Question: Germanium crystal is doped at room temperature with a minute quantity of boron. The charge carriers in the doped semiconductors will be:

Step 1 (Governing Principle & Identification): [0.5 Mark]

Germanium ($Ge$) is a Group 14 element, and Boron ($B$) is a Group 13 element. When a Group 14 semiconductor is doped with a Group 13 impurity (trivalent atom), it creates a p-type semiconductor. In a p-type semiconductor, the trivalent impurity atom creates a vacancy in the covalent bond, known as a hole ($n_h > n_e$).

Step 2 (Step-by-Step Logical Deduction): [0.25 Mark]

• At room temperature, thermal energy is sufficient to excite some electrons from the valence band to the conduction band, generating a small number of electron-hole pairs.
• The doping process significantly increases the concentration of holes ($n_h$) due to the acceptor atoms.
• Therefore, in the doped crystal, the majority charge carriers are holes and the minority charge carriers are electrons.
• Mathematically, the relation holds: $$n_e \cdot n_h = n_i^2$$ where $n_h \gg n_e$.

Step 3 (Final Concluded Answer): [0.25 Mark]

Based on the analysis of p-type semiconductor physics, the charge carriers are holes (majority) and a few electrons (minority).

Correct Option: (C) holes and few electrons

💡 EL10 Concept Insight: Think of the Boron atom as a "missing chair" in a room full of people; because it has one less electron to share, it creates a hole that acts like a positive charge carrier, while the few electrons present are just there because the room is warm enough to shake some loose!