Answer to Question #2151 in Solid State Physics for JOEL LOBO

Question

The fraction of vacancy sites in a metal is 1X10-10 at 500C. What will be the
fraction of vacancy sites at double the temperature?

Answer 1

The fraction of vacant sites is given by the following expression:

$\frac{N_{\nu}}{N} = e^{\left ( -\frac{Q_{\nu}}{kT} \right )}$

At 500C we can find the activation energy :

$Q_{\nu} = \ln {\frac{N_{\nu}}{N}}kT = -\ln{1 \times 10^{-10}}\cdot 8.31\cdot(500+273) = 147909.5 \ J/mol$

At double temperature 1000C = 1000+273 = 1273K the fraction fo vacant sites is:

$\frac{N_{\nu}}{N} = e^{\left ( -\frac{147909.5}{8.31\cdot1273} \right )} = 0.85 \times 10^{-8}$