal centimetres between the levels $n=7$ and $n=6$ of an electron in a box of length $1.50\text{nm}$ is $\underset{\_}{3.49\times 10^{-19}J}$, $\underset{\_}{21.01\times 10^{2}kJmo{l}^{-1}}$, $\underset{\_}{2.16eV}$ and $\underset{\_}{1.756\times 10^{4}c{m}^{-1}}$, respectively.

Explanation of Solution

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

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

Where,

• $n\text{and}n+1$ are the quantum numbers of the energy levels.
• $h$ is the Planck’s constant.
• $m$ is the mass of the electron $\left(9.109\times {10}^{-31}\text{kg}\right)$.
• $L$ is the length of the box $\left(1.50\text{nm}\right)$.

The value of $n$ is $6$.

The value of $n+1$ is $7$.

The value of $h$ is $6.626\times {10}^{-34}\text{kg}{\text{m}}^2{\text{s}}^{-1}$.

The conversion of $\text{nm}$ to $\text{m}$ is done as,

  $\text{1}\text{nm}={10}^{-9}\text{m}$

The value of $L$ is ${10}^{-9}\text{m}$

Substitute the values of $n$, $n+1$, $h$ and $L$ in equation (1).

    $\begin{array}{c}{E}_6-E_5=\left(2\times 6+1\right)\frac{{\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^2{\text{s}}^{-1}\right)}^2}{8\times 9.109\times {10}^{-31}\text{kg}{\left(1.5\times {10}^{-9}\text{m}\right)}^2}\\ =\frac{5.70\times {10}^{-66}}{1.63\times {10}^{-47}}\\ =3.49\times {10}^{-19}\text{kg}{\text{m}}^2{\text{s}}^{-1}\end{array}$

The unit conversion of $\text{kg}{\text{m}}^2{\text{s}}^{-1}$ into $\text{J}$ is,

    $\text{1}\text{kg}{\text{m}}^2{\text{s}}^{-1}=1\text{J}$

The conversion of $3.49\times {10}^{-19}\text{kg}{\text{m}}^2{\text{s}}^{-1}$ into $\text{J}$ is,

    $\begin{array}{c}3.49\times {10}^{-19}\text{kg}{\text{m}}^2{\text{s}}^{-1}=3.49\times {10}^{-19}\text{kg}{\text{m}}^2{\text{s}}^{-1}\times \frac{1\text{J}}{\text{1}\text{kg}{\text{m}}^2{\text{s}}^{-1}}\\ =\underset{\_}{3.49\times 10^{-19}J}\end{array}$

Hence, the energy separation in joules between the levels $n=7$ and $n=6$ of an electron in a box of length $1.50\text{nm}$ is $\underset{\_}{3.49\times 10^{-19}J}$.

The conversion of $\text{J}$ into $\text{kJ}$ is,

    $1\text{kJ}={10}^3\text{J}$

The value of Avogadro number is $6.022\times {10}^{23}{\text{mol}}^{-1}$

The conversion of $3.49\times {10}^{-19}\text{J}$ into $\text{kJ}{\text{mol}}^{-1}$ is,

    $\begin{array}{c}3.49\times {10}^{-19}\text{J}=3.49\times {10}^{-19}\text{J}\times \frac{{10}^{-3}\text{kJ}}{1\text{J}}\times 6.022\times {10}^{23}{\text{mol}}^{-1}\\ =\underset{\_}{21.01\times 10^{2}kJmo{l}^{-1}}\end{array}$

Hence, the energy separation in kilojoules per mole between the levels $n=7$ and $n=6$ of an electron in a box of length $1.50\text{nm}$ is $\underset{\_}{21.01\times 10^{2}kJmo{l}^{-1}}$.

The conversion of $\text{J}$ into $\text{eV}$ is,

    $1\text{J}=6.2\times {10}^{18}\text{eV}$

The conversion of $3.49\times {10}^{-19}\text{J}$ into $\text{eV}$ is,

    $\begin{array}{c}3.49\times {10}^{-19}\text{J}=3.49\times {10}^{-19}\text{J}\times \frac{6.2\times {10}^{18}\text{eV}}{1\text{J}}\\ =\underset{\_}{2.16eV}\end{array}$

Hence, the energy separation in electronvolts between the levels $n=7$ and $n=6$ of an electron in a box of length $1.50\text{nm}$ is $\underset{\_}{2.16eV}$.

The conversion of $\text{J}$ to ${\text{cm}}^{-1}$ is,

    $1\text{J}=5.034\times {10}^{22}{\text{cm}}^{-1}$

The conversion of $3.49\times {10}^{-19}\text{J}$ into ${\text{cm}}^{-1}$ is,

    $\begin{array}{c}3.49\times {10}^{-19}\text{J}=3.49\times {10}^{-19}\text{J}\times \frac{5.034\times {10}^{22}{\text{cm}}^{-1}}{\text{1}\text{J}}\\ =\underset{\_}{1.756\times 10^{4}c{m}^{-1}}\end{array}$

Hence, the energy separation in reciprocal centimetres between the levels $n=7$ and $n=6$ of an electron in a box of length $1.50\text{nm}$ is $\underset{\_}{1.756\times 10^{4}c{m}^{-1}}$