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Semiconductor Electronics question

2011 · Q82
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Semiconductor Electronics question

2011 · Q82

NEETPhysicsSemiconductor ElectronicsMCQ+4 / −1
Pure Si at 500 K has equal number of electron (ne) and hole (nh) concentrations of 1.5 ×\times× 1016 m−-−3. Doping by indium increases nh to 4.5 ×\times× 1022 m−-−3. The doped semiconductor is of
  1. A
    p-type having electron concentration ne = 5 ×\times× 109 m−-−3
  2. B
    n-type with electron concentration ne = 5 ×\times× 1022 m−-−3
  3. C
    p-type with electron concentration ne = 2.5 ×\times× 1010 m−-−3
  4. D
    n-type with electron concentration ne = 2.5 ×\times× 1023 m−-−3
View written solutionFree

Correct answer: A

(ni)2 = ne × nh

(1.5 × 1016)2 = ne (4.5 × 1022)

So ne = 5 × 109

Now nh = 4.5 × 1022

⇒\Rightarrow⇒ nh >>\gt \gt>> ne

Hence, semiconductor is p-type

and ne = 5 × 109 m–3

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