Chapter 8, Problem 8A.1ST

(a)

Interpretation Introduction

Interpretation:

The ionization energy of the lower state of the deuterium atom has to be calculated.

Concept introduction:

Rydberg equation is used to represent the wavenumber or wavelength of the lines present in the atomic spectrum of an element.  The Rydberg equation for a deuterium atom is given by the expression as shown below.

    $\frac{1}{\lambda }=Z^2{R}_{\text{D}}\left(\frac{1}{n_{\text{1}}^2}-\frac{1}{n_{\text{2}}^2}\right)$

Answer to Problem 8A.1ST

The ionization energy of the lower state of the deuterium atom is $\underset{\_}{328.109kJmo{l}^{-1}}$.

Explanation of Solution

The given data is shown below.

$n=3\to n=2$	$15238{\text{ cm}}^{-1}$
$n=4\to n=2$	$20571{\text{ cm}}^{-1}$
$n=5\to n=2$	$23039{\text{ cm}}^{-1}$
$n=6\to n=2$	$24380{\text{ cm}}^{-1}$

The Rydberg equation for a deuterium atom is given by the expression as shown below.

    $\overline{v}=Z^2{R}_{\text{D}}\left(\frac{1}{n_{\text{1}}^2}-\frac{1}{n_{\text{2}}^2}\right)$        (1)

Where,

• $R_{\text{D}}$ is Rydberg constant for deuterium.
• $n_{\text{1}}$ is the lower energy level.
• $n_{\text{2}}$ is the higher energy level.
• $\overline{v}$ is the wavenumber of the photon released.

The atomic number of deuterium is $1$.

Rearrange the equation (1) for the value of $R_{\text{D}}$.

  $R_{\text{D}}=\frac{\overline{v}}{Z^2}{\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)}^{-1}$        (2)

Substitute the value of an atomic number of a deuterium atom, $\overline{v}=15238{\text{ cm}}^{-1}$, $n_{\text{1}}=2$, $Z=1$ and $n_{\text{1}}=3$ in the equation (2).

  $\begin{array}{c}{R}_{\text{D}}=\frac{15238{\text{ cm}}^{-1}}{{\left(1\right)}^2}{\left(\frac{1}{{\left(2\right)}^2}-\frac{1}{{\left(3\right)}^2}\right)}^{-1}\\ =\left(15238{\text{ cm}}^{-1}\right){\left(0.13889\right)}^{-1}{\text{ cm}}^{-1}\\ =109712.7223{\text{ cm}}^{-1}\end{array}$

Therefore, the value of $R_{\text{D}}$ for deuterium atom is $109712.7223{\text{ cm}}^{-1}$.

The lower energy state is $n=2$.

The ionization energy per mole of the deuterium atom is given by the formula as shown below.

  $I=hc{Z}^2{R}_{\text{D}}{N}_{\text{A}}\left(\frac{1}{n^2}\right)$        (3)

Where,

• $h$ is Plank’s constant with a value $6.62608\times {10}^{-34}\text{Js}$.
• $c$ is the speed of light with value $2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}$.
• $N_{\text{A}}$ is the Avogadro’s number $\left(6.022\times {10}^{23}{\text{atoms mol}}^{-1}\right)$.

Substitute the values of $Z$, $h$, $c$, $R_{{}_{\text{D}}}$, and $n=2$ in the equation (3).

  $\begin{array}{c}I=\left(\begin{array}{l}{\left(1\right)}^2\left(6.62608\times {10}^{-34}\text{J s}\right)\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\\ \left(109712.7223{\text{ cm}}^{-1}\right)\left(6.022\times {10}^{23}{\text{mol}}^{-1}\right)\left(\frac{1}{{\left(2\right)}^2}\right)\end{array}\right)\\ =\left(328108.9577{\text{ J mol}}^{-1}\right)\left(\frac{1\text{ kJ}}{{10}^3\text{ J}}\right)\\ =\underset{\_}{328.109kJmo{l}^{-1}}\end{array}$

Therefore, the ionization energy of the lower state of the deuterium atom is $\underset{\_}{328.109kJmo{l}^{-1}}$.

(b)

Interpretation Introduction

Interpretation:

The ionization energy of the ground state of the deuterium atom has to be calculated.

Concept introduction:

The energy required to remove an electron from an ion or isolated atom is known as its ionization energy.  The ionization energy of metals is generally lower than the ionization energy of nonmetals.  The ionization energy per mole of the deuterium atom is given by the formula as shown below.

  $I=hc{Z}^2{R}_{\text{D}}{N}_{\text{A}}\left(\frac{1}{n^2}\right)$

Answer to Problem 8A.1ST

The ionization energy of the ground state of the deuterium atom is $\underset{\_}{1312.4kJmo{l}^{-1}}$.

Explanation of Solution

The atomic number of deuterium is $1$.

