1.

PureSi at 500 Khas equal number of electron (n_(e))andhole (n_h) concentraionsof1.5 xx 10^(16) m^(-3). Dopingby indium incresesn_(h)" to "4.5 xx 10^(22) m^(-3). The doped semiconductor is of

Answer»

n-type with electron concentration ` n_(E)= 5 xx 10^(22) m^(-3)`
p-type with electron - concentration` n_(e) = 2.5 xx 10^(10) m^(-3)`
n- type3 with electron concentration `n_(e) = 2.5 xx 10^(23) m ^(-3)`
p-type having electron concentration `n_(e) = 5 xx 10^(9) m^(-3)`

Solution :As `n_(i)^(2) = n_(e) xx n_(h)`
` (1.5 xx 10^(16)m^(-3))^(2) = n_(e) xx (4.5 xx10^(22) m^(-3)) `
` thereforen_(e) = ((1.5 xx 10^(16) m^(-3))^(2))/ (4.5 xx 10^(22) m^(-3)) = 5XX 10^(9) m^(-3)`
` As n_(n) lt ltn_(e)` So semiconductoris p-typeand ` n_(e) = 5 xx 10^(9) m^(-3) `


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