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 centimetre in the specimen due to doping. Also, give one example of such dopants.

Hint:

In p-type semiconductors, dopants like boron create holes, and the number of holes is equal to the number of dopant atoms introduced into the material.

Answer 1

The number density of silicon atoms is \( 5 \times 10^{28} \, \text{atoms/m}^3 \). The number of dopant atoms per silicon atom is \( 1 / (5 \times 10^7) \). Thus, the number of dopant atoms per unit volume is: \[ \text{Number of dopant atoms} = \frac{1}{5 \times 10^7} \times 5 \times 10^{28} = 10^{21} \, \text{atoms/m}^3 \] For a p-type semiconductor, each dopant atom introduces one hole. Therefore, the number of holes created per unit volume is \( 10^{21} \, \text{holes/m}^3 \). To convert this to the number of holes per cubic centimetre: \[ \text{Number of holes per cm}^3 = 10^{21} \, \text{holes/m}^3 \times \left( \frac{1}{10^6} \right) = 10^{15} \, \text{holes/cm}^3 \] Thus, the number of holes created per cubic centimetre due to doping is \( 10^{15} \, \text{holes/cm}^3 \). One example of such dopants is **boron** (\( \text{B} \)).