First slide

Electrons and holes in semiconductors

Question

In semiconductor, the concentration of electrons and holes are $8 \times 10^{18} / m^{3}$ and $5 \times 10^{18} / m^{3}$ , respectively. If the mobilities of electrons and hole are 2.3m²/ volt-sec and 0.01m²/volt-sec, respectively, then semiconductor is

Difficult

A

N-type and its resistivity is 0.34 ohm-metre
B

P-type and its resistivity is 0.034 ohm-metre
C

N-type and its resistivity is 0.034 ohm-metre
D

P- type and its resistivity is 3.40 ohm-metre

Solution

$\begin{array}{l} n_{e} = 8 \times 10^{18} / m^{3}, n_{h} = 5 \times 10^{18} / m^{3} \\ μ_{e} = 2.3 \frac{m^{2}}{volt - \sec}, μ_{h} = 0.01 \frac{m^{2}}{volt - \sec} \\ ∵ n_{e} > n_{h}, \underline{\underline{so}} semiconductor is N - type. \\ \underline{\underline{Also}} conductivity σ = \frac{1}{Resistivity (ρ)} = e (n_{e} μ_{e} + n_{h} μ_{h}) \\ \Rightarrow \frac{1}{ρ} = 1.6 \times 10^{- 19} [8 \times 10^{18} \times 2.3 + 5 \times 10^{18} \times 0.01] \\ \Rightarrow ρ = 0.34 Ω - m \end{array}$

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