We are given the following values for an n-type semiconductor:

• Electron concentration $n = 4.7 \times 10^8 \, \text{m}^{-3}$
• Electric field $E = 550 \, \text{V/m}$
• Electron drift velocity $v_d = 10^5 \, \text{m/s}$

The goal is to calculate:

1. The electron mobility $\mu_e$
2. The conductivity $\sigma$

Step 1: Calculate Electron Mobility $\mu_e$

Electron mobility $\mu_e$ is given by the relationship between the drift velocity $v_d$ and the electric field $E$:

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

Substitute the given values:

$\mu_e = \frac{10^5 \, \text{m/s}}{550 \, \text{V/m}} = 181.82 \, \text{m}^2/\text{V·s}$

Step 2: Calculate Conductivity $\sigma$

The conductivity $\sigma$ for an n-type semiconductor can be calculated using the formula:

$\sigma = n e \mu_e$

Where:

• $n = 4.7 \times 10^8 \, \text{m}^{-3}$ (electron concentration)
• $e = 1.6 \times 10^{-19} \, \text{C}$ (elementary charge)
• $\mu_e = 181.82 \, \text{m}^2/\text{V·s}$

Substitute the values into the formula:

$\sigma = (4.7 \times 10^8) (1.6 \times 10^{-19}) (181.82)$

$\sigma = 1.36 \times 10^{-8} \, \text{S/m}$

Finally, convert this result to the desired units by multiplying by $10^{12}$:

$\sigma = 1.36 \times 10^{4} \times 10^{-12} \, \text{S/m}$

Thus,

$\sigma = 13.6 \times 10^{-12} \, \text{S/m}$

Final Answers:

1. $\mu_e = 181.82 \, \text{m}^2/\text{V·s}$
2. $\sigma = 13.6 \times 10^{-12} \, \text{S/m}$

Would you like further clarification or details?

5 Relative Questions:

1. What is the significance of electron mobility in semiconductors?
2. How does temperature affect the mobility of charge carriers in semiconductors?
3. What role does the electric field play in determining the drift velocity of electrons?
4. How is conductivity related to both electron mobility and charge carrier concentration?
5. Can you explain how doping affects the conductivity of semiconductors?

Tip: Always check the units of the quantities involved in semiconductor equations to ensure consistency and correctness!