⇦ | ![]() | ⇨ |
Pure Si at 500 K has equal number of electron (nₑ) and hole (nₕ) concentrations of 1.5 x 10¹⁶ m⁻³. Doping by indium increases nₕ to 4.5 x 10²² m⁻³. The doped semiconductor is of
Options
(a) n-type with electron concentration nₑ = 5 x 10²² m⁻³
(b) p-type with electron concentration nₑ = 2.5 x 10¹⁰ m⁻³
(c) n-type with electron concentration nₑ = 2.5 x 10²³ m⁻³
(d) p-type with electron concentration nₑ = 5 x 10⁹ m⁻³
Correct Answer:
p-type with electron concentration nₑ = 5 x 10⁹ m⁻³
Explanation:
nᵢ² = nₑ nₕ
(1.5 x 10¹⁶)² = nₑ (4.5 x 10²²)
⇒ nₑ = 0.5 x 10¹⁰ or nₑ = 5 x 10⁹
Given nₕ = 4.5 x 10²² ⇒ nₕ >> nₑ
semiconductor is p-type and
nₑ = 5 x 10⁹ m⁻³.
Related Questions:
- Masses of three wires of copper are in the ratio of 1:3:5 and their lengths
- The moment of inertia of a uniform circular disc of radius R and mass M about an axis
- A cockroach is moving with a velocity v in anticlockwise direction on the rim
- If energy(E), velocity(V) and time(T) are chosen as fundamental quantities,
- The intensity of magnetisation of a bar magnet is 5×10⁴ Am⁻¹. The magnetic length
Topics: Electronic Devices
(124)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply