Saved Bookmarks
| 1. |
Estimate the porportion of boron impurity whichwill increase the conductivity of a pure silicon sample by a factor of 100.Assume that each boron atom creates a hole and the concentration of holes in pure silicon at the same tempareture is7xx10^15holes per cubic metre. Density of silicon is 5xx10^28atoms per cubic metre. |
|
Answer» Solution :Total number of charge carriers initially `= 2xx7xx10^15` `= 14xx10^15/(Cubic meter)` finally the total number of charge carriers `= 14xx10^(17)/m^(3)` We know the product of the concentrations of holes and conduction electrons remains almost the same. Let 'x' be the number of holes. So `(7XX10^(15))xx(7xx 10^(15))` `= x xx(14^(17)-x)` `rArr 14X xx10^17-x^2 = 49xx10^30` `rArr x^2-14x xx10^17-49xx10^30 = 0` `rArr x = (14xx10^17+-(14)^2xxsqrt(10^34+4xx49xx10^30))/(2)` ` = 10^17+-sqrt(10^34+4xx49xx10^30))/(2)` ` = (28.0007)/(2)xx10^17 = 14.00035xx10^17` = increase in number of boles or the number of atoms of BORON added Now, `1386.035xx10^15 ATOM of `Si` in `1 m^3`. `1` atom of Boron is added per `(5xx10^28)/(1386.035xx10^15)` `3.607xx10^15xx10^13` `3.607xx10^10` |
|