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Germanium has a band gap of 0.67 eV. Doping with arsenic adds donor levels in the gap 0.01 eV below the bottom of the conduction band. At a temperature of 300 K, the probability is 4.4 \(\times\) 10\(^-$$^4\) that an electron state is occupied at the bottom of the conduction band. Where is the Fermi level relative to the conduction band in this case?

Short Answer

Expert verified
The Fermi level is 0.417 eV below the conduction band.

Step by step solution

01

Understand the Problem

We need to calculate the position of the Fermi level relative to the conduction band in doped Germanium at 300 K. Given that the probability of the conduction band being occupied is 4.4 \( \times \) 10\(^{-4}\), we will use the Fermi-Dirac distribution formula to find this position.
02

Recall the Fermi-Dirac Distribution

The probability \( f(E) \) that an electron occupies a state at energy level \( E \) is given by the Fermi-Dirac distribution: \[f(E) = \frac{1}{e^{(E - E_F)/(kT)} + 1}\]where \( E_F \) is the Fermi level, \( k \) is the Boltzmann constant \( 8.6173 \times 10^{-5} \) eV/K, and \( T \) is the temperature in Kelvin.
03

Set Up the Equation

Given that the probability \( f(E) \) at the conduction band edge is 4.4 \( \times \) 10\(^{-4}\), substitute this into the Fermi-Dirac distribution with \( E = E_C \) (bottom of the conduction band), where:- \( E_C \) is 0 (setting conduction band as reference at \( E = 0 \))- \( T = 300 \) KSo the equation becomes:\[0.00044 = \frac{1}{e^{(0 - E_F)/(300 \times 8.6173 \times 10^{-5})} + 1}\]
04

Solve for Fermi Level \( E_F \)

We first isolate the exponential term:- Rearrange to get: \[e^{(E_F)/(300 \times 8.6173 \times 10^{-5})} = \frac{1}{0.00044} - 1\]- Calculate the right side: \[1/0.00044 - 1 \approx 2272.27\]- Take the natural logarithm:\[\ln(2272.27) = \frac{E_F}{300 \times 8.6173 \times 10^{-5}}\]\[E_F = 300 \times 8.6173 \times 10^{-5} \times \ln(2272.27)\]- Calculate \( E_F \):\[E_F \approx 0.417 \, \text{eV}\]
05

Interpret the Result

The Fermi level \( E_F \) is 0.417 eV below the bottom of the conduction band in this case, meaning it is at \( -0.417 \, \text{eV} \) relative to the conduction band.

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Key Concepts

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

Band gap
In semiconductors, the band gap is a vital concept representing the energy difference between the valence band, which houses electrons, and the conduction band, where electrons are free to move, thereby contributing to electrical conduction. The magnitude of the band gap determines the electrical properties of the material. For instance, a small band gap indicates that electrons can easily be excited to the conduction band, thus making the material a good semiconductor. In the context of Germanium, which has a band gap of 0.67 eV, this specific energy gap allows for moderate electrical conductivity. This band gap is crucial in understanding how semiconductors behave when subjected to external influences like temperature or doping.
The band gap affects how well a semiconductor can conduct electricity. If the band gap is too large, it acts more like an insulator, making it difficult for electrons to gain enough energy to jump into the conduction band. Conversely, if it is too small, the semiconductor might act more like a conductor. Therefore, the band gap is a balancing act to achieve desired electrical properties in electronic applications.
Doping in semiconductors
Doping is the intentional introduction of impurities into a semiconductor crystal to change its electrical properties. The impurities added are called dopants. In the case of n-type doping, as seen with Germanium doped with arsenic in our case exercise, the dopants add extra electrons. These additional electrons increase the conductivity of the semiconductor. Specifically, arsenic introduces donor levels just below the conduction band, in this case 0.01 eV below.
When these donor levels are present, they can easily donate electrons to the conduction band. This makes the material more conductive because there are more charge carriers available to move through the semiconductor.
  • n-type doping adds extra electrons to the material.
  • Increases conductivity by providing additional charge carriers.
Doping is essential for creating electronic devices like diodes and transistors because it allows engineers to control the flow of electrons effectively by adjusting the doping level.
Fermi-Dirac distribution
The Fermi-Dirac distribution is a statistical model that describes the probability of occupancy of energy states by electrons in a system. It accounts for the fact that electrons are fermions, which obey the Pauli exclusion principle, meaning no two electrons can occupy the same quantum state simultaneously.
The distribution is given by:\[ f(E) = \frac{1}{e^{(E - E_F)/(kT)} + 1}\]where:
  • \(E\) is the energy of the state.
  • \(E_F\) is the Fermi level, a crucial concept that denotes the energy level at which the probability of an electron occupying that state is 50%.
  • \(k\) is the Boltzmann constant.
  • \(T\) is the absolute temperature in Kelvin.
In our example with Germanium at 300 K, the Fermi level indicates the energy at which electron occupancy probability changes, in this case, it influences how likely electrons are to occupy states in the conduction band. The calculated probability helps predict electronic behavior by showing that even if the system is at room temperature, electrons are already populating certain energy levels, affecting conductivity.

