Question

if a semiconductor, in which intrinsic carrier concentration is 2.5*10^13 /cm^3 is doped with donor impurities of concentration 5*10^15 /cm^3 then the concentration of holes will be (assume that all donor impurities are ionized).

Answer 1

ni is the intrinsic carrier concentration, i.e., the number of electrons in the conduction band. (and also the number of holes in the valence band) per unit volume in a semiconductor.

Answer 2

Dear Students,

● Answer -

p = 1.25×10^11 /cm^3

◆ Explanation -

# Given -

intrinsic carrier concentration (ni) = 2.5×10^13 /cm^3

donor impurity concentration (nd) = 5×10^15 /cm^3

# Solution -

Hole concentration for the semiconductor is given by -

p = ni^2 / nd

p = (2.5×10^13)^2 / (5×10^15)

p = 1.25×10^11 /cm^3

Hence, concentration of holes in semiconductor is 1.25×10^11 /cm^3 .

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