Question

The concentration of hole electron pair in pure silicon at $T=300\text{K}$ is $7\times {10}^{15}$ per cubic meter. Antimony is doped into silicon in proportion $1$ atom in ${10}^7$ atoms. Assuming that half of the impurity atoms contribute electrons in the conduction band. Calculate the factor by which the number of charge carriers increases due to doping. Given that the number of silicon atom per cubic meter is $14\times {10}^{28}.$

• A. $2\times {10}^3$
• B. $2\times {10}^5$
• C. $5\times {10}^3$
• D. $5\times {10}^5$

Answer 1

The correct option is D $5\times {10}^5$

Total number of charge carriers before doping = Number of holes + Number of electron

$=2\times 7\times {10}^{15}=14\times {10}^{15}$

Doping proportion of antimony = $1$ atom in ${10}^7$ atom

$\Rightarrow$ Total number of antimony per cubic meter $=\frac{14\times {10}^{28}}{{10}^7}=14\times {10}^{21}$

Given that, the only half of the impurity atom contributes conduction electrons,

The number of extra conduction electron produced $=\frac{14\times {10}^{21}}{2}$

$=7\times {10}^{21}$ per cubic meter

$\Rightarrow$ Total number of charge carrier after doping

$=7\times {10}^{21}+14\times {10}^{15}$

$=7.000014\times {10}^{21}\approx 7\times {10}^{21}$

The factor by which number of charge carries increased is

$=\frac{7\times {10}^{21}}{14\times {10}^{15}}$

$=5\times {10}^5$

Hence, option $\left(D\right)$ is correct.