Step 1 (Governing Law & Concept): [1 Mark]
In an n-type semiconductor, the electron concentration ($n_e$) is determined by two sources: the thermal excitation of intrinsic charge carriers and the ionization of donor impurity atoms. According to the Mass Action Law for semiconductors in thermal equilibrium:
$$n_e \cdot n_h = n_i^2$$Where $n_e$ is the electron concentration, $n_h$ is the hole concentration, and $n_i$ is the intrinsic carrier concentration.
Step 2 (Step-by-Step Logical Deduction): [0.5 Mark]
- In an n-type semiconductor, pentavalent impurity atoms (donors) are added to the intrinsic semiconductor.
- Each donor atom contributes one free electron to the conduction band without creating a corresponding hole.
- Let $N_D$ be the donor concentration. The total electron concentration becomes: $n_e \approx N_D + n_i$.
- Since $N_D \gg n_i$, the electron concentration increases significantly, while the hole concentration is suppressed because the increased number of electrons enhances the rate of electron-hole recombination.
Step 3 (Final Concluded Answer): [0.5 Mark]
Because the rate of recombination is proportional to the product of electron and hole concentrations ($R \propto n_e \cdot n_h$), the massive increase in $n_e$ forces $n_h$ to decrease to maintain the constant product $n_i^2$. Thus, in an n-type semiconductor:
$$n_e \gg n_h$$The electron concentration remains dominant because the donor atoms provide a surplus of electrons that far outweighs the number of holes generated by thermal agitation.
💡 EL10 Concept Insight: Think of it like a party where you have a few people (intrinsic carriers) and then a busload of extra guests (donor electrons) arrives; because there are so many extra guests now, they keep bumping into the few people already there and "canceling them out," leaving you with way more guests than the original people.