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Home  >>  AIMS  >>  Class12  >>  Chemistry  >>  Solid State
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Find out the ratio of the mole-fraction of the Frenkel’s defect in NaCl crystal at 4000K temperature. The amount of energy needed to form Frenkel’s defects and Schottky defects are respectively 2 eV and 4 eV.Given that $1ev = 1.6 \times 10^{–19}V$ and $k = 1.23 \times 10^{–23}$

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Let the Frenkel defect be n in the ionic crystal of N ions within interstitial space.
Since, n = $\sqrt{N\times Ni}e^{-\Large\frac{E}{2kT}}$
Since for this crystal
Ni = 2N
n = $\sqrt 2\times N\times e^{-{\Large\frac{E}{kT}}}$
$\large\frac{n}{N}=$$\sqrt 2\times e^{-2.8985}$
$x_1 = \large\frac{n}{N} $$= 7.79 \times 10^{–2}$
Now for Schottky defect
Let n defects be present in N-ions of the crystal
Mole fraction =$\large\frac{n}{N}$
$n=e^{-5.797}$
$X_2=3.03\times 10^{-3}$
So the ratio of mole-fraction of Frenkel’s and Schottky defects are
$\large\frac{X_1}{X_2}=\frac{7.79\times 10^{-2}}{3.03\times 10^{-3}}=\frac{25.71}{1}$
Hence (A) is the correct answer.
answered Jun 3, 2014 by sreemathi.v
 
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