askiitian.expert- chandra sekhar
Last Activity: 15 Years ago
Hi Simmi singh,
1 g Germanium------- 150 ppm= 1.5 * 10^-4 g Boron
4% decrease---- now weight is 0.96 g(Ge + B)----0.04 g missing
0.04 missing-------- (0.04-0.00015)g Ge missing------(0.04-0.00015)/72.61 moles Ge missing=5.488*10^-4 moles
Boron replaced= 1.5*10^-4 g = (1.5*10^-4)/10 moles
equvalently replaced = 3/4((1.5*10^-4)/10) moles of Ge ( Valency of B is 3 and Ge is 4)
= 0.1125*10^-4 moles of Ge
% of Missing vacancies filled by B atoms = 0.1125*100/5.488 = 2.05 %