Pure $Si$ at $500\, K$ has equal number of electron $(n_e)$ and hole $(n_h)$ concentrations of $1.5 \times 10^{16}\, m^{-3}$ Doping by indium increases $n_h$ to 4.5 $\times 10^{22}\, m^{-3}$ The doped semiconductor is of

Question

Consider the following statements A and B and identify the correct answer:

A. For a solar-cell, the I-V characteristics lies in the IV quadrant of the given graph.
B. In a reverse biased pn junction diode, the current measured in (µA), is due to majority charge carriers.

Answer 1

The problem states that pure silicon ($Si$) at 500 K has equal electron ($n_e$) and hole ($n_h$) concentrations of $1.5 \times 10^{16}\, \text{m}^{-3}$. After doping with indium, the hole concentration increases to $4.5 \times 10^{22} \, \text{m}^{-3}$. The task is to determine the type of semiconductor and the concentration of electrons.

Step 1: Understanding Semiconductor Doping

In semiconductors, doping introduces additional charge carriers. P-type semiconductors have an excess of holes, while n-type semiconductors have an excess of electrons.
Indium, being a Group III element, is a p-type dopant. It increases the number of holes in the semiconductor.

Step 2: Applying the Mass Action Law

For intrinsic semiconductors, the product of electron and hole concentrations remains constant and equal to the square of the intrinsic carrier concentration ($n_i$): n_i^2 = n_e \cdot n_h.
Given that initially $n_e = n_h = 1.5 \times 10^{16}\, \text{m}^{-3}$, we calculate: n_i^2 = (1.5 \times 10^{16})^2 = 2.25 \times 10^{32}\, \text{m}^{-6}.

Step 3: Calculating Electron Concentration in Doped Semiconductor

After indium doping, the hole concentration becomes $n_h = 4.5 \times 10^{22}\, \text{m}^{-3}$.
Using the mass action law: n_e = \frac{n_i^2}{n_h} = \frac{2.25 \times 10^{32}}{4.5 \times 10^{22}} = 5 \times 10^{9}\, \text{m}^{-3}.

Step 4: Determining Type of Semiconductor

The significantly higher hole concentration ($n_h$) compared to electron concentration ($n_e$) confirms it as a p-type semiconductor.

Conclusion: The doped semiconductor is p-type with an electron concentration of $n_e = 5 \times 10^9\, \text{m}^{-3}$.

Answer 2

The problem states that pure silicon ($Si$) at 500 K has equal electron ($n_e$) and hole ($n_h$) concentrations of $1.5 \times 10^{16}\, \text{m}^{-3}$. After doping with indium, the hole concentration increases to $4.5 \times 10^{22} \, \text{m}^{-3}$. The task is to determine the type of semiconductor and the concentration of electrons.

Step 1: Understanding Semiconductor Doping

In semiconductors, doping introduces additional charge carriers. P-type semiconductors have an excess of holes, while n-type semiconductors have an excess of electrons.
Indium, being a Group III element, is a p-type dopant. It increases the number of holes in the semiconductor.

Step 2: Applying the Mass Action Law

For intrinsic semiconductors, the product of electron and hole concentrations remains constant and equal to the square of the intrinsic carrier concentration ($n_i$): n_i^2 = n_e \cdot n_h.
Given that initially $n_e = n_h = 1.5 \times 10^{16}\, \text{m}^{-3}$, we calculate: n_i^2 = (1.5 \times 10^{16})^2 = 2.25 \times 10^{32}\, \text{m}^{-6}.

Step 3: Calculating Electron Concentration in Doped Semiconductor

After indium doping, the hole concentration becomes $n_h = 4.5 \times 10^{22}\, \text{m}^{-3}$.
Using the mass action law: n_e = \frac{n_i^2}{n_h} = \frac{2.25 \times 10^{32}}{4.5 \times 10^{22}} = 5 \times 10^{9}\, \text{m}^{-3}.

Step 4: Determining Type of Semiconductor

The significantly higher hole concentration ($n_h$) compared to electron concentration ($n_e$) confirms it as a p-type semiconductor.

Conclusion: The doped semiconductor is p-type with an electron concentration of $n_e = 5 \times 10^9\, \text{m}^{-3}$.

Pure $Si$ at $500\, K$ has equal number of electron $(n_e)$ and hole $(n_h)$ concentrations of $1.5 \times 10^{16}\, m^{-3}$ Doping by indium increases $n_h$ to 4.5 $\times 10^{22}\, m^{-3}$ The doped semiconductor is of

The Correct Option is A

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