Chapter 6, Problem 15P

(a)

To determine

The potential energy of the system of all pairs of interaction.

Answer to Problem 15P

Total potential energy of the system is $\frac{\left(-7/3\right)k{e}^2}{d}$ .

Explanation of Solution

Write the expression for potential energy.

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

Here, $q_1$ is charge1, $q_2$ is charge2, $r$ is the distance between charges, $k$ is constant and $U$ is the potential energy.

Conclusion:

Substitute $-e$ for $q_1$, $e$ for $q_2$ and $d$ for $r$ in equation (I) for first nuclei-electron pair.

  $U=\frac{-k{e}^2}{d}$

Similarly, for other pairs, the potential energy is calculated.

The table of potential energy for all pairs is drawn below.

S. No.	Charge 1	Charge 2	Distance b/w charges	Potential Energy
1.	$-e$	$+e$	$d$	$\frac{-k{e}^2}{d}$
2.	$-e$	$-e$	$2d$	$\frac{1}{2}\left(\frac{k{e}^2}{d}\right)$
3.	$-e$	$+e$	$3d$	$\frac{1}{3}\left(\frac{-k{e}^2}{d}\right)$
4.	$+e$	$-e$	$d$	$\frac{-k{e}^2}{d}$
5.	$+e$	$+e$	$2d$	$\frac{1}{2}\left(\frac{k{e}^2}{d}\right)$
6.	$-e$	$+e$	$d$	$\frac{-k{e}^2}{d}$

The total potential of all pairs is the sum of potential energy of each pair.

Calculate the total energy.

  $\begin{array}{l}U=\left(\frac{k{e}^2}{d}\right)\left[-1+\frac{1}{2}-\frac{1}{3}-1+\frac{1}{2}-1\right]\\ =-\frac{7}{3}\left(\frac{k{e}^2}{d}\right)\end{array}$

Thus, total potential energy of the system is $\frac{\left(-7/3\right)k{e}^2}{d}$ .

(b)

To determine

The minimum kinetic energy of two electrons.

Answer to Problem 15P

Minimum kinetic energy of two electrons is $\frac{h^2}{36m{d}^2}$ .

Explanation of Solution

Write the expression for kinetic energy of electron.

  $K=\frac{n^2{h}^2}{8m{L}^2}$        (II)

Here, $n$ is the energy level, $h$ is the Planck’s constant, $m$ is the mass of particle, $L$ is the length of box and $K$ is the  kinetic energy of particle.

Conclusion:

Substitute 1 for $n$ and $3d$ for $L$ in equation (II)

  $\begin{array}{c}K=\frac{h^2}{8m{\left(3d\right)}^2}\\ =\frac{h^2}{72m{d}^2}\end{array}$

Kinetic energy of two electrons would be two times of the above value.

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

(c)

To determine

The value of $d$ for minimum total energy.

Answer to Problem 15P

The value of $d$ is $\text{0}\text{.050}\text{nm}$.

Explanation of Solution

Write the expression for total energy.

  $E=U+K$        (III)

Here, $E$ is total energy.

Substitute $\frac{\left(-7/3\right)k{e}^2}{d}$ for $U$ and $\frac{h^2}{36m{d}^2}$ for $K$  in equation (III).

  $E=\frac{\left(-7/3\right)k{e}^2}{d}+\frac{h^2}{36m{d}^2}$

First derivative of energy should be zero for minimum total energy.

  $\begin{array}{c}\frac{dE}{dd}=\frac{d}{dd}\left(\frac{\left(-7/3\right)k{e}^2}{d}+\frac{h^2}{36m{d}^2}\right)\\ 0=\frac{\left(+7/3\right)k{e}^2}{d^2}-\frac{h^2}{18m{d}^2}\end{array}$

Rearrange the above expression in terms of $d$ .

  $d=\frac{h^2}{42mk{e}^2}$        (IV)

Conclusion:

Substitute $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $9.11\times {10}^{-31}\text{kg}$ for $m$, $9\times {10}^9\text{N}\cdot {\text{m}}^2\cdot {\text{C}}^{-2}$ for $k$ and $1.6\times {10}^{-19}\text{C}$ for $e$ in equation (IV).

  $\begin{array}{c}d=\frac{{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{42\left(9.11\times {10}^{-31}\text{kg}\right)\left(9\times {10}^9\text{N}\cdot {\text{m}}^2\cdot {\text{C}}^{-2}\right){\left(1.6\times {10}^{-19}\text{C}\right)}^2}\\ =0.5\times {10}^{-10}\text{m}\\ =\text{0}\text{.050}\text{nm}\end{array}$

Thus, the value of $d$ is $\text{0}\text{.050}\text{nm}$.

(d)

To determine

To compare the value of $d$ with the atomic spacing of Lithium.

Answer to Problem 15P

Atomic spacing is 2.8 times larger than $2d$ .

Explanation of Solution

Write the expression for volume.

  $V=N{a}^3$

Here, $N$ is Avogadro’s number, $a$ is atomic spacing and $V$ is volume.

Multiply both sides with $m$ above expression.

  $mV=mN{a}^3$

Rearrange the above expression in terms of $a$ .

  $a={\left(\frac{mV}{mN}\right)}^{1/3}$

Substitute $d$ for $\frac{Nm}{V}$ in above expression.

  $a={\left(\frac{m}{d}\right)}^{1/3}$        (V)

Here, $d$ is density and $m$ is the mass of atom.

Conclusion:

Substitute $7\left(1.66\times {10}^{-27}\text{kg}\right)$ for $m$ and $530\text{kg}/{\text{m}}^3$ in equation (V).

  $\begin{array}{c}a={\left(\frac{7\left(1.66\times {10}^{-27}\text{kg}\right)}{530\text{kg}/{\text{m}}^3}\right)}^{1/3}\\ =2.8\times {10}^{-10}\text{m}\\ =0.28\text{nm}\end{array}$

Thus, atomic spacing is 2.8 times larger than $2d$ .