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Mobilities of electrons and holes in a sample of intrinsic germanium at 300K are 0.36 m2V-1 S-1 and 0.17m2V-1S-1 respectively. If the conductivity of the specimen is 2.12 Ω-1 m-1, then the forbidden energy gap is _____ eV.
Correct answer is '0.72'. Can you explain this answer?
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Verified Answer
Mobilities of electrons and holes in a sample of intrinsic germanium a...
0.720
Conductivity of an intrinsic semiconductor is given by
or
= 2.5 x 1019 / m3
But
for C = 4.8 x 1021
and kBT = 1.38 x 10-23 x 300J =
We have
or
and
Eg = 2 x 13.8 x 0.025875 = 0.720 eV.
Conductivity of an intrinsic semiconductor is given by
or
= 2.5 x 1019 / m3
But
for C = 4.8 x 1021
and kBT = 1.38 x 10-23 x 300J =
We have
or
and
Eg = 2 x 13.8 x 0.025875 = 0.720 eV.
Most Upvoted Answer
Mobilities of electrons and holes in a sample of intrinsic germanium a...
The conductivity (σ) of a material is given by the equation:
σ = q(nμn + pμp)
where q is the charge of an electron (1.6 x 10^-19 C), n is the electron concentration, μn is the electron mobility, p is the hole concentration, and μp is the hole mobility.
Intrinsic germanium has equal electron and hole concentrations, so we can assume n = p = n0 (where n0 is the intrinsic carrier concentration).
Given that the conductivity (σ) is 2.12 Ω^-1m^-1, the electron mobility (μn) is 0.36 m^2V^-1S^-1, and the hole mobility (μp) is 0.17 m^2V^-1S^-1, we can rearrange the equation to solve for n0:
σ = q(n0μn + n0μp)
2.12 = (1.6 x 10^-19 C)(n0)(0.36 + 0.17)
2.12 = (1.6 x 10^-19)(n0)(0.53)
n0 = 2.12 / (1.6 x 10^-19)(0.53)
n0 ≈ 7.96 x 10^19 m^-3
Therefore, the intrinsic carrier concentration (n0) of germanium at 300K is approximately 7.96 x 10^19 m^-3.
σ = q(nμn + pμp)
where q is the charge of an electron (1.6 x 10^-19 C), n is the electron concentration, μn is the electron mobility, p is the hole concentration, and μp is the hole mobility.
Intrinsic germanium has equal electron and hole concentrations, so we can assume n = p = n0 (where n0 is the intrinsic carrier concentration).
Given that the conductivity (σ) is 2.12 Ω^-1m^-1, the electron mobility (μn) is 0.36 m^2V^-1S^-1, and the hole mobility (μp) is 0.17 m^2V^-1S^-1, we can rearrange the equation to solve for n0:
σ = q(n0μn + n0μp)
2.12 = (1.6 x 10^-19 C)(n0)(0.36 + 0.17)
2.12 = (1.6 x 10^-19)(n0)(0.53)
n0 = 2.12 / (1.6 x 10^-19)(0.53)
n0 ≈ 7.96 x 10^19 m^-3
Therefore, the intrinsic carrier concentration (n0) of germanium at 300K is approximately 7.96 x 10^19 m^-3.
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Mobilities of electrons and holes in a sample of intrinsic germanium at 300K are 0.36 m2V-1S-1 and 0.17m2V-1S-1 respectively. If the conductivity of the specimen is 2.12 Ω-1 m-1, then the forbidden energy gap is _____ eV.Correct answer is '0.72'. Can you explain this answer?
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Mobilities of electrons and holes in a sample of intrinsic germanium at 300K are 0.36 m2V-1S-1 and 0.17m2V-1S-1 respectively. If the conductivity of the specimen is 2.12 Ω-1 m-1, then the forbidden energy gap is _____ eV.Correct answer is '0.72'. 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 Mobilities of electrons and holes in a sample of intrinsic germanium at 300K are 0.36 m2V-1S-1 and 0.17m2V-1S-1 respectively. If the conductivity of the specimen is 2.12 Ω-1 m-1, then the forbidden energy gap is _____ eV.Correct answer is '0.72'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Mobilities of electrons and holes in a sample of intrinsic germanium at 300K are 0.36 m2V-1S-1 and 0.17m2V-1S-1 respectively. If the conductivity of the specimen is 2.12 Ω-1 m-1, then the forbidden energy gap is _____ eV.Correct answer is '0.72'. Can you explain this answer?.
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