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Chapter 21: Problem 33

A silicon sample is doped with phosphorus at 1 part per \(1.00 \cdot 10^{6} .\) Phosphorus acts as an electron donor, providing one free electron per atom. The density of silicon is \(2.33 \mathrm{~g} / \mathrm{cm}^{3},\) and its atomic mass is \(28.09 \mathrm{~g} / \mathrm{mol}\) a) Calculate the number of free (conduction) electrons per unit volume of the doped silicon. b) Compare the result from part (a) with the number of conduction electrons per unit volume of copper wire (assume each copper atom produces one free (conduction) electron). The density of copper is \(8.96 \mathrm{~g} / \mathrm{cm}^{3},\) and its atomic mass is \(63.54 \mathrm{~g} / \mathrm{mol}\)

Short Answer

Expert verified
Based on our calculations, the number of free (conduction) electrons per unit volume of doped silicon is approximately \(4.99 \cdot 10^{16} \ \text{electrons/cm}^3\), whereas the number of conduction electrons per unit volume for copper wire is approximately \(8.49 \cdot 10^{22} \ \text{electrons/cm}^3\). Comparing these values, we find that the doped silicon has a relative concentration of about \(5.88 \cdot 10^{-7}\) times the number of conduction electrons per unit volume of copper wire.

Step by step solution

01

Finding the number of silicon atoms per unit volume

First, we need to find the number of silicon atoms per unit volume. We can do this by using the density of silicon and its atomic mass. The formula to find the number of atoms per unit volume is: Number of atoms per unit volume = (Density \(\times\) Avogadro's number) / Atomic mass The Avogadro's number is \(6.022 \cdot 10^{23} \ \text{atoms/mol}\), the density of silicon is \(2.33 \ \text{g/cm}^3\), and its atomic mass is \(28.09 \ \text{g/mol}\).
02

Calculate the number of silicon atoms per unit volume

Using the formula and the given values, we can now find the number of silicon atoms per unit volume. Number of silicon atoms per unit volume = \(\frac{2.33 \ \text{g/cm}^3 \times 6.022 \cdot 10^{23} \ \text{atoms/mol}}{28.09 \ \text{g/mol}}\) Number of silicon atoms per unit volume ≈ \(4.99 \cdot 10^{22} \ \text{atoms/cm}^3\)
03

Calculate the number of free electrons per unit volume

Now, we can find the number of free electrons per unit volume as the silicon sample is doped with phosphorus, providing one free electron per phosphorus atom. Since the doping concentration is 1 part per \(1.00 \cdot 10^6\), we can multiply the number of silicon atoms per unit volume by the doping concentration. Number of free electrons per unit volume = \(4.99 \cdot 10^{22} \ \text{atoms/cm}^3 \times \frac{1}{1.00 \cdot 10^6}\) Number of free electrons per unit volume ≈ \(4.99 \cdot 10^{16} \ \text{electrons/cm}^3\) #b) Compare the result from part (a) with the number of conduction electrons per unit volume of copper wire.#
04

Finding the number of copper atoms per unit volume

First, we need to find the number of copper atoms per unit volume. We can use the formula similarly to step 1. The density of copper is \(8.96 \ \text{g/cm}^3\), and its atomic mass is \(63.54 \ \text{g/mol}\). Number of copper atoms per unit volume = \(\frac{8.96 \ \text{g/cm}^3 \times 6.022 \cdot 10^{23} \ \text{atoms/mol}}{63.54 \ \text{g/mol}}\) Number of copper atoms per unit volume ≈ \(8.49 \cdot 10^{22} \ \text{atoms/cm}^3\)
05

Calculate the number of conduction electrons per unit volume of copper wire

As mentioned, each copper atom produces one free conduction electron. Therefore, the number of conduction electrons per unit volume of copper wire is the same as the number of copper atoms per unit volume. Number of conduction electrons per unit volume of copper wire ≈ \(8.49 \cdot 10^{22} \ \text{electrons/cm}^3\) Now, we have the number of conduction electrons per unit volume for both doped silicon and copper wire. We can compare these values to find the relative concentrations.
06

Compare the number of conduction electrons in doped silicon and copper wire

We can compare the values by dividing the number of conduction electrons in doped silicon by the number of conduction electrons in copper wire. Relative concentration = \(\frac{4.99 \cdot 10^{16} \ \text{electrons/cm}^3}{8.49 \cdot 10^{22} \ \text{electrons/cm}^3}\) Relative concentration ≈ \(5.88 \cdot 10^{-7}\) The number of free conduction electrons per unit volume of the doped silicon is about \(5.88 \cdot 10^{-7}\) times the number of conduction electrons per unit volume of copper wire.

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Most popular questions from this chapter

Eight \(1.00-\mu C\) charges are arrayed along the \(y\) -axis located every \(2.00 \mathrm{~cm}\) starting at \(y=0\) and extending to \(y=14.0 \mathrm{~cm} .\) Find the force on the charge at \(y=4.00 \mathrm{~cm} .\)

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