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jeka57 [31]
3 years ago
15

Calculate the fraction of lattice sites that are schottky defects for cesium chloride at its melting temperature (645oc). assume

an energy for defect formation of 1.86 ev.
Chemistry
2 answers:
baherus [9]3 years ago
6 0
Answer : 7.87 X 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 eV 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}.
Brrunno [24]3 years ago
5 0

Answer:

\frac{Ns}{N}=8.398x10^{-6}

Explanation:

Hello,

The fraction lattice sites is computed via:

\frac{Ns}{N}=exp(-\frac{Q}{2kT} )

Whereas Q is the given energy in ev, T the temperature in Kelvins  and k the Boltzmann's constant in ev/K, in this manner, the resulting fraction is shown below:

\frac{Ns}{N} =exp(-\frac{1.85ev}{2*8.62x10^{-5}ev/K*(645+273.15)K})\\ \frac{Ns}{N}=8.398x10^{-6}

Best regards.

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