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

Answers

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 [tex]\rho = \dfrac{1}{\sigma}[/tex]

[tex]\rho = \dfrac{1}{q \mu _n n_n}[/tex]

where;

The number of electron on the charge carriers [tex]n_n[/tex] is unknown??

The charge of the electron q = [tex]1.6 \times 10^{-19} \ C[/tex]

The electron mobility  [tex]\mu_n[/tex] = 0.135 m²/V.s

The electrical conductivity [tex]\sigma[/tex] = 1.75 (Ωm)⁻¹

Making  [tex]n_n[/tex]  the subject from the above equation:

Then;

 [tex]n_n = \dfrac{\sigma }{q \mu_n}[/tex]

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

[tex]n_n =8.1019 \times 10^{19}[/tex] electrons/m³

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

8.1019 × 10¹⁹ electrons/m³


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