Question

A strong current of trivalent gaseous boron passed through a germanium crystal decreases the density of the crystal due to part replacement of germanium by boron and due to interstitial vacancies created by missing Ge atoms. In one such experiment, one gram of germanium is taken and the boron atoms are found to be $150$ ppm by weight when the density of the Ge crystal decreases by $4\%$. Calculate the percentage of missing vacancies due to germanium which are filled up by boron atoms. Atomic weight of Ge $=72.6$ amu and $B=11$ amu.

• A. $2.4\%$
• B. $1.2\%$
• C. $6.6\%$
• D. None of the above

Answer 1

Answer: (A)

$2.4\%$

As given,
$1$ g Germanium $\equiv 150$ ppm of Boron $=1.5\times {10}^{-4}$ g of Boron
$4\%$ decrease in density of Ge occurs in one experiment due to replacement of Ge by B.

The weight becomes $0.96$ g (Ge + B).

Hence, $0.04$ g is missing.

$0.04$ missing means $(0.04-0.00015)$ g Ge missing $=\frac{(0.04-0.00015)}{72.61}$ moles
Ge missing $=$ $5.488\times {10}^{-4}$ moles
Boron replaced $=$ $1.5\times {10}^{-4}g=\frac{(1.5\times {10}^{-4})}{10}$ moles
Equivalently replaced $=\frac{3}{4}\times$ $\frac{1.5\times {10}^{-4}}{10}$ moles of Ge ( Valency of B is $3$ and Ge is $4$.)
$=0.1125\times {10}^{-4}$ moles of Ge
Percentage of missing vacancies filled by B atoms $=\frac{0.1125\times 100}{5.488}$ $=2.05\%$