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
$n_{1}^{2} =n_{e} \times n_{h}$
$\left(1.6 \times 10^{16}\right)^{2} =4.8 \times 10^{20} \times n_{h}$
$n_{h} =\frac{2.56 \times 10^{32}}{4.8 \times 10^{20}}=5.3 \times 10^{11} / m ^{3}$