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Chapter 19: Problem 43

The number densities of electrons and holes in a pure germanium at room temperature are equal and its value is \(3 \times 10^{16}\) per \(\mathrm{m}^{3}\). On doping with aluminium, the whole density increases to \(4.5 \times 10^{22}\) per \(\mathrm{m}^{3}\). Then the electron density in doped germanium is (A) \(2.5 \times 10^{10} \mathrm{~m}^{-3}\) (B) \(2 \times 10^{10} \mathrm{~m}^{-3}\) (C) \(4.5 \times 10^{9} \mathrm{~m}^{-3}\) (D) \(3 \times 10^{9} \mathrm{~m}^{-3}\)

Short Answer

Expert verified
The electron density in doped germanium is approximately \(2 \times 10^{10} \mathrm{m}^{-3}\).

Step by step solution

01

Understand the given data

In this problem, we are given the number densities of electrons and holes in pure germanium, which are equal at room temperature, as \(n_i = p_i = 3 \times 10^{16} \mathrm{m}^{-3}\). After doping with aluminium, the hole density increases to \(n_A = 4.5 \times 10^{22} \mathrm{m}^{-3}\). Note that when germanium is doped with aluminum, which is an acceptor impurity, we have increased the number of holes.
02

Calculate change in hole density due to doping

To find the change in hole density after doping, we will subtract the initial hole density from the hole density after doping: $$\Delta p = n_A - p_i = 4.5 \times 10^{22} \mathrm{m}^{-3} - 3 \times 10^{16} \mathrm{m}^{-3} \approx 4.5 \times 10^{22} \mathrm{m}^{-3}$$
03

Apply the mass-action law

The mass-action law states that the product of the number densities of electrons and holes equals the square of the intrinsic carrier concentration: $$n \cdot p = n_i^2$$ After doping, the number densities become \(n'\) and \(p'\), so the mass-action law becomes: $$n' \cdot p' = n_i^2$$
04

Find the new electron density \(n'\)

Since we know that \(p' = n_i + \Delta p = p_i + \Delta p\), we can substitute this expression into the mass-action law and find the new electron density: $$n' = \frac{n_i^2}{p'} = \frac{n_i^2}{n_i + \Delta p} = \frac{(3 \times 10^{16} \mathrm{m}^{-3})^2}{3 \times 10^{16} \mathrm{m}^{-3} + 4.5 \times 10^{22} \mathrm{m}^{-3}} \approx 2 \times 10^{10} \mathrm{m}^{-3}$$
05

Select the correct answer

The calculated value of the electron density in doped germanium \(n' \approx 2 \times 10^{10} \mathrm{m}^{-3}\) corresponds to the option (B). So, our final answer is: (B) \(2 \times 10^{10} \mathrm{m}^{-3}\)

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