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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 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 :
Option 1)
$2\Omega m\: \:$
Option 2)
$4\Omega m\: \:$
Option 3)
$0.4\Omega m\: \:$
Option 4)
$0.2\Omega m\: \:$

Answers (1)

best_answer

Electrical Conductivity (σ) -

$\sigma =e \left (n_{e}\mu _{e}+n_{h}\mu _{h} \right )$

- wherein

$n_{e}= electron \: density$

$n_{h}= hole \: density$

$\mu _{e}= mobility \: of \: electron$

$\mu _{h}= mobility \: of \: holes$

For intrinsic semiconductors (no impurities), the number of electrons will be equal to the number of holes.

So $n_{e}=n_{h}=n_{i}$

$\sigma =n_{i}e (\mu _{e}+ \mu _{h} )$

For N-type semiconductor electrons are majority carriers .

Conductivity $\sigma \approx ne \mu$

Resistivity $\rho = \frac{1}{\sigma} = \frac{1}{n_{e}e \mu}$

= $\frac{1}{10^{19} \times 1.6 \times \times 10^{-19} \times 1.6 }$

= 0.4 $\Omega m$

Option 1)

$2\Omega m\: \:$

Option 2)

$4\Omega m\: \:$

Option 3)
$0.4\Omega m\: \:$
Option 4)

$0.2\Omega m\: \:$

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