Q: Pure Si at 500 K has equal number of electron (ne ) and hole (nh ) concentration of 1.5 × 1016 m-3. Doping by indium increases nh to 4.5 × 1022 m-3. The doped semiconductor is of
(a) n-type with electron concentration ne = 2.5 × 1023 m-3
(b) p-type having electron concentration ne = 5 × 109 m-3
(c) n-type with electron concentration ne = 5 × 1022 m-3
(d) p-type with electron concentration ne = 2.5 × 1010 m-3
Ans: (b)
Sol: ni2 = ne nh
(1.5 × 1016 )2 = n_e (4.5 × 1022 )
ne = 0.5 × 1010 =5 × 109
nh = 4.5 × 1022
nh > ne (p-type)