Question

The room temperature electrical conductivity of a semiconductor specimen is 2.8 x 104 (?-m)-1. The electron concentration is known to be 2.9 x 1022 m-3. Given that the electron and hole mobilities are 0.14 and 0.023 m2/V-s, respectively, calculate the hole concentration (in m-3)

Answer 1

7.43 × 10²⁴ m⁻³

Step-by-step explanation:

Data provided in the question:

Conductivity of a semiconductor specimen, σ = 2.8 × 10⁴ (Ω-m)⁻¹

Electron concentration, n = 2.9 × 10²² m⁻³

Electron mobility,
`\mu_n` = 0.14 m²/V-s

Hole mobility,
`\mu_p`= 0.023 m²/V-s

Now,

σ =
`nq\mu_n+pq\mu_p`

or

σ =
`q(n\mu_n+p\mu_p)`

here,

q is the charge on electron = 1.6 × 10⁻¹⁹ C

p is the hole density

thus,

2.8 × 10⁴ = 1.6 × 10⁻¹⁹( 2.9 × 10²² × 0.14 + p × 0.023 )

or

1.75 × 10²³ = 0.406 × 10²² + 0.023p

or

17.094 × 10²² = 0.023p

or

p = 743.217 × 10²²

or

p = 7.43 × 10²⁴ m⁻³