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## Q. 10.4

The dissociation energy, that is, the energy needed to break a NaCl crystal into individual sodium and chlorine atoms, is determined experimentally to be 764.4 kJ/mol at standard temperature and pressure (STP) (see Table 10.2). Use this value to calculate the range parameter ρ for NaCl.

Strategy First divide the experimental dissociation energy by Avogadro’s number (the number of ion pairs per mole) to obtain a value of $1.269 \times 10^{-18}$ J/ion pair. Then with the equilibrium position at $r=r_{0}, \text { we have } V\left(r=r_{0}\right)=-1.269\times 10^{-18}$ J. With this numerical value of $V\left(r=r_{0}\right)$, Equation (10.21) can be solved for $\rho / r_{0}$.

$V\left(r=r_{0}\right)=-\frac{\alpha e^{2}}{4 \pi \epsilon_{0} r_{0}}\left[1-\left(\rho / r_{0}\right)\right]$ (10.21)

## Verified Solution

We solve Equation (10.21) for $\rho / r_{0}$ in terms of $V\left(r=r_{0}\right)$ and obtain

\begin{aligned}\frac{\rho}{r_{0}} &=1+\frac{4 \pi \epsilon_{0} r_{0} V\left(r=r_{0}\right)}{\alpha e^{2}} \quad \\&=1+\frac{(0.282 nm )\left(-1.269 \times 10^{-18} J \right)}{\left(8.988 \times 10^{9} N \cdot m ^{2} / C ^{2}\right)(1.7476)\left(1.602 \times 10^{-19} C \right)^{2}} \\&=0.112\end{aligned}

Therefore $\rho=0.112 r_{0}=0.0316$ nm, which is in agreement with the value listed in Table 10.2. Indeed, this shows that the repulsive potential is very short range.

 Table 10.2 Properties of Salt Crystals with the NaCl Structure Nearest- Repulsive Range Dissociation Salt Neighbor Parameter Energy Crystal Separation (nm) $\rho$ (nm) (kJ/mol) LiF 0.214 0.029 1014 LiCl 0.257 0.033 832.6 LiBr 0.275 0.034 794.5 LiI 0.3 0.037 743.8 NaF 0.232 0.029 897.5 NaCl 0.282 0.032 764.4 NaBr 0.299 0.033 726.7 NaI 0.324 0.035 683.2 KF 0.267 0.03 794.5 KCl 0.315 0.033 694 KBr 0.33 0.034 663.5 KI 0.353 0.035 627.5 RbF 0.282 0.03 759.3 RbCl 0.329 0.032 666.8 RbBr 0.345 0.034 638.8 RbI 0.367 0.035 606.6 From C. Kittel, Introduction to Solid State Physics, 5th ed., New York: Wiley (1976), p. 92.