The resistivity of pure germanium at a particular temperature is `0.52Omegam`. If the material is doped with `10^(20)` atoms `m^(-3)` of a trivalent impurity material, determine the new resistivity. The electron and hole mobilities are givenn to be `0.2` and `0.4 m^(2)V^(-1)s^(-1)` respectively.

Answer

To determine the new resistivity of doped germanium, we will follow these steps:

### Step 1: Understand the intrinsic properties of germanium
The resistivity of pure germanium is given as \( \rho_i = 0.52 \, \Omega \cdot m \). For intrinsic semiconductors, the concentration of electrons (\( n_i \)) is equal to the concentration of holes (\( p_i \)), and both are equal to the intrinsic carrier concentration \( N_i \).

### Step 2: Calculate the intrinsic carrier concentration \( N_i \)
Using the formula for resistivity:
\[ \rho_i = \frac{1}{\sigma} = \frac{1}{q (n_i \mu_e + p_i \mu_h)} \]
Since \( n_i = p_i = N_i \), we can rewrite the formula as:
\[ \rho_i = \frac{1}{q N_i (\mu_e + \mu_h)} \]
Where:
- \( q \) is the charge of an electron \( (1.6 \times 10^{-19} \, C) \)
- \( \mu_e \) is the electron mobility \( (0.2 \, m^2/V \cdot s) \)
- \( \mu_h \) is the hole mobility \( (0.4 \, m^2/V \cdot s) \)

### Step 3: Substitute values to find \( N_i \)
Rearranging the equation gives:
\[ N_i = \frac{1}{q \rho_i (\mu_e + \mu_h)} \]
Substituting the known values:
\[ N_i = \frac{1}{(1.6 \times 10^{-19}) (0.52) (0.2 + 0.4)} \]
Calculating:
\[ N_i = \frac{1}{(1.6 \times 10^{-19}) (0.52) (0.6)} \approx 2.003 \times 10^{19} \, m^{-3} \]

### Step 4: Determine the new carrier concentrations after doping
When trivalent impurity is added, it creates holes. The concentration of holes \( p \) becomes:
\[ p = N_a = 10^{20} \, m^{-3} \]
The concentration of electrons \( n \) can be approximated using the mass action law:
\[ n \cdot p = N_i^2 \implies n = \frac{N_i^2}{p} \]
Substituting the values:
\[ n = \frac{(2.003 \times 10^{19})^2}{10^{20}} \approx 4.012 \times 10^{18} \, m^{-3} \]

### Step 5: Calculate the new conductivity \( \sigma \)
Using the formula for conductivity:
\[ \sigma = q (n \mu_e + p \mu_h) \]
Substituting the values:
\[ \sigma = (1.6 \times 10^{-19}) \left( (4.012 \times 10^{18})(0.2) + (10^{20})(0.4) \right) \]
Calculating:
\[ \sigma = (1.6 \times 10^{-19}) \left( 8.024 \times 10^{17} + 4 \times 10^{19} \right) \approx (1.6 \times 10^{-19}) (4.08024 \times 10^{19}) \approx 0.651 \, S/m \]

### Step 6: Calculate the new resistivity \( \rho \)
The resistivity \( \rho \) is the reciprocal of conductivity:
\[ \rho = \frac{1}{\sigma} \approx \frac{1}{0.651} \approx 1.54 \, \Omega \cdot m \]

### Final Answer
The new resistivity of the doped germanium is approximately \( 1.54 \, \Omega \cdot m \).

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