Electronics and Communication Engineering (ECE) Exam > Electronics and Communication Engineering (ECE) Questions > A germanium sample at T = 300 K is doped wit...
Start Learning for Free
A germanium sample at T = 300 K is doped with donor atoms with concentration Nd = 6 x 1018/cm3. At thermal equilibrium, if the intrinsic concentration of the sample is ni = 2.4 x 1013/cm3, then the hole concentration/cm3 is ___ × 106 (Answer up to the nearest integer)
Correct answer is '96'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
A germanium sample at T = 300 K is doped with donor atoms with concen...
Given information:
- Temperature (T) = 300 K
- Donor concentration (Nd) = 6 x 10^18/cm^3
- Intrinsic concentration (ni) = 2.4 x 10^13/cm^3
Introduction:
In a semiconductor material like germanium, doping is done to modify its electrical properties. Doping involves adding impurity atoms to the crystal lattice, which introduces extra charge carriers. These charge carriers can be either electrons (from donor atoms) or holes (resulting from the absence of electrons).
Intrinsic concentration:
The intrinsic concentration (ni) represents the concentration of electrons and holes in an undoped (intrinsic) semiconductor at thermal equilibrium. In an intrinsic semiconductor, the number of electrons and holes is equal, and their concentrations are given by ni.
Doping:
Doping introduces impurity atoms into the semiconductor crystal, which can either donate or accept electrons. In this case, the germanium sample is doped with donor atoms, which means the impurity atoms donate extra electrons to the crystal lattice.
Hole concentration:
In a doped semiconductor, the concentration of holes (p) can be calculated using the equation:
p = ni^2 / Nd
where:
p = hole concentration
ni = intrinsic concentration
Nd = donor concentration
Calculating hole concentration:
Substituting the given values into the equation:
p = (2.4 x 10^13/cm^3)^2 / (6 x 10^18/cm^3)
= (5.76 x 10^26/cm^6) / (6 x 10^18/cm^3)
= 9.6 x 10^7/cm^3
Rounding to the nearest integer:
The calculated hole concentration is 9.6 x 10^7/cm^3. Rounding this to the nearest integer gives the answer as 96 x 10^6/cm^3, which is approximately 96 x 10^6/cm^3.
Final answer:
The hole concentration in the germanium sample is approximately 96 x 10^6/cm^3.
- Temperature (T) = 300 K
- Donor concentration (Nd) = 6 x 10^18/cm^3
- Intrinsic concentration (ni) = 2.4 x 10^13/cm^3
Introduction:
In a semiconductor material like germanium, doping is done to modify its electrical properties. Doping involves adding impurity atoms to the crystal lattice, which introduces extra charge carriers. These charge carriers can be either electrons (from donor atoms) or holes (resulting from the absence of electrons).
Intrinsic concentration:
The intrinsic concentration (ni) represents the concentration of electrons and holes in an undoped (intrinsic) semiconductor at thermal equilibrium. In an intrinsic semiconductor, the number of electrons and holes is equal, and their concentrations are given by ni.
Doping:
Doping introduces impurity atoms into the semiconductor crystal, which can either donate or accept electrons. In this case, the germanium sample is doped with donor atoms, which means the impurity atoms donate extra electrons to the crystal lattice.
Hole concentration:
In a doped semiconductor, the concentration of holes (p) can be calculated using the equation:
p = ni^2 / Nd
where:
p = hole concentration
ni = intrinsic concentration
Nd = donor concentration
Calculating hole concentration:
Substituting the given values into the equation:
p = (2.4 x 10^13/cm^3)^2 / (6 x 10^18/cm^3)
= (5.76 x 10^26/cm^6) / (6 x 10^18/cm^3)
= 9.6 x 10^7/cm^3
Rounding to the nearest integer:
The calculated hole concentration is 9.6 x 10^7/cm^3. Rounding this to the nearest integer gives the answer as 96 x 10^6/cm^3, which is approximately 96 x 10^6/cm^3.
Final answer:
The hole concentration in the germanium sample is approximately 96 x 10^6/cm^3.
Free Test
FREE
| Start Free Test |
Community Answer
A germanium sample at T = 300 K is doped with donor atoms with concen...
N0p0 = n2i
n0 ≈ ND
p0 = (2.42 x 1026)/(6 x 1018) = 0.96 × 108/cm3 = 96 × 106/cm3
Attention Electronics and Communication Engineering (ECE) Students!
