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A semiconductor has equal electron and hole concentration of 6 xx 10^(6) m^(-3) . On doping with certain impurity, electron concentration increases to 9 xx 10^(12)m^(-3). (i) Identify the new semiconductorobtained after doping. (ii) Calculate the new hole concentration. |
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Answer» SOLUTION :Here `n_(i) = 6 XX 10^(8) m^(-3)` and on doping `n_(e) = 9 xx 10^(12) m^(-3)` . (i) As on doping electron concentration has increased, hence the doped SEMICONDUCTOR should behave as n-type semiconductor. (i) We KNOW that for a doped semiconductor `n_(e).n_(h) = n_(i)^(2)` `RARR n_(h) = (n_(i)^(2))/(n_(e)) = ((6 xx 10^(8))^(2))/(9 xx 10^(12)) = 4 xx 10^(4) m^(-3)` . |
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