An intrinsic semiconductor 5×10^28 atoms/m3 with ni=1.5×10^16m^-3. It is doped my 1ppm concentration of pentavalent impurity. Calculate the number of electrons and holes.

To determine the number of electrons and holes in an intrinsic semiconductor that has been doped with a pentavalent impurity, we need to follow a systematic approach. Let's break down the problem step by step.

Understanding the Basics

In semiconductors, the intrinsic carrier concentration (ni) represents the number of charge carriers (electrons and holes) in pure, undoped material. When we introduce a dopant, such as a pentavalent impurity (like phosphorus), it donates extra electrons to the conduction band, increasing the number of free electrons.

Given Data

• Intrinsic carrier concentration, ni = 1.5 × 10¹⁶ m^-3
• Concentration of dopant = 1 ppm = 1 × 10^-6
• Density of atoms in the semiconductor = 5 × 10²⁸ atoms/m³

Calculating the Number of Dopant Atoms

First, we need to find out how many dopant atoms are present in the semiconductor. Since 1 ppm means 1 part per million, we can calculate the number of dopant atoms as follows:

Number of dopant atoms = (Concentration of dopant) × (Density of atoms)

Number of dopant atoms = (1 × 10^-6) × (5 × 10²⁸) = 5 × 10²² atoms/m³

Finding the Number of Electrons

In a semiconductor doped with a pentavalent impurity, each dopant atom contributes approximately one free electron. Therefore, the number of electrons (n) can be approximated as:

n ≈ Number of dopant atoms = 5 × 10²² m^-3

Calculating the Number of Holes

In a doped semiconductor, the number of holes (p) can be calculated using the mass action law, which states that the product of the electron concentration and hole concentration is equal to the square of the intrinsic carrier concentration:

n × p = ni²

Rearranging this gives us:

p = ni² / n

Now, substituting the known values:

p = (1.5 × 10¹⁶)² / (5 × 10²²)

p = (2.25 × 10³²) / (5 × 10²²) = 4.5 × 10⁹ m^-3

Summary of Results

After performing the calculations, we find:

• Number of electrons (n) = 5 × 10²² m^-3
• Number of holes (p) = 4.5 × 10⁹ m^-3

This analysis shows how doping affects the charge carrier concentrations in semiconductors, illustrating the fundamental principles of semiconductor physics. If you have any further questions or need clarification on any part of this process, feel free to ask!