Question

A crystal of intrinsic silicon at room temperature has a carrier concentration of $1.6\times 10^{16}/m^3$ . If the donor concentration level is $4.8\times 10^{20}/m^3$ , then the concentration of holes in the semiconductor is

• A. $53\times 10^{12}/m^3$
• B. $4\times 10^{11}/m^3$
• C. $4\times 10^{12}/m^3$
• D. $5.3\times 10^{11}/m^3$

Answer 1

Answer: (D)

Given
$n_1=1.6\times 10^{16}/m^3$
$n_e=4.8\times 10^{20}/m^3$
$n_h=?$
The concentration of holes in the semiconductor
$n_1^2=n_e\times n_h$
${\left(1.6\times 10^{16}\right)}^2=4.8\times 10^{20}\times n_h$
$n_h=\frac{2.56\times 10^{32}}{4.8\times 10^{20}}=5.3\times 10^{11}/m^3$