In a semiconductor, the number density of intrinsic charge carriers at 27°C is 1.5 x 10^16 / m^3. If

Q: Why does the product of electron and hole densities remain constant?

At a given temperature, the rate of generation of electron-hole pairs equals the rate of recombination. Doping increases one type of carrier, which increases the recombination rate, thereby decreasing the other type of carrier to maintain the constant product $n_i^2$.

Q: Does the mass action law apply to both intrinsic and extrinsic semiconductors?

Yes! The beauty of the mass action law ($n_e n_h = n_i^2$) is that it holds true for both intrinsic and doped (extrinsic) semiconductors under thermal equilibrium.

Q: What happens if we increase the temperature?

Increasing the temperature increases the intrinsic carrier concentration ($n_i$) because more thermal energy is available to break covalent bonds, creating more electron-hole pairs.

Q: Why is the hole density so much higher than the intrinsic density here?

The semiconductor has been doped with an impurity (likely a trivalent atom like Boron or Aluminum) to create a p-type semiconductor, which drastically increases the number of holes.

Q: Can I just subtract the intrinsic density from the hole density to find the electrons?

No, the relationship is multiplicative, not additive. You must use the product rule $n_e n_h = n_i^2$, not an addition or subtraction rule.

Question

In a semiconductor, the number density of intrinsic charge carriers at $27^{\circ}	ext{C}$ is $1.5 	imes 10^{16} / 	ext{m}^3$. If the semiconductor is doped with impurity atom, the hole density increases to $4.5 	imes 10^{22} / 	ext{m}^3$. The electron density in the doped semiconductor is ...... $	imes 10^9 / 	ext{m}^3$.

Sigma Percentile · Accepted Answer

Welcome to a fascinating journey into the microscopic world of semiconductors! Today, we are going to solve a classic problem that tests our understanding of charge carriers and the fundamental laws that govern them.

Imagine a bustling city where electrons and holes are constantly moving, recombining, and being generated. Our goal is to find out exactly how many electrons are left after we heavily populate the city with holes. Let's dive in!

### Analyzing the Setup

First, let's look at what we know. We are given a semiconductor at a steady temperature of $27^{\circ}	ext{C}$. In its pure, intrinsic state, the number density of charge carriers is $n_i = 1.5 	imes 10^{16} 	ext{ m}^{-3}$. This means that before any tampering, the number of electrons perfectly matches the number of holes.

However, the plot thickens! The semiconductor is doped with an impurity atom. This doping process drastically increases the hole density to a staggering $n_h = 4.5 	imes 10^{22} 	ext{ m}^{-3}$. Because the hole concentration is now massively higher than the intrinsic concentration, we are dealing with a **p-type semiconductor**.

Our mission is to find the new electron density, $n_e$, in this doped state.

### The Master Equation

To solve this, we need a powerful tool: the **Mass Action Law**. This law is a beautiful principle of thermal equilibrium. It states that no matter how much you dope a semiconductor, the product of the electron density and the hole density remains constant at a given temperature.

Mathematically, it is expressed as:
$$n_e \cdot n_h = n_i^2$$

This equation tells us that if you increase the holes, the electrons must decrease proportionally to maintain the balance. Since we want to find the electron density, we can easily rearrange this master equation:
$$n_e = \frac{n_i^2}{n_h}$$

### Final Calculation

Now comes the execution phase. Let's carefully substitute our known values into the rearranged equation. 
$$n_e = \frac{(1.5 	imes 10^{16})^2}{4.5 	imes 10^{22}}$$

First, we square the numerator. Squaring $1.5$ gives us $2.25$, and squaring $10^{16}$ gives us $10^{32}$. 
$$n_e = \frac{2.25 	imes 10^{32}}{4.5 	imes 10^{22}}$$

Next, we divide the numbers and the powers of ten separately. Dividing $2.25$ by $4.5$ yields exactly $0.5$. For the powers of ten, we subtract the exponents: $32 - 22 = 10$.
$$n_e = 0.5 	imes 10^{10}$$

To match the format requested in the question, we adjust the decimal point:
$$n_e = 5 	imes 10^9 	ext{ m}^{-3}$$

The question asks for the value that fills in the blank for $...... 	imes 10^9 / 	ext{m}^3$. Comparing our result, the missing integer is **$5$**.

And there we have it! By trusting the Mass Action Law and carefully managing our scientific notation, we've successfully navigated the microscopic world of this doped semiconductor.

Enter Numerical Value:

Visualized Solution

$G i v e n D a t a$

$M a ss A c t i o n L a w$

$R e a r r an g in g t h e F or m u l a$

$S u b s t i t u t in g V a l u es$

$S q u a r in g t h e N u m er a t or$

$F ina l C a l c u l a t i o n$

$C o n c l u s i o n$

The Sigma Insight: Energy Bands in Solids and Semiconductors

Analyzing the Setup

The Master Equation

Final Calculation

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