Chapter 23, Problem 23.1P

Find to three significant digits the charge and the mass of the following particles. Suggestion: Begin by looking up the mass of a neutral atom on the periodic table of the elements in Appendix C. (a) an ionized hydrogen atom, represented as H⁺ (b) a singly ionized sodium atom, Na⁺ (c) a chloride ion Cl⁻ (d) a doubly ionized calcium atom, Ca⁺⁺ = Ca²⁺ (e) the center of an ammonia molecule, modeled as an N³⁻ ion (f) quadruply ionized nitrogen atoms, N⁴⁺, found in plasma in a hot star (g) the nucleus of a nitrogen atom (h) the molecular ion H₂O⁻

To determine

The charge and mass of ionized hydrogen atom up to three significant digits.

Answer to Problem 23.1P

The charge and mass of ionized hydrogen atom is $+1.60\times {10}^{-19}\text{C}$ and $1.67\times {10}^{-27}\text{kg}$.

Explanation of Solution

The ionized hydrogen atom is ${\text{H}}^{\text{+}}$. Hydrogen atom has the atomic number one that is equal to the number of electrons and protons that means hydrogen atom consists of one electron and one proton.

Hydrogen atom is ionized by emission of electron or absorption of electron. In emission process the electron is removed from the atom and in absorption process electron is gained. Thus, ${\text{H}}^{\text{+}}$ an ion only has one proton.

The charge is calculated as,

  $\begin{array}{c}\text{Charge of}{\text{H}}^{\text{+}}=+e\\ \approx 1.60\times {10}^{-19}\text{C}\end{array}$

$e$ is the fundamental unit of charge.

The mass is calculated as,

  $M_{{\text{H}}^{\text{+}}}=M_{\text{H}}-m_{\text{e}}$

Here,

$M_{\text{H}}$ is the mass of the neutral hydrogen atom.

$M_{{\text{H}}^{\text{+}}}$ is the mass of the ionized hydrogen atom.

$m_{\text{e}}$ is the mass of the electron.

Substitute $1.0079\text{u}$ for $M_{\text{H}}$ and $9.11\times {10}^{-31}\text{}\text{kg}$ for $m_{\text{e}}$ in the above equation

  $\begin{array}{c}{M}_{{\text{H}}^{\text{+}}}=1.0079\text{u}-9.11\times {10}^{-31}\text{}\text{kg}\\ \text{=}1.0079\left(1.660\times {10}^{-27}\text{Kg}\right)-9.11\times {10}^{-31}\text{}\text{kg}\\ =1.67\times {10}^{-27}\text{kg}\end{array}$

Conclusion:

Therefore, the charge of ionized hydrogen atom is $+1.60\times {10}^{-19}\text{C}$ and mass $1.67\times {10}^{-27}\text{Kg}$

To determine

The charge and mass of singly ionized sodium atom $N{a}^{+}$ upto three significant digits.

Answer to Problem 23.1P

The charge and mass of singly ionized sodium atom ${\text{Na}}^{\text{+}}$ is $+1.6\times {10}^{-19}\text{C}$ and $3.82\times {10}^{-26}\text{kg}$.

Explanation of Solution

A singly ionized sodium atom is ${\text{Na}}^{+}$. As the electronic configuration of sodium atom is $2,8,1$ , there is one electron in the outermost shell so to become stable or achieve noble gas electronic configuration it has to lose one electron or gain seven electrons. Seven electrons cannot be gained due to energy considerations. So it will lose one electron by losing the electron the number of electrons decreases compare to the number of protons. Sodium atom always forms the positive ion.

The charge of an ionized atom is given by the number of electron is gained or lost by the neutral atom as the atom forms an ion by losing a electron so its charge is $+e$.

The charge of sodium ion

    ${\text{Na}}^{\text{+}}=+e=+1.6\times {10}^{-19}\text{C}$

The mass is calculated as,

  $M_{{\text{Na}}^{\text{+}}}=M_{\text{Na}}-m_{\text{e}}$

Here,

$M_{\text{Na}}$ is the mass of the neutral sodium atom.

$M_{{\text{Na}}^{\text{+}}}$ is the mass of the ionized sodium atom.

Substitute $22.99\text{u}$ for $M_{\text{Na}}$ and $9.11\times {10}^{-31}\text{}\text{kg}$ for $m_{\text{e}}$ in the above equation

  $\begin{array}{c}{M}_{{\text{Na}}^{\text{+}}}=22.99\text{u}-9.11\times {10}^{-31}\text{}\text{kg}\\ =22.99\left(1.660\times {10}^{-27}\text{Kg}\right)-9.11\times {10}^{-31}\text{}\text{kg}\\ =3.82\times {10}^{-26}\text{kg}\end{array}$

Conclusion:

Therefore, the charge of ionized sodium atom is $+1.6\times {10}^{-19}\text{C}$ and mass is $3.82\times {10}^{-26}\text{Kg}$

To determine

The charge and mass of ionized chlorine atom ${\text{Cl}}^{-}$ up to three significant digits.

