Question

The resistivity of a pure semiconductor is
0.5 ohm-m. If the electron and hole mobility be 0.39
m²/V-s and 0.19 m^2/V-s respectively then calculate
the intrinsic carrier concentration.
(1) 2.16 x 10^19/m^3
(2) 4.32 x 10^19/m^3
(3) 10^20/m^3
(4) None of these​

Answer 1

Given :

- Electron and hole mobility is 0.39

m²/V-s and 0.19 m^2/V-s respectively

- Resistivity of a pure semiconductor is 0.5 ohm-m

To find :

• The intrinsic carrier concentration.

Solution :

• We know that, resistivity is reciprocal of conductivity

R = 1 / σ

And,

σ = n×e [ ue + uh ]

Where n = intrinsic concentration

ue = mobility of electrons

uh= mobility of holes

• R = 1 / ne[ ue + uh ]

n = 1 / R×e[ ue + uh ]

= 1 / 0.5×1.6×10^-19×[ 0.39+ 0.19 ]

= 1 / 0.08×10^-19×0.58

= 10^19 / 0.08×0.58

= 2.16 × 10^19 /m^3

Hence, the intrinsic carrier concentration is 2.16 × 10^19 /m^3

Answer 2

Given :

Resistivity of pure semiconductor :

$0.5 \: \Omega m$

Electron mobility :

$\mu _{e} \: = 0.39 \: {m}^{2} {V}^{ - 1} {s}^{ - 1}$

Hole mobility :

$\mu _{h} \: = 0.19 \: {m}^{2} {V}^{ - 1} {s}^{ - 1}$

To find :

Intrinsic carrier concentration

Solution:

We have given the values of resistivity, electron mobility and hole mobility of pure semiconductor.

The formula of conductivity of pure semiconductor :

$\sigma = e(( n_i \times \mu_e) + (n_i \times \mu_h )) \:$

we know that

Resistivity is inverse of conductivity,so

putting its value as,

$\bf \: \frac{1}{ \rho} = \sigma = e( n_i \times \mu_e + n_i \times \mu_h ) \: \:$

So,

Intrinsic carrier concentration :

$\bf n_i \: = \: \frac{1}{\rho \times e( \mu_e + \mu_h )}$

Now putting values of all the given values lf resistivity, hole concentration and electron concentration in the formula of intrinsic carrier concentration,

we get:

$\bf n_i \: = \: \frac{1}{0.5 \times (1.6 \times {10}^{ - 19} )( 0.39 + 0.19 )}$

so,

$\bf n_i \: = \: \frac{1}{0.5 \times (1.6 \times {10}^{ - 19} )( 0.58)}$

$\bf n_i \: = \: \frac{ {10}^{19} }{0.464}$

So,

Intrinsic carrier concentration :

$\bf n_i \: = \: 2.16 \times {10}^{19} \: \: {m}^{ - 3}$

Correct option ( 1 )