Question

A certain n-type semiconductor has a resistivity of 0.25 Ωm and the mobility of electrons in them is 0.3 m² per V-s. Calculate the number of donor atoms per m³ of this n-type semiconductor.
1. 2.35 × 1019 
2. 4.58 × 1019 
3. 8.34 × 10¹⁹ 
4. 6.76 × 1019 

Answer 1

Correct Answer - Option 3 : 8.34 × 10¹⁹ 

The correct answer is option 3) i.e. 8.34 × 1019 

CONCEPT:

• The conductivity of a semiconductor is given by the formula

Conductivity, σ = q(nμe + pμp)

Where q is the electrical charge, n is the density of electrons, p is the density of holes, μe is the mobility of electrons and μp is the mobility of holes.

CALCULATION:

• We know that conductivity (σ) is the reciprocal of resistivity (ρ).

Conductivity, \(σ = \frac{1}{ρ}\)

• Each donor atom that makes a semiconductor into n-type semiconductor leaves behind a free electron. So, the number of donor atoms per m3 is equal to the number of free electrons per m3 (n) produced. 
• In an n-type semiconductor, the concentration of electrons is more than that of holes.

⇒ n >> p

⇒ σ ≈ qnμe

Given that:

Charge of the electron, q = 1.6 × 10-19 C

Mobility of electrons, μe = 0.3 m2 per V-s

Resistivity, ρ = 0.25 Ωm

⇒ Conductivity, \(σ = \frac{1}{ρ} =\frac{1}{0.25} = 4\: \Omega^{-1} m^{-1}\)

⇒ σ = qnμ_e = (1.6 × 10-19) ×  n × 0.3

⇒ 4 = (1.6 × 10-19) ×  n × 0.3

⇒ n = 8.34 × 1019 per m3

Therefore, the number of donor atoms per m3 is 8.34 × 1019