Chapter 12, Problem 6P

(a)

To determine

Show that the force on an ion in ionic solid.

Answer to Problem 6P

The force on an ion in ionic solid is proved has $F=-k\alpha \frac{e^2}{r^2}\left[1-{\left(\frac{r_0}{r}\right)}^{m-1}\right]$.

Explanation of Solution

The total potential energy stored in the solid object is,

  $U_T=-\frac{k\alpha e^2}{r}+\frac{B}{r^m}$        (I)

Here, $k$ is the coulomb’s constant, $\alpha$ is the Madelung constant, $e$ is the charge of an electron, $r$ is the distance, $r^m$ is the distance of the solid displaced from an order, and $B$ is the magnetic field.

From problem (1) the value of the magnetic field is,

  $B=\left(\frac{k\alpha e^2}{m}\right)r_0^{m-1}$        (II)

Here, $m$ is an integer and $r_0$ is the equilibrium separation distance.

Conclusion:

Substitute equation (II) in equation (I).

  $\begin{array}{c}{U}_T=-\frac{k\alpha e^2}{r}+\frac{\left(\frac{k\alpha e^2}{m}\right)r_0^{m-1}}{r^m}\\ =-\frac{k\alpha e^2}{r}+\left(\frac{k\alpha e^2}{m}\right)\left(\frac{r_0^{m-1}}{r^m}\right)\end{array}$

The force on an ion in ionic solid is,

  $\begin{array}{c}F=-\frac{d{U}_T}{dr}\\ =-\frac{d}{dr}\left[-\frac{k\alpha e^2}{r}+\left(\frac{k\alpha e^2}{m}\right)\left(\frac{r_0^{m-1}}{r^m}\right)\right]\\ =-\frac{k\alpha e^2}{r^2}+\left(\frac{k\alpha e^2}{m}\right)\left(m\right)\left(\frac{r_0^{m-1}}{r^{m+1}}\right)\\ =-\left(\frac{k\alpha e^2}{r^2}\right)\left[1-{\left(\frac{r_0}{r}\right)}^{m-1}\right]\end{array}$

Therefore, the force on an ion in ionic solid is proved has $F=-k\alpha \frac{e^2}{r^2}\left[1-{\left(\frac{r_0}{r}\right)}^{m-1}\right]$.

(b)

To determine

Show that the ion experiences a restoring force.

Answer to Problem 6P

The ion experiences a restoring force is $K=\frac{k\alpha e^2}{r_0^3}\left(m-1\right)$.

Explanation of Solution

From part (a),

Consider that atom is displaced by a small distance $x$ from $r_0$ and replace $r$ with $r_0+x$, so that the force on an ion becomes,

  $F=-\left[\frac{k\alpha e^2}{{\left(r_0+x\right)}^2}\right]\left[1-{\left(\frac{r_0}{r_0+x}\right)}^{m-1}\right]$        (III)

Here, $x$ is the displaced distance of the solid.

Conclusion:

Substitute the condition $x/r_0\ll 1$ in above relation.

  $\begin{array}{c}F=-\left[\frac{k\alpha e^2}{r_0^2{\left(1+x/r_0\right)}^2}\right]\left[1-\left(\frac{r_0^{m-1}}{{\left(r_0+x\right)}^{m-1}}\right)\right]\\ =-\left[\frac{k\alpha e^2}{r_0^2{\left(1+x/r_0\right)}^2}\right]\left[1-\frac{1}{{\left(1+x/r_0\right)}^{m-1}}\right]\\ =-\left(\frac{k\alpha e^2}{r_0^2}\right)\left(1-\frac{2x}{r_0}\right)\left[1-\frac{1}{\left(1+\left(m-1\right)\left(x/r_0\right)+\mathrm{..}\right)}\right]\\ =-\left(\frac{k\alpha e^2}{r_0^2}\right)\left(1-\frac{2x}{r_0}\right)\frac{1+\left(m-1\right)\left(x/r_0\right)-1}{1+\left(m-1\right)\left(x/r_0\right)+\mathrm{..}}\end{array}$

Neglect the term $x/r_0$ in the denominator.

  $\begin{array}{c}F=-\frac{\left(k\alpha e^2/r_0^2\right)\left(1-2x/r_0\right)\left(m-1\right)x}{r_0}\\ =-\left(\frac{k\alpha e^2}{r_0^3}\right)\left(m-1\right)x\\ =-Kx\end{array}$

Therefore, the ion experience a restoring force $F=-Kx$, here the effective force constant K is $K\approx \left(\frac{k\alpha e^2}{r_0^3}\right)\left(m-1\right)$.

(c)

To determine

The frequency of vibration of ${\text{Na}}^{+}$ ion in NaCl.

Answer to Problem 6P

The frequency of vibration of ${\text{Na}}^{+}$ ion in NaCl is $9.15\times {10}^{12}\text{Hz}$.

Explanation of Solution

From part (b),

The effective force constant is,

  $K\approx \left(\frac{k\alpha e^2}{r_0^3}\right)\left(m-1\right)$        (IV)

Write the expression for frequency of vibration.

  $f=\frac{1}{2\pi }{\left(\frac{K}{m}\right)}^{1/2}$        (V)

Here, $f$ is the frequency, $K$ is the effective force constant, and $m$ is the mass sodium ion (Na⁺).

Conclusion:

Substitute $9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k$, $1.7476$ for $\alpha$, $1.6\times {10}^{-19}\text{C}$ for $e$, $0.281\text{nm}$ for $r_0$, and $8$ for $m$ in equation (IV) to find $K$.

  $\begin{array}{c}K\approx \left(\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(1.7476\right){\left(1.6\times {10}^{-19}\text{C}\right)}^2}{{\left[\left(0.281\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right]}^3}\right)\left(8-1\right)\\ \approx 127\text{N/m}\end{array}$

Substitute $127\text{N/m}$ for $K$ and $23\left(1.67\times {10}^{-27}\text{kg}\right)$ for $m$ in equation (V) to find $f$.

  $\begin{array}{c}f=\frac{1}{2\left(3.14\right)}{\left(\frac{127\text{N/m}}{23\left(1.67\times {10}^{-27}\text{kg}\right)}\right)}^{1/2}\\ =9.15\times {10}^{12}\text{Hz}\end{array}$

Therefore, the frequency of vibration of ${\text{Na}}^{+}$ ion in NaCl is $9.15\times {10}^{12}\text{Hz}$.

(d)

To determine

The wavelength of the Na⁺ ion absorb incident radiation and is this wavelength is UV, visible, or IR of the spectrum.

Answer to Problem 6P

The wavelength of the Na⁺ ion absorb incident radiation is $3.16\times {10}^4\text{nm}$ and this wavelength is IR of the spectrum.

Explanation of Solution

Write the expression for wavelength of the radiation.

  $\lambda =\frac{c}{f}$        (VI)

Here, $\lambda$ is the wavelength and $c$ is the velocity of light.

Conclusion:

Substitute $3.0\times {10}^8\text{m/s}$ for $c$ and $9.15\times {10}^{12}\text{Hz}$ for $f$ in equation (VI) to find $\lambda$.

  $\begin{array}{c}\lambda =\frac{3.0\times {10}^8\text{m/s}}{9.15\times {10}^{12}\text{Hz}}\\ =3.16\times {10}^{-5}\text{m}\left(\frac{1\text{nm}}{{10}^{-9}\text{m}}\right)\\ =3.16\times {10}^4\text{nm}\end{array}$

The above wavelength >> 800nm, therefore absorption is in infrared spectrum.