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Q.

Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is $10^{19} m^{- 3}$ and their mobility is $1.6 m^{2} (V . s)$ then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close

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a

$0.2 Ω m$

b

$2 Ω m$

c

$0.4 Ω m$

d

$4 Ω m$

answer is C.

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Detailed Solution

As we know, current density, $j = σ E = n e v_{d} \Rightarrow σ = n e \frac{v_{d}}{E} = n e μ$

$\frac{1}{σ} = ρ = \frac{1}{n_{e} e μ_{e}} =$ Resistivity $= \frac{1}{10^{19} \times 1.6 \times 10^{- 19} \times 19 \times 1.6}$ or $ρ = 0.4 Ω m$

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Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019m−3 and their mobility is 1.6m2V.s then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close