Chapter 11, Problem 1P

(a)

To determine

The energy needed to transfer an electron from ${\rm K}$ to ${\rm I}$, to form ${{\rm K}}^{+}$ and ${{\rm I}}^{-}$ ions from neutral atoms.

Answer to Problem 1P

The energy needed to transfer an electron from ${\rm K}$ to ${\rm I}$, to form ${{\rm K}}^{+}$ and ${{\rm I}}^{-}$ ions from neutral atoms is $1.28\text{eV}$.

Explanation of Solution

Ionization energy is the energy needed to remove an electron from the atom and electron affinity is the amount of energy released when the atom gains an electron.

Ionization energy: ${\rm K}+4.34\text{ eV}\to {{\rm K}}^{+}+e^{-}$

Electron affinity: ${\rm I}+e^{-}\to {{\rm I}}^{-}+3.06\text{ eV}$

Therefore, the activation energy or the minimum energy needed to transfer an electron from ${\rm K}$ to ${\rm I}$ is

$\begin{array}{l}=4.34\text{eV}-3.06\text{eV}\\ =1.28\text{eV}\end{array}$

Conclusion:

Thus, the energy needed to transfer an electron from ${\rm K}$ to ${\rm I}$, to form ${{\rm K}}^{+}$ and ${{\rm I}}^{-}$ ions from neutral atoms is $1.28\text{eV}$.

(b)

To determine

The adjustable constants $\sigma$ and $\epsilon$.

Answer to Problem 1P

The adjustable constants $\sigma$ is $0.273\text{ nm}$ and $\epsilon$ is $4.65\text{ eV}$.

Explanation of Solution

Given, a model potential energy function for the $\text{KI}$ molecule is

    $U\left(r\right)=4\epsilon \left[{\left(\frac{\sigma }{r}\right)}^{12}-{\left(\frac{\sigma }{r}\right)}^6\right]+E_a$        (I)

Here, $U\left(r\right)$ is the potential energy, $r$ is the internuclear separation distance, $\sigma$ and $\epsilon$ are the adjustable constants, $E_a$ is the activation energy.

Differentiate equation (I) with respect to $r$

$\begin{array}{c}\frac{dU}{dr}=\frac{d}{dr}\left[4\epsilon \left[{\left(\frac{\sigma }{r}\right)}^{12}-{\left(\frac{\sigma }{r}\right)}^6\right]+E_a\right]\\ =4\epsilon \left[-12\frac{{\sigma }^{12}}{r^{13}}-\left(-6\right)\frac{{\sigma }^6}{r^7}\right]\\ =\frac{4\epsilon }{\sigma }\left[-12{\left(\frac{\sigma }{r}\right)}^{13}+6{\left(\frac{\sigma }{r}\right)}^7\right]\end{array}$        (II)

At $r=r_0$ , $\frac{dU}{dr}=0$. The above equation becomes,

$\begin{array}{c}\frac{4\epsilon }{\sigma }\left[-12{\left(\frac{\sigma }{r_0}\right)}^{13}+6{\left(\frac{\sigma }{r_0}\right)}^7\right]=0\\ -2{\left(\frac{\sigma }{r_0}\right)}^{13}+{\left(\frac{\sigma }{r_0}\right)}^7=0\\ 2\left(\frac{{\sigma }^{13}}{r_0^{13}}\right)=\left(\frac{{\sigma }^7}{r_0^7}\right)\\ {\sigma }^6=\frac{1}{2}{r}_0^6\end{array}$

Further solving,

$\begin{array}{l}\sigma ={\left(\frac{1}{2}{r}_0^6\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\\ \sigma ={\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{r}_0\end{array}$

Substitute $0.305\text{ nm}$ for $r_0$ in the above equation

$\begin{array}{c}\sigma ={\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\times 0.305\text{ nm}\\ =0.273\text{ nm}\end{array}$

The adjustable constant $\sigma$ is $0.273\text{ nm}$.

