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The Conductivity of a p-type Germanium at 300 k is 100(ohm-cm)-1. The intrinsic concentration of germanium at 300 k is 2.5 x 1013/cm3 & its mobility is given as 1800 cm2/v-s & atoms/cm3 = 4.4 x 1022. Which of the following statement is/are correct, If a voltage source of 0.010V is applied across the semiconductor & the dimension of the p-type germanium piece is (1 x 1) m:-
  • a)
    The Concentration of holes in Ge is 3.468  x 1017/cm3.
  • b)
    Concentration of electrons will be 1.802 x 10-4/cm3.
  • c)
    Current through the Ge semiconductor is 100 amp.
  • d)
    Resistance of the Ge is 10-2Ω.
Correct answer is option 'A,C'. Can you explain this answer?
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The Conductivity of a p-type Germanium at 300 k is 100(ohm-cm)-1. The ...

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The Conductivity of a p-type Germanium at 300 k is 100(ohm-cm)-1. The ...
To calculate the conductivity of a p-type germanium at 300 K, we can use the formula:

conductivity = charge carrier mobility * charge carrier concentration

Since the germanium is p-type, the charge carriers are holes. The mobility of holes in germanium is typically around 400 cm2/Vs.

Let's assume the charge carrier concentration is equal to the intrinsic concentration of germanium at 300 K, which is given as 2.5 x 10^13/cm3.

Substituting the values into the formula:

conductivity = 400 cm2/Vs * 2.5 x 10^13/cm3

The units of cm^2/Vs cancel out with the units of cm^3, leaving us with the units of conductivity (ohm-cm)^-1.

conductivity = 100(ohm-cm)^-1

Therefore, the conductivity of the p-type germanium at 300 K is 100 (ohm-cm)^-1.
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The Conductivity of a p-type Germanium at 300 k is 100(ohm-cm)-1. The intrinsic concentration of germanium at 300 k is 2.5 x 1013/cm3 & its mobility is given as 1800 cm2/v-s & atoms/cm3 = 4.4 x 1022. Which of the following statement is/are correct, If a voltage source of 0.010V is applied across the semiconductor & the dimension of the p-type germanium piece is (1 x 1) m:-a)The Concentration of holes in Ge is 3.468 x 1017/cm3.b)Concentration of electrons will be 1.802 x 10-4/cm3.c)Current through the Ge semiconductor is 100 amp.d)Resistance of the Ge is 10-2Ω.Correct answer is option 'A,C'. Can you explain this answer?
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The Conductivity of a p-type Germanium at 300 k is 100(ohm-cm)-1. The intrinsic concentration of germanium at 300 k is 2.5 x 1013/cm3 & its mobility is given as 1800 cm2/v-s & atoms/cm3 = 4.4 x 1022. Which of the following statement is/are correct, If a voltage source of 0.010V is applied across the semiconductor & the dimension of the p-type germanium piece is (1 x 1) m:-a)The Concentration of holes in Ge is 3.468 x 1017/cm3.b)Concentration of electrons will be 1.802 x 10-4/cm3.c)Current through the Ge semiconductor is 100 amp.d)Resistance of the Ge is 10-2Ω.Correct answer is option 'A,C'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about The Conductivity of a p-type Germanium at 300 k is 100(ohm-cm)-1. The intrinsic concentration of germanium at 300 k is 2.5 x 1013/cm3 & its mobility is given as 1800 cm2/v-s & atoms/cm3 = 4.4 x 1022. Which of the following statement is/are correct, If a voltage source of 0.010V is applied across the semiconductor & the dimension of the p-type germanium piece is (1 x 1) m:-a)The Concentration of holes in Ge is 3.468 x 1017/cm3.b)Concentration of electrons will be 1.802 x 10-4/cm3.c)Current through the Ge semiconductor is 100 amp.d)Resistance of the Ge is 10-2Ω.Correct answer is option 'A,C'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The Conductivity of a p-type Germanium at 300 k is 100(ohm-cm)-1. The intrinsic concentration of germanium at 300 k is 2.5 x 1013/cm3 & its mobility is given as 1800 cm2/v-s & atoms/cm3 = 4.4 x 1022. Which of the following statement is/are correct, If a voltage source of 0.010V is applied across the semiconductor & the dimension of the p-type germanium piece is (1 x 1) m:-a)The Concentration of holes in Ge is 3.468 x 1017/cm3.b)Concentration of electrons will be 1.802 x 10-4/cm3.c)Current through the Ge semiconductor is 100 amp.d)Resistance of the Ge is 10-2Ω.Correct answer is option 'A,C'. Can you explain this answer?.
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