Question

A p-type Si semiconductor is made by doping an average of one dopant atom per \( 5 \times 10^7 \) silicon atoms. If the number density of silicon atoms in the specimen is \( 5 \times 10^{28} \, \text{atoms m}^{-3} \), find the number of holes created per cubic centimeter in the specimen due to doping. Also give one example of such dopants.

Hint:

In a p-type semiconductor, the dopant atoms create holes by accepting electrons. The dopant concentration determines the number of holes.

Answer 1

The number of dopant atoms per silicon atom is \( \frac{1}{5 \times 10^7} \). The number of holes created in the specimen per cubic meter is the number of dopant atoms per cubic meter, which is: \[ n_{\text{holes}} = \left( \frac{1}{5 \times 10^7} \right) \times (5 \times 10^{28}) = 10^{21} \, \text{holes m}^{-3} \] To find the number of holes per cubic centimeter, we convert from cubic meters to cubic centimeters. Since \( 1 \, \text{m}^3 = 10^6 \, \text{cm}^3 \): \[ n_{\text{holes}} = \frac{10^{21}}{10^6} = 10^{15} \, \text{holes cm}^{-3} \] Thus, the number of holes created per cubic centimeter due to doping is \( 10^{15} \, \text{holes cm}^{-3} \). An example of a dopant for p-type semiconductors is Boron.

Answer 2

We are given that a p-type silicon semiconductor is doped with an average of one dopant atom per \( 5 \times 10^7 \) silicon atoms. The number density of silicon atoms in the specimen is \( 5 \times 10^{28} \, \text{atoms/m}^3 \). We are tasked with finding the number of holes created per cubic centimeter in the specimen due to doping. 

1. Understanding the Concepts:

- P-type Semiconductor: A p-type semiconductor is created by doping silicon with acceptor atoms, such as boron (B). These dopants have one less valence electron than silicon, creating "holes" in the semiconductor. Each hole can act as a positive charge carrier.

- Doping Concentration: The doping concentration determines the number of dopant atoms in the semiconductor. Here, one dopant atom is added for every \( 5 \times 10^7 \) silicon atoms.

- Number Density: The number density of silicon atoms is given as \( 5 \times 10^{28} \, \text{atoms/m}^3 \). This means there are \( 5 \times 10^{28} \) silicon atoms in one cubic meter of the specimen.

2. Calculating the Number of Holes:

- The fraction of dopant atoms is \( \frac{1}{5 \times 10^7} \). This represents the number of dopant atoms per silicon atom.

- The number of dopant atoms per cubic meter is calculated as:

\[ \text{Number of dopant atoms per cubic meter} = \text{Number density of silicon atoms} \times \frac{1}{5 \times 10^7} \]

\[ \text{Number of dopant atoms per cubic meter} = 5 \times 10^{28} \times \frac{1}{5 \times 10^7} = 10^{21} \, \text{atoms/m}^3 \]

- To convert this to holes per cubic centimeter, we use the conversion factor: \( 1 \, \text{m}^3 = 10^6 \, \text{cm}^3 \). So, the number of holes per cubic centimeter is:

\[ \text{Number of holes per cubic centimeter} = \frac{10^{21}}{10^6} = 10^{15} \, \text{holes/cm}^3 \]

3. Example of a Dopant:

An example of a dopant for creating a p-type silicon semiconductor is boron (B). Boron has one fewer valence electron than silicon, which creates a hole in the crystal structure of silicon, thus making it p-type.

Final Answer:

The number of holes created per cubic centimeter in the specimen due to doping is \( 10^{15} \, \text{holes/cm}^3 \). An example of such a dopant is boron (B).