Question

Pure Si at 500K has equal number of electron (nₑ) and hole
(nₕ) concentrations of 1.5 × 10¹⁶ m⁻³. Doping by indium
increases nₕ to 4.5 × 10²² m⁻³. The doped semiconductor is
of
(a) n–type with electron concentration
nₑ = 5 × 10²² m⁻³
(b) p–type with electron concentration
nₑ = 2.5 ×10¹⁰ m⁻³
(c) n–type with electron concentration
nₑ = 2.5 × 10²³ m⁻³
(d) p–type having electron concentration
nₑ = 5 × 10⁹ m⁻³

Answer 1

Explanation:

Prove the law of areas ie.. kepler's 2nd law

provide elaborate explanation, and also a list of the assumptions that are made, make sure to provide mathematical expressions.

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Answer 2

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.