In a doped semiconductor, the hole concentration $p$ (density of holes) can be calculated using the mass action law. The mass action law relates the intrinsic carrier concentration $n_i$, the electron concentration $n$, and the hole concentration $p$.

The mass action law states:

$n_i^2 = n \cdot p$

Where:

• $n_i$ is the intrinsic carrier concentration (in this case, $n_i = 1.5 \times 10^{10} \, \text{cm}^{-3}$),
• $n$ is the electron concentration (which is equal to the donor concentration in this case),
• $p$ is the hole concentration (which is what we are solving for).

For a semiconductor with an acceptor concentration $N_A$, the hole concentration $p$ can be approximated (in steady-state conditions) by the formula:

$p \approx N_A$

Where:

• $N_A$ is the acceptor concentration.

In this case, the acceptor concentration is given as $N_A = 10^{18} \, \text{cm}^{-3}$. Therefore, the density of holes $p$ is approximately equal to the acceptor concentration $N_A$.

Thus, the density of holes is:

$p \approx 10^{18} \, \text{cm}^{-3}$

So the correct answer is:

A. $10^{18} \, \text{cm}^{-3}$

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Would you like more details on how this calculation is performed or have any further questions?

Here are 5 related questions:

1. What is the electron concentration in this doped semiconductor?
2. How do donor and acceptor impurities affect the carrier concentrations in semiconductors?
3. How can we calculate the mobility of carriers in a semiconductor?
4. What is the relationship between carrier mobility and electrical conductivity in semiconductors?
5. What is the intrinsic carrier concentration, and how is it affected by temperature?

Tip: The mobility of charge carriers is a key factor in determining the speed at which carriers move in response to an electric field, affecting the overall conductivity of a semiconductor.