To solve this problem, we are tasked with calculating two quantities:

1. Electron mobility $\mu_e$
2. Conductivity $\sigma$

Given dataTo solve this problem, we are tasked with calculating two quantities:

1. Electron mobility $\mu_e$
2. Conductivity $\sigma$

Given

• Electron concentration $n = 6.2 \times 10^8 \, \text{m}^{-3}$
• Electric field $E = 800 \, \text{V/m}$
• Drift velocity $v_d = 145 \, \frac{\text{m}}{\text{s}}$
• Elementary charge $e = 1.6 \times 10^{-19} \, \text{C}$

Step 1: Calculate Electron Mobility $\mu_e$

The formula for electron mobility is:

$\mu_e = \frac{v_d}{E}$

Substituting the known values:

$\mu_e = \frac{145 \, \text{m/s}}{800 \, \text{V/m}} = 0.18125 \, \frac{\text{m}^2}{\text{V} \cdot \text{s}}$

Thus, the electron mobility $\mu_e = 0.18125 \, \frac{\text{m}^2}{\text{V} \cdot \text{s}}$.

Step 2: Calculate Conductivity $\sigma$

The formula for conductivity is:

$\sigma = n e \mu_e$

Substituting the known values:

$\sigma = (6.2 \times 10^8 \, \text{m}^{-3})(1.6 \times 10^{-19} \, \text{C})(0.18125 \, \frac{\text{m}^2}{\text{V} \cdot \text{s}})$

$\sigma = 1.798 \times 10^{-10} \, \text{S/m}$

Thus, the conductivity $\sigma = 1.798 \times 10^{-10} \, \text{S/m}$.

Final Answers:

• $\mu_e = 0.18125 \, \frac{\text{m}^2}{\text{V} \cdot \text{s}}$
• $\sigma = 1.798 \times 10^{-10} \, \text{S/m}$

Let me know if you need further clarifications or details!

Further questions:

1. What factors affect the electron mobility in a semiconductor?
2. How does temperature affect the conductivity of an n-type semiconductor?
3. Can the same formulas be used for p-type semiconductors?
4. How does the drift velocity of electrons change with increasing electric field?
5. What is the significance of the elementary charge in these calculations?

Tip:

The mobility of charge carriers in a semiconductor typically decreases as temperature increases due to increased scattering.