Question

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.

Answer 1

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 $10^{-6}$.

Answer 2

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.