Question

For a p-type semiconductor, the relation between the number density ' \(\mathrm{n}_{\mathrm{h}}\) ' and ' \(\mathrm{n}_{\mathrm{e}}\) ' is

• A. \(\mathrm{n}_{\mathrm{e}}=\mathrm{n}_{\mathrm{h}}\)
• B. \(n_{h}\gg n_{e}\)
• C. \(\mathrm{n}_{\mathrm{h}}\ll \mathrm{n}_{\mathrm{e}}\)
• D. \(\mathrm{n}_{\mathrm{h}}=\mathrm{n}_{\mathrm{e}}=0\)

Hint:

In a p-type semiconductor holes are the majority carriers.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
A p-type semiconductor is doped with acceptor impurities, which create many holes.

Step 2: Carrier Densities
Holes are the majority carriers and electrons the minority carriers, so \(n_h \gg n_e\).

Step 3: Result
Option B.

Final Answer:
In p-type material hole density is much greater than electron density. \[ \boxed{\text{(B) }n_{h}\gg n_{e}} \]