# Interatomic distance and lattice constant of NaCl

1) The interaction potential energy between two ions of an ionic crystal can be approximated by the relation:

$E_p =A*exp[-(r/\rho_0)] -\frac{\alpha*e^2}{4\pi\epsilon_0}*\frac{1}{r}$

where $\rho_0$ is the parameter for the repulsive energy, $\alpha$ is the parameter of the electrostatic attraction, $\epsilon_0$ is the vacuum permittivity and $A$ is a constant. For the NaCl ionic crystal one has $\rho_0=0.321*10^{-10} m$, $\alpha =1.747$ and the distance between two ions at equilibrium is  $r_0 =2.82*10^{-10}m$. Find the following:

a) The value of A constant

b) The total energy of interaction for the crystal with mass $m=1 Kmol$.

c) The distance $r_l$ where the electrostatic attraction energy is equal to the repulsive energy between ions

(The electron charge $e$ and the vacuum permittivity $\epsilon_0$ are known).

a) We find the value of the constant $A$ from the condition of minimum interaction potential energy:

$\frac {d E_p}{d r} = 0$
or equivalent

$-\frac {A}{\rho_0}*exp[-(r_0/\rho_0)] +\frac{\alpha*e^2}{4\pi\epsilon_0}*\frac{1}{r_0^2} =0$

Therefore:

$A =\frac{\alpha*e^2}{4\pi\epsilon_0}*\frac{\rho_0}{r_0^2}*exp(r_0/\rho_0)$

Doing the math one gets

$A =6.63*10^3 eV$

b) In a mass $m=1 Kmol$ there are $N_A$ atoms, therefore the total interaction energy is simply:

$E_{tot} =-N_A\left (A*exp[-(r_0/\rho_0)] -\frac{\alpha*e^2}{4\pi\epsilon_0}*\frac{1}{r_0} \right ) =7.608*10^8 J$

c) The condition of equality between repulsive energy and attractive (electrostatic) energy is written as:

$A*exp(-r/\rho_0) =\frac {\alpha*e^2}{4\pi\epsilon_0}*\frac{1}{r}$

By taking the logarithm and rearranging one has

$r =\rho_0*ln(r) -\rho_0*ln \frac{\alpha*e^2}{4\pi\epsilon0*A}$

This is a transcendental equation and as such can be solved only using graphical methods. The value obtained for its solution is

$r_l =2.01*10^{-10} m$

2) By knowing the molecular mass $M =58.44 kg/Kmol$ and the mass density $\rho=2.165 kg/m^3$ for NaCl, find the crystal lattice constant $\alpha_0$.

(The Avogadro number $N_A$ is being given).

The mass of a NaCl molecule is $m =M/N_A$. In the unit cell of NaCl enters 4 molecules. Thus:

$\rho =\frac {4m}{a_0^3} =\frac {4M}{N_A*a_0^3}$

from above one obtains the lattice constant $a_0$:

$a_0 = \left (\frac{4M}{\rho N_A} \right )^{1/3} =5.63*10^{-10} m$