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The problem asks to determine the conductivity of doped Germanium (Ge) with given parameters. Here's a step-by-step breakdown:

### Given Data:
- **Intrinsic carrier concentration (n_i)**: \( 2.5 \times 10^{19} \, \text{m}^{-3} \)
- **Donor concentration (N_d)** (As atoms): \( 5 \times 10^{23} \, \text{m}^{-3} \)
- **Electron mobility (\mu_n)**: \( 0.38 \, \text{m}^2/\text{V·s} \)
- **Hole mobility (\mu_p)**: \( 0.18 \, \text{m}^2/\text{V·s} \)
- Assume all As atoms are ionized.

### Key Formula for Conductivity:
The conductivity (\( \sigma \)) of a semiconductor can be determined using the following formula:
\[
\sigma = q \left( n \mu_n + p \mu_p \right)
\]
Where:
- \( q \) is the elementary charge (\( 1.6 \times 10^{-19} \, \text{C} \))
- \( n \) is the concentration of electrons (equal to the ionized donor concentration \( N_d \))
- \( p \) is the concentration of holes, which in n-type material is approximately:
  \[
  p = \frac{n_i^2}{n}
  \]
  since the semiconductor is doped.

### Steps to Calculate Conductivity:

1. **Electron concentration**: 
   Since all donor atoms are ionized, the electron concentration is \( n = N_d = 5 \times 10^{23} \, \text{m}^{-3} \).

2. **Hole concentration**: 
   The hole concentration can be calculated using:
   \[
   p = \frac{n_i^2}{n}
   \]
   Substituting the values:
   \[
   p = \frac{(2.5 \times 10^{19})^2}{5 \times 10^{23}} = \frac{6.25 \times 10^{38}}{5 \times 10^{23}} = 1.25 \times 10^{15} \, \text{m}^{-3}
   \]

3. **Conductivity**: 
   Now use the formula for conductivity:
   \[
   \sigma = q \left( n \mu_n + p \mu_p \right)
   \]
   Substituting values:
   \[
   \sigma = (1.6 \times 10^{-19}) \left( (5 \times 10^{23})(0.38) + (1.25 \times 10^{15})(0.18) \right)
   \]

   Calculating each part:
   - \( (5 \times 10^{23})(0.38) = 1.9 \times 10^{23} \)
   - \( (1.25 \times 10^{15})(0.18) = 2.25 \times 10^{14} \)

   Now, summing these terms:
   \[
   1.9 \times 10^{23} + 2.25 \times 10^{14} \approx 1.9 \times 10^{23}
   \]

   Finally:
   \[
   \sigma = (1.6 \times 10^{-19})(1.9 \times 10^{23}) = 3.04 \times 10^4 \, \text{S/m}
   \]

Thus, the **conductivity** of the doped Germanium is approximately \( 3.04 \times 10^4 \, \text{S/m} \).

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Do you have any questions or need further clarification on any step?

### 5 Related Questions:
1. How does temperature affect the conductivity of semiconductors?
2. What is the difference between intrinsic and extrinsic semiconductors?
3. How do donor and acceptor impurities affect carrier concentrations in semiconductors?
4. What is the significance of mobility in determining semiconductor conductivity?
5. How do you calculate the intrinsic carrier concentration for different materials?

### Tip:
In semiconductor problems, always check if the doping concentration is significantly higher than the intrinsic carrier concentration. If it is, you can assume the electron concentration equals the doping level for an n-type semiconductor.

Germanium has an intrinsic concentration of 2.5 x 10^19 /m^3 at 300K. It is doped with 5 x 10^23 As atoms per m^3. Assume that all the As atoms are ionized. If the electron and hole mobilities are 0.38 and 0.18 m^2/V·s, respectively, determine the conductivity of the doped Ge.

Learn how to calculate the conductivity of doped Germanium (Ge) with As atoms. This problem involves key concepts in semiconductor physics, including carrier concentrations, electron and hole mobilities, and the conductivity formula. Suitable for undergraduate students.

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