Answer to Problem 23.1P

The charge and mass of ionized chlorine atom ${\text{Cl}}^{-}$ is $-1.6\times {10}^{-19}\text{C}$ and $5.89\times {10}^{-26}\text{kg}$.

Explanation of Solution

A ionized Chlorine atom that is represented is ${\text{Cl}}^{\text{-}}$. Chlorine atom has the atomic number $17$ which is equal to the number of electrons and protons that means Chlorine atom consists of $17$ electrons and $17$ protons. Electrons are revolving in the orbit and proton is residing inside the nucleus

When the Chlorine atom is ionized by emission of electron as the electronic configuration of Chlorine atom is $2,8,7$, there are seven electrons in the outermost shell so to become stable or achieve noble gas electronic configuration it has to gain one electron or lose seven electrons. Seven electrons cannot lose due to energy considerations. So it will gain one electron by gaining the electron the number of electrons increases compare to the number of protons. Chlorine atom always forms the negative ion.

The charge of an ionized atom is given by the number of electron is gained or lost by the neutral atom as the atom forms an ion by losing a electron so its charge is $-e$.

The charge of Chlorine ion

    $\begin{array}{c}{\text{Cl}}^{-}=-e\\ \approx -1.60\times {10}^{-19}\text{C}\end{array}$

The mass is calculated as,

  $M_{{\text{Cl}}^{\text{-}}}=M_{\text{Cl}}+m_{\text{e}}$

Here,

$M_{\text{Cl}}$ is the mass of the neutral Chlorine atom.

$M_{{\text{Cl}}^{\text{-}}}$ is the mass of the ionized Chlorine atom.

Substitute $35.543\text{u}$ for $M_{\text{H}}$ and $9.11\times {10}^{-31}\text{}\text{kg}$ for $m_{\text{e}}$ in the above equation

  $\begin{array}{c}{M}_{{\text{Cl}}^{\text{-}}}=35.453\text{u}+9.11\times {10}^{-31}\text{}\text{kg}\\ =35.453\left(1.660\times {10}^{-27}\text{Kg}\right)+9.11\times {10}^{-31}\text{}\text{kg}\\ =5.89\times {10}^{-26}\text{kg}\end{array}$

Conclusion:

The charge of ionized Chlorine atom is $-1.6\times {10}^{-19}\text{C}$ and mass $5.89\times {10}^{-26}\text{Kg}$.

To determine

The charge and mass of doubly ionized Calcium atom ${\text{Ca}}^{++}={\text{Ca}}^{2+}$ up to three significant digits.

Answer to Problem 23.1P

The charge and mass of doubly ionized Calcium atom ${\text{Ca}}^{++}={\text{Ca}}^{2+}$ is $+3.20\times {10}^{-19}\text{C}$ and $6.65\times {10}^{-26}\text{kg}$

Explanation of Solution

A doubly ionized Calcium atom is ${\text{Ca}}^{++}={\text{Ca}}^{2+}$. When the Calcium atom is ionized by emission of electron as the electronic configuration of sodium atom is $2,8,8,2$, there is two electron in the outermost shell so to become stable or achieve noble gas electronic configuration it has to lose two electron or gain six electrons. Six electrons cannot be gain due to energy considerations so it will lose two electron by losing the electron the number of electrons decreases compare to the number of protons. Calcium atom always forms the positive ion.

The charge of an ionized atom is given by the number of electron is gained or lost by the neutral atom as the atom forms an ion by losing two electron so its charge is $+2e$.

The charge of Calcium ion

  $\begin{array}{c}{\text{Ca}}^{2+}=+2e\\ =2\left(1.6\times {10}^{-19}\text{C}\right)\\ \approx +3.20\times {10}^{-19}\text{C}\end{array}$

The mass is calculated as,

  $M_{{\text{Ca}}^{\text{2+}}}=M_{\text{Ca}}-2m_{\text{e}}$

Here,

$M_{\text{Ca}}$ is the mass of the neutral Calcium atom.

$M_{{\text{Ca}}^{\text{2+}}}$ is the mass of the ionized Calcium atom.

Substitute $40.078\text{u}$ for $M_{\text{Ca}}$ and $9.11\times {10}^{-31}\text{}\text{kg}$ for $m_{\text{e}}$ in the above equation

  $\begin{array}{c}{M}_{{\text{Ca}}^{\text{2+}}}=40.078\text{u}-2\left(9.11\times {10}^{-31}\text{}\text{kg}\right)\\ =40.078\left(1.660\times {10}^{-27}\text{Kg}\right)-2\left(9.11\times {10}^{-31}\text{}\text{kg}\right)\\ =6.65\times {10}^{-26}\text{kg}\end{array}$

Conclusion:

Therefore, the charge of ionized Chlorine atom is $+3.20\times {10}^{-19}\text{C}$ and mass is $6.65\times {10}^{-26}\text{Kg}$.