Substitute ${\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{r}_0$ for $\sigma$ and $r_0$ for $r$ in equation (I) and solve for $\epsilon$

$\begin{array}{c}U\left(r_0\right)=4\epsilon \left[{\left(\frac{{\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{r}_0}{r_0}\right)}^{12}-{\left(\frac{{\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{r}_0}{r_0}\right)}^6\right]+E_a\\ =4\epsilon \left[{\left({\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\right)}^{12}-{\left({\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\right)}^6\right]+E_a\\ =4\epsilon \left[{\left(\frac{1}{2}\right)}^2-\frac{1}{2}\right]+E_a\\ =4\epsilon \left[\frac{1}{4}-\frac{1}{2}\right]+E_a\end{array}$

Further solving,

$\begin{array}{c}U\left(r_0\right)=4\epsilon \left[-\frac{1}{4}\right]+E_a\\ U\left(r_0\right)=-\epsilon +E_a\\ \epsilon =E_a-U\left(r_0\right)\end{array}$

Substitute  $1.28\text{ eV}$ for $E_a$ and $-3.37\text{ eV}$ for $U\left(r_o\right)$ in the above equation and solve further

$\begin{array}{c}\epsilon =1.28\text{ eV}+3.37\text{ eV}\\ =4.65\text{ eV}\end{array}$

The adjustable constant $\epsilon$ is $4.65\text{ eV}$.

Conclusion:

Thus, the adjustable constants $\sigma$ is $0.273\text{ nm}$ and $\epsilon$ is $4.65\text{ eV}$.

(c)

To determine

The force needed to rupture the molecule.

Answer to Problem 1P

The force needed to rupture the molecule is $+6.55\text{ nN}$.

Explanation of Solution

Write the expression for the force of attraction of the molecule

    $F\left(r\right)=-\frac{dU}{dr}$        (III)

Here, $F\left(r\right)$ is the force.

The rupture of the molecule can be determined using $\frac{dF}{dr}$ .

Substitute equation (II) in (III)

$F\left(r\right)=\frac{4\epsilon }{\sigma }\left[12{\left(\frac{\sigma }{r}\right)}^{13}-6{\left(\frac{\sigma }{r}\right)}^7\right]$        (IV)

Differentiate the above equation with respect to $r$ and solve for $\frac{\sigma }{r_{rupture}}$

$\begin{array}{c}\frac{dF}{dr}=\frac{d}{dr}\left[\frac{4\epsilon }{\sigma }\left[12{\left(\frac{\sigma }{r}\right)}^{13}-6{\left(\frac{\sigma }{r}\right)}^7\right]\right]\\ =\frac{4\epsilon }{{\sigma }^2}\left[\left(12\right)\left(-13\right){\left(\frac{\sigma }{r}\right)}^{14}-\left(6\right)\left(-7\right){\left(\frac{\sigma }{r}\right)}^8\right]\\ =\frac{4\epsilon }{{\sigma }^2}\left[-156{\left(\frac{\sigma }{r}\right)}^{14}+42{\left(\frac{\sigma }{r}\right)}^8\right]\end{array}$

Equating $\frac{dF}{dr}=0$

$\begin{array}{c}\frac{4\epsilon }{{\sigma }^2}\left[-156{\left(\frac{\sigma }{r}\right)}^{14}+42{\left(\frac{\sigma }{r}\right)}^8\right]=0\\ -156{\left(\frac{\sigma }{r}\right)}^{14}+42{\left(\frac{\sigma }{r}\right)}^8=0\\ 156{\left(\frac{\sigma }{r}\right)}^{14}=42{\left(\frac{\sigma }{r}\right)}^8\\ {\left(\frac{\sigma }{r}\right)}^6=\frac{42}{156}\end{array}$

$\frac{\sigma }{r}={\left(\frac{42}{156}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}$

Substitute ${\left(\frac{42}{156}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}$ for $\frac{\sigma }{r}$ in equation (IV)

$\begin{array}{c}{F}_{\max }=\frac{4\epsilon }{\sigma }\left[12{\left({\left(\frac{42}{156}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\right)}^{13}-6{\left({\left(\frac{42}{156}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\right)}^7\right]\\ =\frac{4\epsilon }{\sigma }\left[12{\left(\frac{42}{156}\right)}^{\raisebox{1ex}{$13$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}-6{\left(\frac{42}{156}\right)}^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\right]\end{array}$

Substitute $0.273\text{ nm}$ for $\sigma$ and $4.65\text{ eV}$ for $\epsilon$ in the above equation to find the value of $F_{\max }$

$\begin{array}{c}{F}_{\max }=\frac{4\left(4.65\text{ eV}\right)}{0.273\text{ nm}}\left[12{\left(\frac{42}{156}\right)}^{\raisebox{1ex}{$13$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}-6{\left(\frac{42}{156}\right)}^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\right]\\ =-41.0\text{eV}/\text{nm}\times \frac{1.6\times {10}^{-19}\text{Nm}}{{10}^{-9}\text{m}}\\ =-6.55\text{nN}\end{array}$

Conclusion:

Therefore, the force needed to rupture the molecule is $+6.55\text{ nN}$.