Physics - Semiconductors Question with Solution | TestHub

Given:

Ratio of dopant atoms to silicon atoms = $1$ dopant atom per $5\times 10^7$ silicon atoms

Number density of silicon atoms ($N_{Si}$) = $5\times 10^{28}$ atoms m⁻³

Step 1: Calculate the number density of dopant atoms ($N_A$).

The number density of dopant atoms can be found by dividing the number density of silicon atoms by the given ratio.

$$N_A=\frac{N_{Si}}{Ratio}$$

$$N_A=\frac{5\times 10^{28}atoms{m}^{-3}}{5\times 10^7}$$

$$N_A=1\times 10^{21}atoms{m}^{-3}$$

In a p-type semiconductor, each dopant atom (acceptor) creates one hole. Therefore, the number of holes created per cubic centimeter is equal to the number density of dopant atoms.

Number of holes per cubic centimeter = $1\times 10^{15}holesc{m}^{-3}$.

An example of such a dopant is Boron (B).