Question

The hole concentration in an intrinsic semiconductor is \(5\times10^{8}\,\text{m}^{-3}\). When it is doped with certain impurity, the electron concentration becomes \(4\times10^{12}\,\text{m}^{-3}\). Find the new value of the hole concentration. Also identify the type of new semiconductor formed after doping.

Hint:

For semiconductor numericals involving carrier concentrations, always use \[ np=n_i^2. \] If \(n>p\), the semiconductor is n-type. If \(p>n\), the semiconductor is p-type.

Answer 1

Concept: In an intrinsic semiconductor, \[ n_i=p_i, \] where \[ n_i \] is the intrinsic electron concentration and \[ p_i \] is the intrinsic hole concentration. For any semiconductor in thermal equilibrium, the law of mass action states that \[ np=n_i^2, \] where \[ n \] is the electron concentration after doping and \[ p \] is the hole concentration after doping. This relation remains valid even after doping.

Step 1: Determine the intrinsic carrier concentration. Given intrinsic hole concentration, \[ p_i=5\times10^8\,\text{m}^{-3}. \] For an intrinsic semiconductor, \[ n_i=p_i. \] Therefore, \[ n_i=5\times10^8\,\text{m}^{-3}. \]

Step 2: Apply the mass action law. The law of mass action gives \[ np=n_i^2. \] Substituting the given values, \[ (4\times10^{12})p = (5\times10^8)^2. \]

Step 3: Calculate the square of the intrinsic concentration. \[ (5\times10^8)^2 = 25\times10^{16} = 2.5\times10^{17}. \] Hence, \[ (4\times10^{12})p = 2.5\times10^{17}. \]

Step 4: Calculate the new hole concentration. \[ p = \frac{2.5\times10^{17}} {4\times10^{12}}. \] \[ p = 0.625\times10^5. \] \[ p = 6.25\times10^4\,\text{m}^{-3}. \] Thus, the new hole concentration is \[ \boxed{6.25\times10^4\,\text{m}^{-3}}. \]

Step 5: Identify the type of semiconductor. After doping, \[ n=4\times10^{12}\,\text{m}^{-3} \] whereas \[ p=6.25\times10^4\,\text{m}^{-3}. \] Clearly, \[ n \gg p. \] Therefore, electrons are the majority charge carriers. Hence, the doped semiconductor is an \[ \boxed{\text{n-type semiconductor}}. \]

Step 6: Write the final answer. The new hole concentration is \[ \boxed{6.25\times10^4\,\text{m}^{-3}} \] and the semiconductor formed is \[ \boxed{\text{n-type semiconductor}}. \]