Calculate the fraction of lattice sites that are Schottky defects for cesium chloride at its melting temperature (645°C). Assume an energy for defect formation of 1.86 eV
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Answer:
Answer : 7.87 X 10^{-6}10−6 .
Explanation :
To find the fraction of schottky defects in the given lattice of CsCl,
we use the formula, \frac{N_{s} }{N}} = exp ( \frac{-Q_{s}}{2KT} )
on solving with the given values ,Q_{s}= 1.86 eVQs=1.86eV and T as 645 + 273 K and rest are the constants.
\frac{N_{s} }{N}} = exp ( \frac{-1.86 eV}{2 X (8.62 X 10^{-5}) X (645 +273)} )
we get the answer as 7.87 X 10^{-6}10−6
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