To determine

The charge and mass of the centre of ammonia molecule ${\text{N}}^{3-}$ up to three significant digits.

Answer to Problem 23.1P

The charge and mass of the centre of ammonia molecule ${\text{N}}^{3-}$ is $-4.80\times {10}^{-19}\text{C}$ and $2.33\times {10}^{-26}\text{kg}$

Explanation of Solution

The centre of ammonia molecule is ${\text{N}}^{3-}$. When the Nitrogen atom is ionized by emission of electron as the electronic configuration of sodium atom is $2,5$, there are five electrons in the outermost shell so to become stable or achieve noble gas electronic configuration it has to lose five electrons or gain three electrons. Five electrons cannot lose as the atom collapse due to the removal of the orbit. So it will gain two electrons by gaining the electron the number of electrons increases compare to the number of protons. Nitrogen atom always forms the negative ion.

The charge of an ionized atom is given by the number of electron is gained or lost by the neutral atom as the atom forms an ion by losing a electron so its charge is $+2e$.

The charge of Nitrogen ion

  $\begin{array}{c}{\text{N}}^{3-}=-3e\\ =3\left(1.6\times {10}^{-19}\text{C}\right)\\ \approx -4.80\times {10}^{-19}\text{C}\end{array}$

The mass is calculated as,

  $M_{{\text{N}}^{\text{3-}}}=M_{\text{N}}+3m_{\text{e}}$

Here,

$M_{\text{N}}$ is the mass of the neutral Nitrogen atom.

$M_{{\text{N}}^{\text{3-}}}$ is the mass of the ionized Nitrogen atom.

Substitute $14.007\text{u}$ for $M_{\text{N}}$ and $9.11\times {10}^{-31}\text{}\text{kg}$ for $m_{\text{e}}$ in the above equation

  $\begin{array}{c}{M}_{{\text{N}}^{\text{3-}}}=14.007\text{u}+3\left(9.11\times {10}^{-31}\text{}\text{kg}\right)\\ =14.007\left(1.660\times {10}^{-27}\text{Kg}\right)+3\left(9.11\times {10}^{-31}\text{}\text{kg}\right)\\ =2.33\times {10}^{-26}\text{kg}\end{array}$

Conclusion:

The charge of centre of Ammonia molecule is $-4.80\times {10}^{-19}\text{C}$ and mass $2.33\times {10}^{-26}\text{Kg}$.

To determine

The charge and mass of the quadruple ionized Nitrogen ${\text{N}}^{4+}$ up to three significant digits.

Answer to Problem 23.1P

The charge and mass of the quadruple ionized Nitrogen ${\text{N}}^{4+}$ is $+6.40\times {10}^{-19}\text{C}$ and $2.32\times {10}^{-26}\text{Kg}$

Explanation of Solution

The quadruple ionized Nitrogen atoms found in plasma of hot star is ${\text{N}}^{4+}$.

When the Nitrogen atom is ionized by emission of electron as the electronic configuration of sodium atom is $2,5$, there is five electrons in the outermost shell so to become stable or achieve noble gas electronic configuration it has to lose four electrons. As the nucleus of nitrogen is very strong which works as a plasma in hot star so that property of plasma does tends to collapse the atom although it has lose four of its electrons.

The charge of an ionized atom is given by the number of electron is gained or lost by the neutral atom as the atom forms an ion by losing four electrons so its charge is $+4e$.

The charge of quadruple Nitrogen ion

    $\begin{array}{c}{\text{N}}^{4+}=+4e\\ =4\left(1.6\times {10}^{-19}\text{C}\right)\\ \approx +6.40\times {10}^{-19}\text{C}\end{array}$

The mass is calculated as,

  $M_{{\text{N}}^{\text{4+}}}=M_{\text{N}}-4m_{\text{e}}$

Here,

$M_{{\text{N}}^{\text{4+}}}$ is the mass of the quadruply ionized Nitrogen atom.

Substitute $14.007\text{u}$ for $M_{\text{N}}$ and $9.11\times {10}^{-31}\text{}\text{kg}$ for $m_{\text{e}}$ in the above equation

  $\begin{array}{c}{M}_{{\text{N}}^{\text{4+}}}=14.007\text{u}-4\left(9.11\times {10}^{-31}\text{}\text{kg}\right)\\ =14.007\left(1.660\times {10}^{-27}\text{Kg}\right)-4\left(9.11\times {10}^{-31}\text{}\text{kg}\right)\\ =2.32\times {10}^{-26}\text{kg}\end{array}$

Conclusion:

The charge and mass of the quadruple ionized Nitrogen ${\text{N}}^{4+}$ is $+6.40\times {10}^{-19}\text{C}$ and mass $2.32\times {10}^{-26}\text{Kg}$.

