Question

If a semiconductor has an intrinsic carrier concentration of $\frac{1.41\times {10}^{16}}{m^3}$ when doped with $\frac{{10}^{21}}{m^3}$ phosphorous atoms, then the concentration of holes per cubic meter at room temperature will be

• A. $2\times {10}^{21}$
  
• B. $2\times {10}^{11}$
  
• C. $1.41\times {10}^{10}$
  
• D. $1.41\times {10}^{16}$

Answer 1

The correct option is D $1.41\times {10}^{16}$

The number of holes will be equal to number of intrinsic carrier concentration.
$=\frac{1.41\times {10}^6}{m^3}$