Question

Phosphorous is added to make an n-type silicon semiconductor with an electrical conductivity of 1.75 (Ωm)-1 . Calculate the necessary number of charge carriers required

Answer 1

Answer:

The necessary number of electron charge carriers required is:

8.1019 × 10¹⁹ electrons/m³

Explanation:

The necessary number of charge carriers required can be determined from the resistivity. Given that, the phosphorus  make an n-type of silicon semiconductor;

Resistivity $\rho = \dfrac{1}{\sigma}$

$\rho = \dfrac{1}{q \mu _n n_n}$

where;

The number of electron on the charge carriers $n_n$ is unknown??

The charge of the electron q = $1.6 \times 10^{-19} \ C$

The electron mobility  $\mu_n$ = 0.135 m²/V.s

The electrical conductivity $\sigma$ = 1.75 (Ωm)⁻¹

Making  $n_n$  the subject from the above equation:

Then;

 $n_n = \dfrac{\sigma }{q \mu_n}$

$n_n = \dfrac{1.75 \ \Omega .m^{-1} }{1.6 \times 10^{-19} \times 0.135 \ m^2/V.s}$

$n_n =8.1019 \times 10^{19}$ electrons/m³

Thus; the necessary number of electron charge carriers required is:

8.1019 × 10¹⁹ electrons/m³