To make sure you are not studying endlessly, EduRev has designed Electronics and Communication Engineering (ECE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Electronics and Communication Engineering (ECE).
|
Explore Courses for Electronics and Communication Engineering (ECE) exam
|
|
Similar Electronics and Communication Engineering (ECE) Doubts
Top Courses for Electronics and Communication Engineering (ECE)View all
A germanium sample at T = 300 K is doped with donor atoms with concentration Nd = 6 x 1018/cm3. At thermal equilibrium, if the intrinsic concentration of the sample is ni = 2.4 x 1013/cm3, then the hole concentration/cm3 is ___ × 106 (Answer up to the nearest integer)Correct answer is '96'. Can you explain this answer?
Question Description
A germanium sample at T = 300 K is doped with donor atoms with concentration Nd = 6 x 1018/cm3. At thermal equilibrium, if the intrinsic concentration of the sample is ni = 2.4 x 1013/cm3, then the hole concentration/cm3 is ___ × 106 (Answer up to the nearest integer)Correct answer is '96'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A germanium sample at T = 300 K is doped with donor atoms with concentration Nd = 6 x 1018/cm3. At thermal equilibrium, if the intrinsic concentration of the sample is ni = 2.4 x 1013/cm3, then the hole concentration/cm3 is ___ × 106 (Answer up to the nearest integer)Correct answer is '96'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A germanium sample at T = 300 K is doped with donor atoms with concentration Nd = 6 x 1018/cm3. At thermal equilibrium, if the intrinsic concentration of the sample is ni = 2.4 x 1013/cm3, then the hole concentration/cm3 is ___ × 106 (Answer up to the nearest integer)Correct answer is '96'. Can you explain this answer?.
A germanium sample at T = 300 K is doped with donor atoms with concentration Nd = 6 x 1018/cm3. At thermal equilibrium, if the intrinsic concentration of the sample is ni = 2.4 x 1013/cm3, then the hole concentration/cm3 is ___ × 106 (Answer up to the nearest integer)Correct answer is '96'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A germanium sample at T = 300 K is doped with donor atoms with concentration Nd = 6 x 1018/cm3. At thermal equilibrium, if the intrinsic concentration of the sample is ni = 2.4 x 1013/cm3, then the hole concentration/cm3 is ___ × 106 (Answer up to the nearest integer)Correct answer is '96'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A germanium sample at T = 300 K is doped with donor atoms with concentration Nd = 6 x 1018/cm3. At thermal equilibrium, if the intrinsic concentration of the sample is ni = 2.4 x 1013/cm3, then the hole concentration/cm3 is ___ × 106 (Answer up to the nearest integer)Correct answer is '96'. Can you explain this answer?.
Solutions for A germanium sample at T = 300 K is doped with donor atoms with concentration Nd = 6 x 1018/cm3. At thermal equilibrium, if the intrinsic concentration of the sample is ni = 2.4 x 1013/cm3, then the hole concentration/cm3 is ___ × 106 (Answer up to the nearest integer)Correct answer is '96'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electronics and Communication Engineering (ECE).
Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by signing up for free.
Here you can find the meaning of A germanium sample at T = 300 K is doped with donor atoms with concentration Nd = 6 x 1018/cm3. At thermal equilibrium, if the intrinsic concentration of the sample is ni = 2.4 x 1013/cm3, then the hole concentration/cm3 is ___ × 106 (Answer up to the nearest integer)Correct answer is '96'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A germanium sample at T = 300 K is doped with donor atoms with concentration Nd = 6 x 1018/cm3. At thermal equilibrium, if the intrinsic concentration of the sample is ni = 2.4 x 1013/cm3, then the hole concentration/cm3 is ___ × 106 (Answer up to the nearest integer)Correct answer is '96'. Can you explain this answer?, a detailed solution for A germanium sample at T = 300 K is doped with donor atoms with concentration Nd = 6 x 1018/cm3. At thermal equilibrium, if the intrinsic concentration of the sample is ni = 2.4 x 1013/cm3, then the hole concentration/cm3 is ___ × 106 (Answer up to the nearest integer)Correct answer is '96'. Can you explain this answer? has been provided alongside types of A germanium sample at T = 300 K is doped with donor atoms with concentration Nd = 6 x 1018/cm3. At thermal equilibrium, if the intrinsic concentration of the sample is ni = 2.4 x 1013/cm3, then the hole concentration/cm3 is ___ × 106 (Answer up to the nearest integer)Correct answer is '96'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A germanium sample at T = 300 K is doped with donor atoms with concentration Nd = 6 x 1018/cm3. At thermal equilibrium, if the intrinsic concentration of the sample is ni = 2.4 x 1013/cm3, then the hole concentration/cm3 is ___ × 106 (Answer up to the nearest integer)Correct answer is '96'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.
|
|
Explore Courses for Electronics and Communication Engineering (ECE) 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.
x
![]()
For Your Perfect Score in Electronics and Communication Engineering (ECE)
The Best you need at One Place
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup