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 1019 m-3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to:

Question

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 1019 m-3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to: (A)  2 Ω m  (B)  0.4 Ω m  (C)  4 Ω m  (D)  0.2 Ω m

Tardigrade Admin · Accepted Answer

j = σ E = nevd σ =ne (vd/E) =ne μ (1/σ) = ρ = (1/neeμe) = (1/1019 ×1.6 ×10-19 ×1.6) = 0.4 Ω m

$2Ω m$

$0.4Ω m$

$4Ω m$

$0.2Ω m$

$j = σ E = n e v_{d}$
$σ = n e \frac{v _{d}}{E}$
$= n e μ$
$\frac{1}{σ} = ρ = \frac{1}{n _{e} e μ _{e}} = \frac{1}{1 0 ^{19} \times 1.6 \times 1 0 ^{- 19} \times 1.6}$
$= 0.4 Ω m$

2Ωm

0.4Ωm

4Ωm

0.2Ωm

j=σE=nevd​ σ=neEvd​​ =neμ σ1​=ρ=ne​eμe​1​=1019×1.6×10−19×1.61​ =0.4Ωm

$2Ω m$

$0.4Ω m$

$4Ω m$

$0.2Ω m$

$j = σ E = n e v_{d}$
$σ = n e \frac{v _{d}}{E}$
$= n e μ$
$\frac{1}{σ} = ρ = \frac{1}{n _{e} e μ _{e}} = \frac{1}{1 0 ^{19} \times 1.6 \times 1 0 ^{- 19} \times 1.6}$
$= 0.4 Ω m$