To determine

The charge and mass of the nucleus of nitrogen atom ${\text{N}}^{7+}$ upto three significant digits.

Answer to Problem 23.1P

The charge and mass of the nucleus of nitrogen atom ${\text{N}}^{7+}$ is $+11.2\times {10}^{-19}\text{C}$ and $2.32\times {10}^{-26}\text{kg}$.

Explanation of Solution

The nucleus of nitrogen ion is ${\text{N}}^{7+}$. When the Nitrogen atom is ionized by emission of electron as the electronic configuration of sodium atom is $2,5$, there are five electrons in the outermost shell so to become stable or achieve noble gas electronic configuration it has to lose seven electrons. As the nucleus of nitrogen is very strong which works as a plasma in hot star so that property of plasma does tends to collapse the atom although it has lose all its electrons.

The charge of an ionized atom is given by the number of electron is gained or lost by the neutral atom as the atom forms an ion by losing a electron so its charge is $+7e$.

The charge of nucleus of Nitrogen ion is;

  $\begin{array}{c}{\text{N}}^{7+}=+7e\\ =7\left(1.6\times {10}^{-19}\text{C}\right)\\ \approx +1.12\times {10}^{-18}\text{C}\end{array}$

The mass is calculated as,

  $M_{{\text{N}}^{\text{7+}}}=M_{\text{N}}-7m_{\text{e}}$

Here,

$M_{{\text{N}}^{\text{7+}}}$ is the mass of the quadruply ionized Nitrogen atom.

Substitute $14.007\text{u}$ for $M_{\text{N}}$ and $9.11\times {10}^{-31}\text{}\text{kg}$ for $m_{\text{e}}$ in the above equation

  $\begin{array}{c}{M}_{{\text{N}}^{\text{7+}}}=14.007\text{u}-7\left(9.11\times {10}^{-31}\text{}\text{kg}\right)\\ =14.007\left(1.660\times {10}^{-27}\text{Kg}\right)-7\left(9.11\times {10}^{-31}\text{}\text{kg}\right)\\ =2.32\times {10}^{-26}\text{kg}\end{array}$

Conclusion:

The charge of quadruple is $+1.12\times {10}^{-18}\text{C}$ and mass $2.32\times {10}^{-26}\text{Kg}$.

To determine

The charge and mass of the Molecular ion ${\text{H}}_{\text{2}}{\text{O}}^{-}$ up to three significant digits.

Answer to Problem 23.1P

The charge and mass of the Molecular ion ${\text{H}}_{\text{2}}{\text{O}}^{-}$ is $-1.6\times {10}^{-19}\text{C}$ and $2.99\times {10}^{-26}\text{kg}$.

Explanation of Solution

The molecular ion of water is ${\text{H}}_{\text{2}}{\text{O}}^{-}$. The atomic mass of water is eighteen in that the oxygen atom is attached to the two hydrogen atom. Oxygen has the valency of two it can share two electrons to the hydrogen atom and for a molecule as a whole it gains one electron and become a negative ion.

The charge of molecule of ${\text{H}}_{\text{2}}{\text{O}}^{-}$ ion

    $\begin{array}{c}{\text{H}}_{\text{2}}{\text{O}}^{-}=-e\\ \approx -1.6\times {10}^{-19}\text{C}\end{array}$

The mass is calculated as,

  $M_{{\text{H}}_2{\text{O}}^{-}}=2M_{\text{H}}+2M_{\text{O}}+m_{\text{e}}$

Here,

$M_{\text{O}}$ is the mass of the neutral oxygen atom.

$M_{{\text{H}}_2{\text{O}}^{\text{-}}}$ is the mass of the ionized water molecule.

Substitute $1.0079\text{u}$ for $M_{\text{H}}$, $15.999\text{u}$ for $M_{\text{O}}$ and $9.11\times {10}^{-31}\text{}\text{kg}$ for $m_{\text{e}}$ in the above equation

  $\begin{array}{c}{M}_{{\text{H}}^{\text{+}}}=2\left(1.0079\text{u}\right)+15.999\text{u}-9.11\times {10}^{-31}\text{}\text{kg}\\ =\left[\text{2}\left(1.0079\right)+15.999\text{u}\right]\left(1.660\times {10}^{-27}\text{Kg}\right)-9.11\times {10}^{-31}\text{}\text{kg}\\ =2.99\times {10}^{-27}\text{kg}\end{array}$

Conclusion:

Therefore, the charge of molecule of ${\text{H}}_{\text{2}}{\text{O}}^{-}$ is $-1.6\times {10}^{-19}\text{C}$ and mass $2.99\times {10}^{-26}\text{Kg}$.