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Calculate the number of free electrons and holes (in m-3) in an intrinsic semiconductor that has electron and hole mobilities of 0.35 and 0.17 m2/V-s, respectively, and a conductivity of 1.7 (Ω-m)-1. Use scientific notation.

Respuesta :

Answer:

ni = 2.04e19

Explanation:

we know that in semiconductor like intrinsic, when electron leave the band, it leave a hole in valence band so we have

n = p = ni

from intrinsic carrier concentration

[tex]\sigma = n\left | e \right | \mu_e + n\left | e \right | \mu_h[/tex]

[tex] \sigma = ni\left | e \right | \mu_e  + ni\left | e \right | \mu_h[/tex]

[tex]\sigma = ni \left | e \right | ( \mu_e + \mu_h)[/tex]

1.7 = ni * 1.6*10^{-19} * (.35 + .17)

ni = 2.014 *10^{19} m^{-3}

ni = 2.04e19

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