Chapter 29, Problem 1E

(a)

To determine

The energy required to transfer one electron from $Na$ to $Cl$.

Answer to Problem 1E

The  energy  required to transfer one electron from $Na$ to $Cl$ is $1.3\text{eV}$.

Explanation of Solution

The energy required to transfer one electron from $Na$ to $Cl$ is equal to the difference between the required energy to remove an electron from the sodium and the energy release to bind an electron to chlorine.

Write the expression for transfer on electron from $Na$ to $Cl$.

  $\Delta E=E_{Na}-E_{Cl}$        (I)

Here, $\Delta E$ is the energy required to transfer one electron from $Na$ to $Cl$, $E_{Na}$ is the ionization energy of sodium and $E_{Cl}$ is the energy to bind the electron to chlorine.

Conclusion:

Substitute $5.12\text{eV}$ for $E_{Na}$ and $3.82\text{eV}$ for $E_{Cl}$ in equation (I).

  $\begin{array}{l}\Delta E=5.12\text{eV-3}\text{.82}\text{eV}\\ \Delta E=1.3\text{eV}\end{array}$

Thus, the energy required to transfer one electron from $Na$ to $Cl$ is $1.3\text{eV}$.

(b)

To determine

The electric potential energy.

Answer to Problem 1E

The electric potential energy is $-6.1\text{eV}$.

Explanation of Solution

Coulomb force is the amount of force between two point charges, if the point charges are equal in magnitude and sign then the forces are repulsive in nature and if the charges are opposite in nature, the force between the two charges will be attractive in nature.

Eelctric potential energy is the potential energy that results from coulombic forces.

Write the expression for the electric potential energy for the charge $+e$ and $-e$.

  $U=\frac{-k{e}^2}{r_0}$        (II)

Here, $U$ is the electric potential energy, $k$ is the coulomb constant, $+e$ and $-e$

are the point charges and $r_0$ is the distance between the two point charges.

Conclusion:

Substitute $9.0\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$, $1.6\times {10}^{-19}\text{C}$ for $+e$ and $-e$ and $2.36\times {10}^{-10}\text{m}$ for $r_0$  in equation (II).

    $\begin{array}{l}U=\frac{-9.0\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^{\text{2}}{\left(1.6\times {10}^{-19}\text{C}\right)}^2}{2.36\times {10}^{-10}\text{m}}\times \frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\\ U=-6.1\text{eV}\end{array}$

Thus, the electric potential energy is $-6.1\text{eV}$.

(c)

To determine

The energy required to separate the sodium chloride molecule to its constituents.

Answer to Problem 1E

The  energy required to separate the sodium chloride molecule to its constituents is $-4.8\text{eV}$.

Explanation of Solution

The total energy of the molecule is the sum of the required energy to transfer one electron from sodium to chlorine and the potential energy.

Write the expression for total energy that is the sum of transfer on electron from $Na$ to $Cl$ and the potential energy.

  $E\text{'}=\Delta E+U$        (III)

Here, $E\text{'}$ is the total energy and $\Delta E$ is the energy required to transfer one electron from $Na$ to $Cl$ and $U$ is the electric potential energy.

Conclusion:

Substitute $1.3\text{eV}$ for $\Delta E$ and $\text{-6}\text{.1}\text{eV}$ for $U$ in equation (III).

  $\begin{array}{l}E\text{'}=1.3\text{eV+}\left(\text{-6}\text{.1}\text{eV}\right)\\ E\text{'}=-4.8\text{eV}\end{array}$

Thus, the  energy required to separate the sodium chloride molecule to its constituents is $-4.8\text{eV}$.