Silicon ( $M = 28 g mo l^{- 1}$ , density $2.33 g c m^{- 3}$ ) is doped with phosphorus at 1 atom per $1 0^{7}$ Si atoms. Assuming each P atom donates one free electron and $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , the donated electron concentration per $c m^{3}$ is

Answer: A

Si atoms per

c m^{3}

= \frac{2.33}{28} \times 6.022 \times 1 0^{23} = 5.01 \times 1 0^{22}

. Dopant atoms (and electrons)

= 5.01 \times 1 0^{22} /1 0^{7} = 5.01 \times 1 0^{15} c m^{- 3}