Semiconductor Resistivity Calculation

The problem asks for the resistivity ($\rho$) of a Germanium (Ge) semiconductor doped with an acceptor impurity.

Given Parameters

• Dopant Type: Acceptor (making it a p-type semiconductor)
• Acceptor Concentration ($N_A$): $1 \times 10^{15}$ atoms/cm$^3$
• Hole Mobility ($\mu_p$): $1800 \text{ cm}^2/\text{V-s}$
• Elementary Charge ($q$): $1.6 \times 10^{-19} \text{ C}$

Resistivity Formula

For a p-type semiconductor, the majority charge carriers are holes. The concentration of holes ($p$) is approximately equal to the acceptor concentration ($N_A$), i.e., $p \approx N_A$. The electrical conductivity ($\sigma$) is given by:

$ \sigma = q p \mu_p $

Resistivity ($\rho$) is the reciprocal of conductivity:

$ \rho = \frac{1}{\sigma} = \frac{1}{q p \mu_p} $

Since $p \approx N_A$, the formula becomes:

$ \rho = \frac{1}{q N_A \mu_p} $

Calculation

Substitute the given values into the resistivity formula:

$ \rho = \frac{1}{(1.6 \times 10^{-19} \text{ C}) \times (1 \times 10^{15} \text{ cm}^{-3}) \times (1800 \text{ cm}^2/\text{V-s})} $

First, calculate the denominator:

$ q N_A \mu_p = (1.6 \times 10^{-19}) \times (1 \times 10^{15}) \times 1800 $

$ q N_A \mu_p = (1.6 \times 1800) \times 10^{-19+15} $

$ q N_A \mu_p = 2880 \times 10^{-4} \text{ S/cm} $

$ q N_A \mu_p = 0.2880 \text{ S/cm} $

Now, calculate the resistivity:

$ \rho = \frac{1}{0.2880 \text{ S/cm}} $

$ \rho \approx 3.4722 \text{ } \Omega \text{ cm} $

Conclusion

The calculated resistivity is approximately $3.472 \text{ } \Omega \text{ cm}$, which matches Option C.