Question

A pure silicon crystal at temperature \(300\) K has electron and hole concentration (\(n_i\)) \(10^{16}\) per m\(^3\) each and \(10^{21}\) phosphorus atoms (\(n_e\)) are added per cubic metre. The new hole concentration in silicon crystal is

• A. \(10^{19}\) per m\(^3\)
• B. \(10^{21}\) per m\(^3\)
• C. \(10^5\) per m\(^3\)
• D. \(10^{11}\) per m\(^3\)

Hint:

Use the mass action law \(n_en_h=n_i^2\).

Answer 1

The Correct Option is D

Step 1: Plan:
Think about how doping pushes down the hole count.

Step 2: Steps:
Doping raises electrons from $10^{16}$ to about $10^{21}$, by a factor of $10^5$. The product $n_en_h$ stays at $10^{32}$, so holes fall by the same factor, from $10^{16}$ to $10^{11}$ per m$^3$.

Final Answer:
The new hole concentration is $10^{11}$ per m$^3$, option (D). \[ \boxed{10^{11}\ \text{m}^{-3}} \]