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Most popular questions from this chapter

(a) The equilibrium separation of the two nuclei in an NaCl molecule is 0.24 nm. If the molecule is modeled as charges \(+e\) and \(-e\) separated by 0.24 nm, what is the electric dipole moment of the molecule (see Section 21.7)? (b) The measured electric dipole moment of an NaCl molecule is \(3.0 \times 10^{-29}\) \(C \cdot m\). If this dipole moment arises from point charges \(+q\) and \(-q\) separated by 0.24 nm, what is \(q\)? (c) A definition of the \(fractional\) \(ionic\) \(character\) of the bond is \(q/e\). If the sodium atom has charge \(+e\) and the chlorine atom has charge \(-e\), the fractional ionic character would be equal to 1. What is the actual fractional ionic character for the bond in NaCl? (d) Theequilibrium distance between nuclei in the hydrogen iodide (HI) molecule is 0.16 nm, and the measured electric dipole moment of the molecule is \(1.5 \times 10^{-30}\) \(C \cdot m\). What is the fractional ionic character for the bond in HI? How does your answer compare to that for NaCl calculated in part (c)? Discuss reasons for the difference in these results.

Compute the Fermi energy of potassium by making the simple approximation that each atom contributes one free electron. The density of potassium is \(851 kg/m^{3}\), and the mass of a single potassium atom is \(6.49 \times 10^{-26}\) kg.

A hypothetical diatomic molecule of oxygen \((mass = 2.656 \times 10^{-26} kg)\) and hydrogen \((mass = 1.67 \times 10^{-27} kg)\) emits a photon of wavelength 2.39 \(\mu\)m when it makes a transition from one vibrational state to the next lower state. If we model this molecule as two point masses at opposite ends of a massless spring, (a) what is the force constant of this spring, and (b) how many vibrations per second is the molecule making?

The hydrogen iodide (HI) molecule has equilibrium separation 0.160 nm and vibrational frequency \(6.93 \times 10^{13}\) Hz. The mass of a hydrogen atom is \(1.67 \times 10^{-27}\) kg, and the mass of an iodine atom is 2.11 \(\times\) 10\(^-$$^2$$^5\) kg. (a) Calculate the moment of inertia of HI about a perpendicular axis through its center of mass. (b) Calculate the wavelength of the photon emitted in each of the following vibrationrotation transitions: (i) \(n = 1\), \(l = 1 \rightarrow n = 0\), \(l = 0\); (ii) \(n = 1\), \(l = 2\rightarrow n = 0\), \(l = 1\); (iii) \(n = 2\), \(l = 2\rightarrow n = 1\), \(l = 3\).

Consider a system of \(N\) free electrons within a volume \(V\). Even at absolute zero, such a system exerts a pressure \(p\) on its surroundings due to the motion of the electrons. To calculate this pressure, imagine that the volume increases by a small amount \(dV\). The electrons will do an amount of work \(p\) \(dV\) on their surroundings, which means that the total energy \(E_{tot}\) of the electrons will change by an amount \(dE_{tot} = -p dV\). Hence \(p = -dE_{tot}/dV\). (a) Show that the pressure of the electrons at absolute zero is \(p = \frac{3^{2/3}\pi^{4/3}\hbar^{2}}{5m} \lgroup \frac{N}{V}\ \rgroup^{5/3}\) (b) Evaluate this pressure for copper, which has a freeelectron concentration of \(8.45 \times 10^{28} m^{-3}\). Express your result in pascals and in atmospheres. (c) The pressure you found in part (b) is extremely high. Why, then, don't the electrons in a piece of copper simply explode out of the metal?

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