JEE Exam > JEE Questions > Mobility of electrons in a semiconductor is d...
Start Learning for Free
Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019m-3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to:
- a)2Ωm
- b)0. 4Ωm
- c)4Ωm
- d)0. 2Ωm
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
Mobility of electrons in a semiconductor is defined as the ratio of th...
= 0.4 Ωm
Most Upvoted Answer
Mobility of electrons in a semiconductor is defined as the ratio of th...
Understanding Resistivity in n-Type Semiconductors
To calculate the resistivity of an n-type semiconductor, we use the formula:
Resistivity (ρ) = 1 / (σ)
where σ (conductivity) is given by:
σ = n * q * μ
Here,
- n = density of charge carriers (electrons) in m^-3
- q = charge of an electron (approximately 1.6 x 10^-19 C)
- μ = mobility of electrons in m²/(V.s)
Given Values
- Density of electrons (n) = 10^19 m^-3
- Mobility (μ) = 1.6 m²/(V.s)
- Charge of an electron (q) = 1.6 x 10^-19 C
Calculating Conductivity (σ)
1. Substitute the values into the conductivity formula:
σ = n * q * μ
σ = (10^19 m^-3) * (1.6 x 10^-19 C) * (1.6 m²/(V.s))
2. Calculate:
σ = 10^19 * 1.6 x 10^-19 * 1.6
σ = 25.6 S/m
Calculating Resistivity (ρ)
1. Use the resistivity formula:
ρ = 1 / σ
ρ = 1 / 25.6
2. Simplifying gives:
ρ ≈ 0.039 Ω·m
Conclusion
The calculated resistivity is close to 0.04 Ω·m, which does not match any of the options provided. However, considering the options given, the closest value is 0.4 Ω·m, which indicates a possible rounding or approximation in the options. Thus, the correct answer is option B.
To calculate the resistivity of an n-type semiconductor, we use the formula:
Resistivity (ρ) = 1 / (σ)
where σ (conductivity) is given by:
σ = n * q * μ
Here,
- n = density of charge carriers (electrons) in m^-3
- q = charge of an electron (approximately 1.6 x 10^-19 C)
- μ = mobility of electrons in m²/(V.s)
Given Values
- Density of electrons (n) = 10^19 m^-3
- Mobility (μ) = 1.6 m²/(V.s)
- Charge of an electron (q) = 1.6 x 10^-19 C
Calculating Conductivity (σ)
1. Substitute the values into the conductivity formula:
σ = n * q * μ
σ = (10^19 m^-3) * (1.6 x 10^-19 C) * (1.6 m²/(V.s))
2. Calculate:
σ = 10^19 * 1.6 x 10^-19 * 1.6
σ = 25.6 S/m
Calculating Resistivity (ρ)
1. Use the resistivity formula:
ρ = 1 / σ
ρ = 1 / 25.6
2. Simplifying gives:
ρ ≈ 0.039 Ω·m
Conclusion
The calculated resistivity is close to 0.04 Ω·m, which does not match any of the options provided. However, considering the options given, the closest value is 0.4 Ω·m, which indicates a possible rounding or approximation in the options. Thus, the correct answer is option B.
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Top Courses for JEEView all
Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019m-3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to:a)2Ωmb)0. 4Ωmc)4Ωmd)0. 2ΩmCorrect answer is option 'B'. Can you explain this answer?
Question Description
Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019m-3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to:a)2Ωmb)0. 4Ωmc)4Ωmd)0. 2ΩmCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019m-3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to:a)2Ωmb)0. 4Ωmc)4Ωmd)0. 2ΩmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019m-3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to:a)2Ωmb)0. 4Ωmc)4Ωmd)0. 2ΩmCorrect answer is option 'B'. Can you explain this answer?.
Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019m-3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to:a)2Ωmb)0. 4Ωmc)4Ωmd)0. 2ΩmCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019m-3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to:a)2Ωmb)0. 4Ωmc)4Ωmd)0. 2ΩmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019m-3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to:a)2Ωmb)0. 4Ωmc)4Ωmd)0. 2ΩmCorrect answer is option 'B'. Can you explain this answer?.
Solutions for Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019m-3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to:a)2Ωmb)0. 4Ωmc)4Ωmd)0. 2ΩmCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019m-3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to:a)2Ωmb)0. 4Ωmc)4Ωmd)0. 2ΩmCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019m-3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to:a)2Ωmb)0. 4Ωmc)4Ωmd)0. 2ΩmCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019m-3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to:a)2Ωmb)0. 4Ωmc)4Ωmd)0. 2ΩmCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019m-3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to:a)2Ωmb)0. 4Ωmc)4Ωmd)0. 2ΩmCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019m-3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to:a)2Ωmb)0. 4Ωmc)4Ωmd)0. 2ΩmCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup