Answer
$c_{Ge} $=11.75 wt%
Work Step by Step
Given:
Germanium (Ge) forms a substantial solid solution with silicon (Si)
$N_{Ge} = 2.43 \times 10^{21} atoms/cm^{3}$
$p_{Ge} = 5.32 g/cm^{3}$
$p_{Si} = 2.33 g/cm^{3}$
Required:
wt% of Ge
Solution:
Using Equation 4.19:
$N_{1} = N_{Ge} = 2.43 \times 10^{21} atoms/cm^{3}$
$p_{1} = p_{Ge} = 5.32 g/cm^{3}$
$p_{2}= p_{Si} = 2.33 g/cm^{3}$
$A_{1} = A_{Ge} = 72.64 g/mol$
$A_{2} = A_{Si} = 28.09 g/mol$
$c_{Ge} = \frac{100}{1 + \frac{N_{A}p_{Si}}{N_{Ge}A_{Ge}} - \frac{p_{Si}}{p_{Ge}}} = \frac{100}{1 + \frac{(6.022 \times 10^{23} atoms/mol)(2.33 g/cm^{3}) }{2.43 \times 10^{21} atoms/cm^{3}(72.64 g/mol)} - \frac{2.33 g/cm^{3}}{5.32 g/cm^{3}}} $=11.75 wt%
