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What mass of phosphorus is needed to dope \(1.0\) \(\mathrm{g}\) of silicon so that the number density of conduction electrons in the silicon is increased by a multiply factor of \(10^{6}\) from the \(10^{16} \mathrm{~m}^{-3}\) in pure silicon.

Short Answer

Expert verified
You need approximately \(2.21 \times 10^{-7} \, \text{g}\) of phosphorus.

Step by step solution

01

Determine the Number Density of Doped Electrons

The problem states that pure silicon has a number density of conduction electrons as \(10^{16} \, \text{m}^{-3}\). To increase it by a factor of \(10^{6}\), the new number density becomes \(10^{22} \, \text{m}^{-3}\).
02

Calculate the Total Number of Electrons Needed

Given the volume of silicon, we use the mass (1g) and density of silicon (\(2.33 \, \text{g/cm}^3\)) to find its volume:\[V = \frac{1 \, \text{g}}{2.33 \, \text{g/cm}^3} = 0.429 \, \text{cm}^3 = 4.29 \times 10^{-7} \, \text{m}^3.\]Now calculate the total number of conduction electrons needed:\[N = 10^{22} \, \text{electrons/m}^3 \times 4.29 \times 10^{-7} \, \text{m}^3 = 4.29 \times 10^{15} \, \text{electrons}.\]
03

Calculate the Moles of Phosphorus Atoms Required

Each phosphorus atom contributes one electron to the conduction band. Therefore, the moles of phosphorus atoms needed equals the number of electrons, \(4.29 \times 10^{15}\), divided by Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\)):\[n = \frac{4.29 \times 10^{15}}{6.022 \times 10^{23}} \approx 7.123 \times 10^{-9} \, \text{mol}.\]
04

Convert Moles of Phosphorus to Grams

The molar mass of phosphorus is approximately \(30.97 \, \text{g/mol}\). Hence, the mass \(m\) of phosphorus required is:\[m = 7.123 \times 10^{-9} \, \text{mol} \times 30.97 \, \text{g/mol} = 2.206 \times 10^{-7} \, \text{g}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conduction Electrons
In the world of semiconductors, conduction electrons play a vital role. These are the electrons that are free to move and enable electrical conduction within a material. In pure silicon, the number of conduction electrons is relatively low, which limits its electrical conductivity.

When a material has more conduction electrons, its ability to conduct electricity improves significantly. Doping—a process where foreign atoms are introduced to a semiconductor—alters the number of conduction electrons. This can change how the semiconductor behaves. By adding donor atoms, such as phosphorus, more electrons are available for conduction, increasing the semiconductor's efficiency.

In brief:
  • Conduction electrons are free electrons contributing to electrical conduction.
  • Increased conduction electrons result in better conductivity.
  • Doping allows control over the number of conduction electrons.
Number Density
Number density refers to the number of conduction electrons present per unit volume in a semiconductor. It's an essential measure that tells us how many charge carriers are available within a given space to support electric current flow.

In silicon, a typical number density for conduction electrons is about \(10^{16} \, \text{m}^{-3}\). By adjusting the number density through processes such as doping, the electrical properties can be significantly altered. For example, phosphorus doping can increase the electron density in silicon by a factor of \(10^6\), resulting in a much denser field of electrons.

Here are a few points to remember about number density:
  • It's a measure of electron concentration within a material.
  • Higher number density means more electrons are present to conduct electricity.
  • Processes like doping can dramatically change the number density.
Phosphorus Doping
Phosphorus doping is a method used to enhance silicon's conductivity by introducing phosphorus atoms into the silicon crystal lattice. Silicon itself is a semiconductor with limited electrical conductivity, but when phosphorus atoms are added, they release free electrons.

Phosphorus atoms have five valence electrons—one more than silicon. When phosphorus replaces some of the silicon atoms in the lattice, the extra electrons become free conduction electrons. This increase in conduction electrons improves the material's overall conductivity.

Some key points about phosphorus doping:
  • It's a process that boosts electron availability by adding donor atoms.
  • Phosphorus atoms contribute extra electrons compared to silicon.
  • Doping changes the electrical properties of silicon, making it more conductive.
Silicon Material Properties
Silicon is a crucial element in the electronics industry, mainly due to its semiconductor properties. It has a unique atomic structure, which makes it ideal for forming the base material for integrated circuits and other electronic components.

Silicon has four valence electrons, which allow it to form stable covalent bonds within a crystal lattice. Its intrinsic form is not highly conductive as a semiconductor. However, its properties can be modified through doping, making it an excellent material for electronic devices.

Why silicon is so important:
  • It has a stable crystal structure that supports effective doping.
  • Its semiconductor nature makes it versatile for electronic purposes.
  • Doping allows for tailoring its conductivity to fit various needs in electronic components.

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Most popular questions from this chapter

What is the probability that a state \(0.0620 \mathrm{eV}\) above the Fermi energy will be occupied at (a) \(T=0 \mathrm{~K}\) and (b) \(T=320 \mathrm{~K}\) ?

Copper, a monovalent metal, has molar mass \(63.54 \mathrm{~g} / \mathrm{mol}\) and density \(8.96 \mathrm{~g} / \mathrm{cm}^{3}\). What is the number density \(n\) of conduction electrons in copper?

A silicon-based MOSFET has a square gate \(0.50 \mu \mathrm{m}\) on edge. The insulating silicon oxide layer that separates the gate from the \(p\) -type substrate is \(0.20 \mu \mathrm{m}\) thick and has a dielectric constant of 4.5. (a) What is the equivalent gate-substrate capacitance (treating the gate as one plate and the substrate as the other plate)? (b) Approximately how many elementary charges \(e\) appear in the gate when there is a gate-source potential difference of \(1.0 \mathrm{~V}\) ?

In a simplified model of an undoped semiconductor, the actual distribution of energy states may be replaced by one in which there are \(N_{v}\) states in the valence band, all these states having the same energy \(E_{v}\), and \(N_{c}\) states in the conduction band, all these states having the same energy \(E_{c} .\) The number of electrons in the conduction band equals the number of holes in the valence band. (a) Show that this last condition implies that $$ \frac{N_{c}}{\exp \left(\Delta E_{c} / k T\right)+1}=\frac{N_{v}}{\exp \left(\Delta E_{v} / k T\right)+1}, $$ in which $$ \Delta E_{c}=E_{c}-E_{\mathrm{F}} \quad \text { and } \quad \Delta E_{v}=-\left(E_{v}-E_{\mathrm{F}}\right). $$ (b) If the Fermi level is in the gap between the two bands and its distance from each band is large relative to \(k T\), then the exponentials dominate in the denominators. Under these conditions, show that $$ E_{\mathrm{F}}=\frac{\left(E_{c}+E_{v}\right)}{2}+\frac{k T \ln \left(N_{v} / N_{c}\right)}{2} $$ and that, if \(N_{v} \approx N_{c}\), the Fermi level for the undoped semiconductor is close to the gap's center.

The Fermi energy of copper is \(7.0 \mathrm{eV}\). Verify that the corresponding Fermi speed is \(1600 \mathrm{~km} / \mathrm{s}\).

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