Question

For intrinsic semiconductor, if \(n_h\) and \(n_e\) represent the number of holes per unit volume and the number of free electrons per unit volume respectively then

• A. \(n_h < n_e\)
• B. \(n_h > n_e\)
• C. \(n_h = n_e\)
• D. \(n_h\neq n_e\)

Hint:

In a pure semiconductor each electron that leaves its bond leaves behind one hole, so the two numbers are equal.

Answer 1

The Correct Option is C

Step 1: Count carriers created by heat:
A photon or thermal vibration lifts one electron from the valence band to the conduction band. That leaves exactly one empty state in the valence band, which behaves as a hole.

Step 2: Result:
One electron and one hole appear together, so their concentrations are equal in a pure crystal: $n_e = n_h = n_i$, the intrinsic carrier concentration.

Step 3: Compare with doped types:
Adding donor atoms makes $n_e > n_h$, and adding acceptor atoms makes $n_h > n_e$. That is why (A), (B) and (D) fail for the intrinsic case.

Final Answer:
Option (C). \[ \boxed{n_h=n_e \text{ (C)}} \]