The value of $R_{\text{D}}$ for deuterium atom is $109712.7223{\text{ cm}}^{-1}$.

The ground energy level is $n=1$.

The ionization energy per mole of the deuterium atom is given by the formula as shown below.

  $I=hc{Z}^2{R}_{\text{D}}{N}_{\text{A}}\left(\frac{1}{n^2}\right)$        (3)

Where,

• $h$ is Plank’s constant with a value $6.62608\times {10}^{-34}\text{Js}$.
• $c$ is the speed of light with value $2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}$.
• $N_{\text{A}}$ is the Avogadro’s number $\left(6.022\times {10}^{23}{\text{atoms mol}}^{-1}\right)$.

Substitute the values of $Z$, $h$, $c$, $R_{{}_{\text{D}}}$, and $n=2$ in the equation (3).

  $\begin{array}{c}I=\left(\begin{array}{l}{\left(1\right)}^2\left(6.62608\times {10}^{-34}\text{J s}\right)\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\\ \left(109712.7223{\text{ cm}}^{-1}\right)\left(6.022\times {10}^{23}{\text{mol}}^{-1}\right)\left(\frac{1}{{\left(1\right)}^2}\right)\end{array}\right)\\ =\left(\text{1312435}{\text{.831 J mol}}^{-1}\right)\left(\frac{1\text{ kJ}}{{10}^3\text{ J}}\right)\\ =\underset{\_}{1312.4kJmo{l}^{-1}}\end{array}$

Therefore, the ionization energy of the ground state of the deuterium atom is $\underset{\_}{1312.4kJmo{l}^{-1}}$.

(c)

Interpretation Introduction

Interpretation:

The mass of deuteron has to be stated.

Concept introduction:

An atom is made up of three subatomic particles-neutrons, protons, and electrons. Neutron and protons are present in the nucleus of the atom, whereas electrons are revolving outside the nucleus in an atom.  The ground state energy of atom corresponds to the energy level for which the value of the principal quantum number is $1$.

Answer to Problem 8A.1ST

The mass of deuteron is $\underset{\_}{3.7037\times 10^{-27}kg}$.

Explanation of Solution

The value of $R_{\text{D}}$ for deuterium atom is $109712.7223{\text{ cm}}^{-1}$.

The value of Rydberg constant for an atom is given by the expression as shown below.

  ${\tilde{R}}_{\text{N}}=\frac{\mu }{m_{\text{e}}}\times {\tilde{R}}_{\infty }$

Where,

• ${\tilde{R}}_{\infty }$ is the Rydberg constant for an electron with a value of $109737{\text{cm}}^{-1}$.
• $m_{\text{e}}$ is the mass of an electron with a value of $9.10938\times {10}^{-31}\text{kg}$.
• $\mu$ is the reduced mass of the atom.

Substitute the value of ${\tilde{R}}_{\infty }$, Rydberg constant for deuterium atom and $m_{\text{e}}$ in the above equation.

    $\begin{array}{c}109712.7223{\text{ cm}}^{-1}=\frac{\mu }{9.10938\times {10}^{-31}\text{kg}}\left(109737{\text{cm}}^{-1}\right)\\ 109712.7223{\text{ cm}}^{-1}=1.2047\times {10}^{35}\text{kg}{\text{cm}}^{-1}\mu \end{array}$

Rearrange the above equation for the value of $\mu$.

    $\begin{array}{c}\mu =\frac{109712.7223{\text{ cm}}^{-1}}{1.2047\times {10}^{35}\text{kg}{\text{cm}}^{-1}}\\ =9.1071\times {10}^{-31}\text{kg}\end{array}$

The reduced mass of a deuterium atom is given by the expression as shown below.

  $\frac{1}{\mu }=\frac{1}{m_{\text{e}}}+\frac{1}{m_{\text{D}}}$

Where,

• $m_{\text{e}}$ is the mass of an electron with a value of $9.10938\times {10}^{-31}\text{kg}$.
• $m_{\text{D}}$ is the mass of the deuterium nucleus (deuteron).

Substitute the values of $m_{\text{e}}$ and $\mu$ in the above equation.

  $\begin{array}{c}\frac{1}{9.1071\times {10}^{-31}\text{kg}}=\frac{1}{9.10938\times {10}^{-31}\text{kg}}+\frac{1}{m_{\text{D}}}\\ 1.09804\times {10}^{30}{\text{kg}}^{-1}=1.09777\times {10}^{30}{\text{kg}}^{-1}+\frac{1}{m_{\text{D}}}\end{array}$

Rearrange the above equation for the value of $m_{\text{D}}$.

    $\begin{array}{c}{m}_{\text{D}}=\frac{1}{1.09804\times {10}^{30}-1.09777\times {10}^{30}}\text{kg}\\ =\underset{\_}{3.7037\times 10^{-27}kg}\end{array}$

Therefore, the mass of deuteron is $\underset{\_}{3.7037\times 10^{-27}kg}$.