Chapter 41, Problem 60CP

(a)

To determine

The potential energy of the system.

Answer to Problem 60CP

The potential energy of the system is $-\frac{7k_e{e}^2}{3d}$.

Explanation of Solution

Write the equation for potential energy.

    $U=\frac{k_e{q}_1{q}_2}{r}$                                                                                             (I)

Here, $U$ is the potential energy, $k_e$ is the Coulomb constant, $q_1$ is the first charge, $q_2$ is the second charge and $r$ is the distance between the charges.

Substitute $e$ for $q_1$ and $q_2$ and $d$ for $r$ in the equation (I).

    $U=\frac{k_e{e}^2}{d}$                                                                                                         (II)

Here, $e$ is the charge and $d$ is the distance between the adjacent charges.

Write the equation for the potential energy of the system.

    $U=U_{12}+U_{13}+U_{14}+U_{23}+U_{24}+U_{34}$                                                          (III)

Here, $U_{12}$ is the potential energy between the first electron and the first ion, $U_{13}$ is the potential energy between the two electrons, $U_{14}$ is the potential energy between the first electron and the second ion, $U_{23}$ is the potential energy between the first ion and the second electron, $U_{24}$ is the potential energy between the two ions, and  $U_{34}$ is the potential energy between the second electron and the second ion.

Conclusion:

Substitute equation (II) in equation (III) for each pair of energy.

    $\begin{array}{c}U=\left(\frac{-k_e{e}^2}{d}+\frac{k_e{e}^2}{2d}-\frac{k_e{e}^2}{3d}\right)+\left(-\frac{k_e{e}^2}{d}+\frac{k_e{e}^2}{2d}\right)-\frac{k_e{e}^2}{d}\\ =\frac{k_e{e}^2}{d}\left(-1+\frac{1}{2}-\frac{1}{3}-1+\frac{1}{2}-1\right)\\ =-\frac{7k_e{e}^2}{3d}\end{array}$                                       (IV)

Thus, the potential energy of the system is $-\frac{7k_e{e}^2}{3d}$.

(b)

To determine

The minimum kinetic energy of the two electrons.

Answer to Problem 60CP

The minimum kinetic energy of the two electrons is $\frac{h^2}{36m_e{d}^2}$.

Explanation of Solution

Let the minimum energy of the electron be $E_1$.

    $E_1=\frac{h^2}{8m_e{L}^2}$

Here, $h$ is the Plank’s constant, $m_e$ is the mass of electron and $L$ is the length of the box.

Substitute $3d$ for $L$ in the above equation.

    $E_1=\frac{h^2}{8m_e{\left(3d\right)}^2}$                                                                                                (V)

Write the equation for the minimum kinetic energy of the two electrons.

    $K=2E_1$                                                                                                (VI)

Here, $K$ is the minimum kinetic energy of the two electrons.

Conclusion:

Substitute equation (V) in equation (VI) to find $K$.

    $\begin{array}{c}K=2\left(\frac{h^2}{8m_e{\left(3d\right)}^2}\right)\\ =\frac{h^2}{4m_e\left(9d^2\right)}\\ =\frac{h^2}{36m_e{d}^2}\end{array}$                                                                                        (VII)

Thus, the minimum kinetic energy of the two electrons is $\frac{h^2}{36m_e{d}^2}$.

(c)

To determine

The value of $d$ when the total energy is minimum.

Answer to Problem 60CP

The value of $d$ when the total energy minimum is $49.9\text{pm}$.

Explanation of Solution

Write the equation for the total energy.

    $E=K+U$

Here, $E$ is the total energy.

Substitute equation (IV) and (VII) in the above equation to find $E$.

    $E=\frac{h^2}{36m_e{d}^2}-\frac{7k_e{e}^2}{3d}$                                                                             (VIII)

Write the condition for the energy to be minimum.

    $\frac{dE}{d\left(d\right)}=0$

Conclusion:

Substitute equation (VIII) in the above equation to find $d$.

    $\begin{array}{c}\frac{d\left(\frac{h^2}{36m_e{d}^2}-\frac{7k_e{e}^2}{3d}\right)}{d\left(d\right)}=0\\ \left(-2\right)\left(\frac{h^2}{36m_e{d}^3}\right)-\left(-1\right)\left(\frac{7k_e{e}^2}{3d^2}\right)=0\\ \frac{h^2}{18m_e{d}^3}=\frac{7k_e{e}^2}{3d^2}\end{array}$

Rearrange the above equation for $d$.

  $\begin{array}{c}d=\frac{3h^2}{7\left(18m_e\right)k_{e}e}\\ =\frac{h^2}{42m_e{k}_e{e}^2}\end{array}$                                                                                                (IX)

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k_e$, $9.11\times {10}^{-31}\text{kg}$ for $m_e$ and $1.60\times {10}^{-19}\text{C}$ for $e$ in the above equation to find $d$.

    $\begin{array}{c}d=\frac{{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{42\left(9.11\times {10}^{-31}\text{kg}\right)\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right){\left(1.60\times {10}^{-19}\text{C}\right)}^2}\\ =4.99\times {10}^{-11}\text{m}\\ =\left(49.9\times {10}^{-12}\text{m}\right)\left(\frac{{10}^{12}\text{pm}}{1\text{m}}\right)\\ =49.9\text{pm}\end{array}$

Thus, the value of $d$ when the total energy is minimum is $49.9\text{pm}$.

(d)

To determine

Compare the value of $d$ to the spacing of atoms in lithium.

Answer to Problem 60CP

The lithium inter-atomic spacing is in the same order of magnitude as the interatomic spacing of $2d$.

Explanation of Solution

Write the equation for volume.

    $V=N{d}^3$                                                                                                      (X)

Here, $V$ is the volume and $N$ is the number of atoms.

Write the equation for density.

    $\rho =\frac{Nm}{V}$

Here, $m$ is the mass of one atom and $\rho$ is the density.

Substitute equation (X) in the above equation and rearrange to find $d$.

    $\begin{array}{c}\rho =\frac{Nm}{N{d}^3}\\ =\frac{m}{d^3}\\ d={\left(\frac{m}{\rho }\right)}^{\frac{1}{3}}\end{array}$

Conclusion:

Substitute $6.94\text{g/mol}$ for $m$ and $0.530{\text{g/cm}}^3$ for $\rho$ in the above equation to find $d$.

    $\begin{array}{c}d={\left(\frac{6.94\text{g/mol}}{0.530{\text{g/cm}}^3}\right)}^{\frac{1}{3}}\\ ={\left(\frac{\left(6.94\text{g/mol}\right)\left(\frac{1\text{mol}}{6.022\times {10}^{23}\text{atoms}}\right)}{0.530{\text{g/cm}}^3}\right)}^{\frac{1}{3}}\\ =\left(279\times {10}^{-12}\text{m}\right)\left(\frac{{10}^{12}\text{pm}}{1\text{m}}\right)\\ =279\text{pm}\end{array}$

The lithium interatomic spacing is 5.59 times larger than $d$.

Thus, the lithium interatomic spacing is in the same order of magnitude as the interatomic spacing $2d$.