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

The necessary number of electron charge carriers required is:

8.1019 × 10¹⁹ electrons/m³

Step-by-step 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 = (1)/(\sigma)`

`\rho = (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 * 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 = (\sigma )/(q \mu_n)`

`n_n = (1.75 \ \Omega .m^(-1) )/(1.6 * 10^(-19) * 0.135 \ m^2/V.s)`

`n_n =8.1019 * 10^(19)` electrons/m³

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

8.1019 × 10¹⁹ electrons/m³