Chapter 43, Problem 4P

(a)

To determine

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

Answer to Problem 4P

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

Explanation of Solution

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

The ionization of energy of $\text{K}$ is $4.34\text{eV}$ and electron affinity of $\text{I}$ is $3.06\text{eV}$.

Write the expression that describes the ionization of K.

  $\text{K+}4.34\text{eV}\to {\text{K}}^{+}+{\text{e}}^{-}$

Here, $\text{K}$ is the potassium atom and ${\text{K}}^{+}$ is the potassium ion.

Write the expression that describes the ionization of I.

  $\text{I}+{\text{e}}^{-}\to {\text{I}}^{-}+3.06\text{eV}$

Here, $\text{I}$ is the iodine atom and ${\text{I}}^{-}$ is the iodine ion.

Add equation (I) and (II) to get activation energy.

  $\left(\text{K+}4.34\text{eV}\right)+\left(\text{I}+{\text{e}}^{-}\right)\to \left({\text{K}}^{+}+{\text{e}}^{-}\right)+\left({\text{I}}^{-}+3.06\text{eV}\right)$

Rearrange above equation to get activation energy.

  $\begin{array}{l}\left(\text{K+I}+\left(4.34\text{eV}-3.06\text{eV}\right)\right)\to \left({\text{K}}^{+}+{\text{I}}^{-}\right)\\ \text{K+I}+\left(1.28\text{eV}\right)\to {\text{K}}^{+}+{\text{I}}^{-}\end{array}$

The above relation indicates the $1.28\text{eV}$ energy is required to transfer an electron from $\text{K}$ to $\text{I}$, to form ${\text{K}}^{+}$ and ${\text{I}}^{-}$ ions from neutral atoms. Thus, $1.28\text{eV}$ represents the activation energy, $\left(E_a\right)$ .

Conclusion:

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

(b)

To determine

The values of $\sigma$ and $\in$.

Answer to Problem 4P

The values of $\sigma$ is $0.272\text{nm}$ and value of $\in$ is $4.65\text{eV}$.

Explanation of Solution

Write the expression of Lennard-Jones potential.

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

Here, $U\left(r\right)$ is the Lennard-Jones potential, $\in$ and $\sigma$ are adjustable parameters, $r$ is the internuclear separation distance.

Differentiate above equation.

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

Write the value of derivative of potential at equilibrium distance.

  ${\frac{dU}{dr}|}_{r=r_0}=0$                                                                                                             (III)

Here, $r_0$ is the equilibrium distance.

Use equation (III) in equation (II) to get value of $\sigma$.

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

Rearrange above equation to get $\sigma$.

  $\begin{array}{c}{\left(\frac{\sigma }{r_0}\right)}^{13}=\frac{1}{2}{\left(\frac{\sigma }{r_0}\right)}^7\\ {\left(\frac{\sigma }{r_0}\right)}^6=\frac{1}{2}\\ \sigma =\frac{r_0}{2^{\frac{1}{6}}}\end{array}$                                                                                                     (IV)

It is given that $U\left(r_0\right)=-3.37\text{eV}$.

Substitute $r_0$ for $r$ in equation (I).

  $\begin{array}{c}U\left(r_0\right)=4\in \left[{\left(\frac{\frac{r_0}{2^{\frac{1}{6}}}}{r_0}\right)}^{12}-{\left(\frac{\frac{r_0}{2^{\frac{1}{6}}}}{r_0}\right)}^6\right]+E_a\\ =4\in \left[{\left(\frac{1}{2^{\frac{1}{6}}}\right)}^{12}-{\left(\frac{1}{2^{\frac{1}{6}}}\right)}^6\right]+E_a\\ =4\in \left[\left(\frac{1}{4}\right)-\left(\frac{1}{2}\right)\right]+E_a\\ =-\in +E_a\end{array}$                                                                       (V)

Conclusion:

Substitute $0.305\text{nm}$ for $r_0$ in equation (IV) to get $\sigma$.

  $\begin{array}{c}\sigma =\frac{0.305\text{nm}}{2^{\frac{1}{6}}}\\ =0.272\text{nm}\end{array}$

Substitute $-3.37\text{eV}$ for $U\left(r_0\right)$ and $1.28\text{eV}$ for $E_a$ in equation (V) to get $\in$.

$\begin{array}{c}-3.37\text{eV}=-\in +1.28\text{eV}\\ \in =1.28\text{eV+}3.37\text{eV}\\ \in =\text{4}\text{.65}\text{eV}\end{array}$

Therefore, the values of $\sigma$ is $0.272\text{nm}$ and value of $\in$ is $4.65\text{eV}$.

(c)

To determine

The force needed to break up KI molecule.

Answer to Problem 4P

The force needed to break up KI molecule is $+6.55\text{nN}$.

Explanation of Solution

Write the expression for the force of attraction between the atoms.

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

Use equation (II) in equation (VI) to get $F\left(r\right)$.

  $\begin{array}{c}F\left(r\right)=-\left(\frac{4\in }{\sigma }\left[-12{\left(\frac{\sigma }{r}\right)}^{13}+6{\left(\frac{\sigma }{r}\right)}^7\right]\right)\\ =\frac{4\in }{\sigma }\left[12{\left(\frac{\sigma }{r}\right)}^{13}-6{\left(\frac{\sigma }{r}\right)}^7\right]\end{array}$                                                                (VII)

Write the expression for the maximum force.

