Question

An n-type semiconductor is known to have an electron concentration of 3 ´ 1018 m-3. If the electron drift velocity is 100 m/s in an electric field of 500 V/m. Calculate the mobility and conductivity of this material

Answer 1

Answer:

Electron Mobility of the material= 0.20 m2/V-s

Conductivity of the material= 0.096 (Ω-m)-1

Explanation:

Electron mobility = Vd/E

= 100/500

= 0.20 m2/V-s

Conductivity = σ = n |e|μ

= 3 x 1018 x 1.602 x 10 -19 x 0.20

= 0.096 (Ω-m)-1

Answer 2

1. The electron mobility of this n-type semiconductor is equal to 0.2 $m^2/Vs^{-1}$  

2. The conductivity of this n-type semiconductor is equal to 0.096 $(\Omega-m)^{-1}$.

Given the following data:

• Electron concentration = $3 \times 10^{18}\;m^3$

• Drift velocity = 100 m/s

• Electric field = 500 V/m

Scientific data:

• Charge of an electron = $1.6 \times 10^{-19}\;C$

To calculate the mobility and conductivity of this n-type semiconductor:

For the mobility:

Mathematically, electron mobility is given by this formula:

$\mu = \frac{V_d}{E}$

Where:

• $V_d$ is the electron drift velocity.

• E is the electric field.

Substituting the given parameters into the formula, we have;

$\mu = \frac{100}{500}$

Electron mobility = 0.2 $m^2/Vs^{-1}$  

For the conductivity:

Mathematically, conductivity is given by this formula:

$\sigma = qn_e \mu$

Where:

• q is the electron charge.

• $\mu$ is the electron mobility.

• $n_e$ is the electron concentration.

Substituting the given parameters into the formula, we have;

$\sigma = 1.6 \times 10^{-19}\times 3 \times 10^{18}\times 0.2$

Conductivity = 0.096 $(\Omega-m)^{-1}$

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