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# Find out the ratio of the mole-fraction of the Frenkel’s defect in $NaCl$ crystal at $4000\;K$ temperature. The amount of energy needed to form Frenkel’s defects and Schottky defects are respectively $2\;eV$ and $4\;eV$.

$Note$: $1eV=1.6×10^{–19} \;V, \; k=1.23×10^{–23}$

$\begin{array}{1 1} 25.71:1 \\ 51.42 :1 \\ 12.85:1 \\ 22.35 :1 \end{array}$

Let the Frenkel defect be $n$ in the ionic crystal of $N$ ions within interstitial space, where $n = (NN_i)^{1/2}\;$$e^{\large\frac{-E}{2kT}} Since for this crystal, N_i = 2N \rightarrow n = (2N^2)^{1/2}\;$$ e^{\large\frac{-E}{2kT}} = \sqrt 2 N e^{\large\frac{-E}{2kT}}$
$\big (\large\frac{n}{N} $$\big )$$_\text{Frenkel} = \sqrt 2 \times e ^ {\large\frac{-2 \times 1.6 \times 10^{-19}} {2 \times 1.23 \times 10^{-23} \times 4000}}$$= \sqrt 2 e^{-2.8985} It is given that the amount of energy needed to form Schottky defects (4\;eV) is twice that of what's needed for Frenkel defect (2\; eV). We can use what we just calculated and write the ratio as follows: \big (\large\frac{n}{N}$$\big )$$_\text{Schottky} = \sqrt 2 \times e ^ {\large\frac{-4 \times 1.6 \times 10^{-19}} {2 \times 1.23 \times 10^{-23} \times 4000}}$$ = \sqrt 2 e^{-5.797}$
Therefore the ratio of mole fractios of Frenkel and Schottky's defects is: $\large\frac{\sqrt 2 e^{-2.8985}} { \sqrt 2 e^{-5.797}}$$= 25.71$