  ${\frac{dF}{dr}|}_{r=r_{\text{break}}}=0$                                                                                                        (VIII)

Here, $r_{\text{break}}$ is the distance at which force is maximum.

Put equation (VII) in equation (VIII).

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

Use equation (VIII) in equation (IX) to get $\frac{\sigma }{r_{\text{break}}}$.

  $\begin{array}{c}\frac{4\in }{{\sigma }^2}\left[-156{\left(\frac{\sigma }{r_{\text{break}}}\right)}^{14}+42{\left(\frac{\sigma }{r_{\text{break}}}\right)}^8\right]=0\\ 156{\left(\frac{\sigma }{r_{\text{break}}}\right)}^{14}=42{\left(\frac{\sigma }{r_{\text{break}}}\right)}^8\\ {\left(\frac{\sigma }{r_{\text{break}}}\right)}^6=\frac{42}{156}\\ \frac{\sigma }{r_{\text{break}}}={\left(\frac{42}{156}\right)}^{\frac{1}{6}}\end{array}$

Conclusion:

Substitute $4.65\text{eV}$ for $\in$, $0.272\text{nm}$ for $\sigma$ and ${\left(\frac{42}{156}\right)}^{\frac{1}{6}}$ for $\frac{\sigma }{r}$ in equation (IX) to get maximum force.

  $\begin{array}{c}F\left(r\right)=\frac{4\left(4.65\text{eV}\right)}{\left(0.272\text{nm}\right)}\left[12\left({\left(\frac{42}{156}\right)}^{\frac{13}{6}}\right)-6\left({\left(\frac{42}{156}\right)}^{\frac{7}{6}}\right)\right]\times \frac{{10}^9\text{nm}}{1\text{m}}\\ =-6.55\times {10}^{-9}\text{N}\times \frac{{10}^9\text{nN}}{1\text{N}}\\ =-6.55\text{nN}\end{array}$

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

(d)

To determine

The force constant for small oscillations about $r=r_0$.

Answer to Problem 4P

The force constant for small oscillations about $r=r_0$ is $576\text{N}/\text{m}$.

Explanation of Solution

Rewrite expression of Lennard-Jones potential.

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

Substitute $r_0+s$ for $r$ in above equation.

  $U\left(r_0+s\right)=4\in \left[{\left(\frac{\sigma }{r_0+s}\right)}^{12}-{\left(\frac{\sigma }{r_0+s}\right)}^6\right]+E_a$

Substitute $\frac{r_0}{2^{\frac{1}{6}}}$ for $\sigma$ in above equation to get $U\left(r_0+s\right)$.

  $U\left(r_0+s\right)=4\in \left[{\left(\frac{\frac{r_0}{2^{\frac{1}{6}}}}{r_0+s}\right)}^{12}-{\left(\frac{\frac{r_0}{2^{\frac{1}{6}}}}{r_0+s}\right)}^6\right]+E_a$

Expand above equation using binomial expansion.

  $\begin{array}{c}U\left(r_0+s\right)=4\in \left[{\left(\frac{\frac{r_0}{2^{\frac{1}{6}}}}{r_0+s}\right)}^{12}-{\left(\frac{\frac{r_0}{2^{\frac{1}{6}}}}{r_0+s}\right)}^6\right]+E_a\\ =4\in \left[\frac{1}{4}{\left(1+\frac{s}{r_0}\right)}^{-12}--\frac{1}{2}{\left(1+\frac{s}{r_0}\right)}^{-6}\right]+E_a\\ =4\in [\frac{1}{4}\left(1-12\frac{s}{r_0}+78\frac{s^2}{r_0^2}-\cdots \right)-\frac{1}{2}\left(1-6\frac{s}{r_0}+21\frac{s^2}{r_0^2}-\cdots \right)]+E_a\\ =\in -12\in \frac{s}{r_0}+78\in \frac{s^2}{r_0^2}-2\in +12\in \frac{s}{r_0}-42\in \frac{s^2}{r_0^2}+E_a+\cdots \end{array}$

Simplify above equation up to second order terms of $s$

  $U\left(r_0+s\right)=-\in +E_a+0\left(\frac{s}{r_0}\right)+36\in \frac{s^2}{r_0^2}+\cdots$

Use equation (V) in above equation.

  $U\left(r_0+s\right)=U\left(r_0\right)+36\in \frac{s^2}{r_0^2}$                                                                                    (X)

The above equation is similar to equation of potential of small oscillations.

Write the general potential equation.

  $U\left(r_0+s\right)=U\left(r_0\right)+\frac{1}{2}k{s}^2$

Compare above equation with (X) to get $k$.

  $k=\frac{72\in }{r_0^2}$                                                                                                                 (XI)

Conclusion:

Substitute $4.65\text{eV}$ for $\in$ and $0.305\text{nm}$ for $r_0$ in equation (XI) to get $k$.

  $\begin{array}{c}k=\frac{72\left(4.65\text{eV}\right)}{{\left(0.305\text{nm}\right)}^2}\\ =3599\text{eV}/{\text{nm}}^2\times \frac{1.6\times {10}^{-19}\text{N}\cdot {\text{m}}^2}{1\text{eV}}\times \frac{{\left({10}^9\text{nm}\right)}^2}{{\text{m}}^2}\\ =576\text{N}/\text{m}\end{array}$

Therefore, the force constant for small oscillations about $r=r_0$ is $576\text{N}/\text{m}$.