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 \(5 \times 10^{28}\) 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:

Doping with trivalent impurities like Boron introduces holes in the semiconductor, making it p-type.

Answer 1

The number density of silicon atoms is \( 5 \times 10^{28} \, \text{atoms/m}^3 \). The doping ratio is 1 dopant atom per \( 5 \times 10^7 \) silicon atoms. Therefore, the number of dopant atoms per cubic metre is: \[ \text{Dopant density} = \frac{5 \times 10^{28}}{5 \times 10^7} = 10^{21} \, \text{atoms/m}^3 \] Since each dopant atom creates one hole, the number of holes per cubic metre is \( 10^{21} \). Converting to per cubic centimetre: \[ \text{Holes per cm}^3 = 10^{21} \times 10^{-6} = 10^{15} \, \text{holes/cm}^3 \] An example of such a dopant is Boron (B), which is a trivalent impurity and creates p-type semiconductors.