First slide

Classification of semi-conductors

Question

In semiconductor the concentrations of electrons and holes are $8 x 10^{18} / m^{3}$ and $5 x 10^{18} / m^{3}$ , respectively. If the mobilities of electrons and holes are 2.3 $m^{2}$ /V-s and 0.01 $m^{2}$ /V-s, respectively, then resistivity of semiconductor is

Moderate

A

N-type and its resistivity is 0.34 $Ω$ m
B

P-type and its resistivity is 0.034 $Ω$ m
C

N-type and its resistivity is 0.034 $Ω$ m
D

P-type and its resistivity is 3.40 $Ω$ m

Solution

$\begin{array}{l} n_{e} = 8 \times 10^{18} / m^{3}, n_{h} = 5 \times 10^{18} / m^{3} \\ μ_{e} = 2 .3 \frac{m^{2}}{V - s}, μ_{h} = 0 .01 \frac{m^{2}}{V - s} \\ ∵ n_{e} > n_{h}, So semiconductor is N - type \\ Also, conductivity, σ = \frac{1}{Resistivity (ρ)} = e (n_{e} μ_{e} + n_{h} μ_{h}) \\ \Rightarrow \frac{1}{ρ} = 1 .6 \times 10^{- 19} [8 \times 10^{18} \times 2 .3 + 5 \times 10^{18} \times 0 .01] \\ \Rightarrow ρ = 0 .34 Ω - m \end{array}$

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