The electrical conductivity of a semiconductor specimen at room temperature is (

Question

The electrical conductivity of a semiconductor specimen at room temperature is (2.9690x10^4) (?-m)-1. The hole concentration is known to be (2.4130x10^22) m-3. Given that the electron and hole mobilities are (4.90x10^-1) and (7.8000x10^-2) m2/V-s, respectively, calculate the electron concentration (in m^-3). As always, use scientific notation in the form X.YZ x 10^n.

The electrical conductivity of a semiconductor specimen at room temperature is (2.9690x10 4) (2-m The hole concentration is known to be (2.4130x10 22) m3. Given that the electron and hole mobilities are (4.90x10-1) and (7.8000x10-2) m2N-s, respectively, calculate the electron The hole concentration is known to be (24130x10-22) m concentration (in m-3) As always, use scientific notation in the form X.YZ x 10n. Note: Your answer is assumed to be reduced to the highest power possible. Your Answer: x10 Answer

Explanation / Answer

we know that the conductivity of semidconductot can be calculated as

conductivity = q(nXMn + pXMp)

Where q = charge = 1.6x10^-19 C

n = electron density = ?

Mn = mobility of electrons = 4.9X10^-1

p = holes density = 2.413 X 10^22

Mp = mobility of holes = 7.8000x10^-2

conductivity = 2.9690x10^4 = 1.6x10^-19 ( n X 4.9X10^-1 + 2.413 X 10^22 X 7.8000x10^-2)

2.9690x10^4 / 1.6x10^-19 = 0.49n + 18.8214 X 10^20

1.85563 X 10^23 = 0.49n + 18.8214 X 10^20

(18.5563 - 0.18821) X 10^22 / 0.49 = n

37.4858 X 10^22 = n

n = electron density = 37.49 X 10^22 m-3

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