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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 \(\mathrm{K}\) , 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 approximately 0.01 eV below the conduction band.

Step by step solution

01

Understand the concept

The Fermi level is a concept in solid state physics which represents the top of the collection of electron energy levels at absolute zero temperature. In semiconductors like germanium, the Fermi level can change due to external factors like temperature and doping.
02

Apply the Fermi-Dirac distribution formula

In semiconductors, the probability that an energy state is occupied by an electron is described by the Fermi-Dirac distribution: \[ f(E) = \frac{1}{1 + e^{(E - E_F)/kT}} \] where \(E\) is the energy of the state, \(E_F\) is the Fermi level, \(k\) is the Boltzmann constant (\(8.617 \times 10^{-5} \text{ eV/K}\)), and \(T\) is the temperature in Kelvin. Given \(T = 300\) \(\text{K}\) and \(f(E) = 4.4 \times 10^{-4}\), we'll set \(E\) as the conduction band minimum, which is \(0.01 \text{ eV}\) below the actual conduction band node.
03

Rearrange and solve for the Fermi level

Given the known values, rearrange the equation to solve for \(E_F\):\[ 4.4 \times 10^{-4} = \frac{1}{1 + e^{(0.01 - E_F)/kT}} \] Solving for \(E_F\), convert the exponential term:\[ e^{(0.01 - E_F)/kT} = \frac{1}{4.4 \times 10^{-4}} - 1 \]Calculating \( e^{(0.01 - E_F)/kT} \), find the corresponding exponent's value. Use logs if necessary to further simplify and find \(E_F\).
04

Calculate specific values

Specify the Boltzmann constant in appropriate units and perform numerical estimation:\[ e^{(0.01 - E_F)/(8.617 \times 10^{-5} \times 300)} \approx \frac{1}{4.4 \times 10^{-4}} - 1 \] This yields \( e^{(0.01 - E_F)/(0.025851)} \approx 2272 \).Using logarithms to solve,\[ 0.01 - E_F = kT \log(2272) \]Finally calculate the value of \(E_F\):\[ E_F = 0.01 - 0.025851 \times \log(2272) \].
05

Conclusion

Perform the above calculations, leading to an approximation with logarithms applied. These computations outline the finding of \(E_F\) or the Fermi level's depiction relative to the band structure. Typically it results less than 0.01 eV below the conduction band's bottom, signifying its minute positioning inside the conduction band gap as required.

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Key Concepts

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

Band Gap Energy
In the world of semiconductors, **band gap energy** is a key concept. It represents the energy difference between the top of the valence band and the bottom of the conduction band. This gap must be overcome for an electron to move from a bound state (within the atom) to a conductive state (across the material). Germanium, specifically, has a band gap of 0.67 eV. This relatively small gap means electrons need only a small amount of energy to become conductive at room temperature.
  • A smaller band gap means better conductivity at lower temperatures.
  • A larger band gap requires more energy for electrons to jump across.
Having a precise understanding of band gap energy helps in designing and tuning electronic devices like transistors and diodes.
Fermi-Dirac Distribution
The **Fermi-Dirac distribution** is a statistical model that describes the probability of an electron occupying a certain energy state at thermal equilibrium. This probability is influenced by the temperature and the energy level of the electrons compared to the Fermi level.The distribution function is given by:\[ f(E) = \frac{1}{1 + e^{(E - E_F)/kT}} \]Where:
  • \(E\) is the energy of the electron state.
  • \(E_F\) is the Fermi level.
  • \(k\) is Boltzmann's constant.
  • \(T\) is the absolute temperature.
At absolute zero, all states below the Fermi level are filled, and none above it are. As temperature increases, electrons gain enough thermal energy to occupy states at higher energies than the Fermi level. Calculating these probabilities is crucial in studying semiconductor behaviors, especially to determine the position of the Fermi level relative to the conduction band.
Doping in Semiconductors
**Doping** is a process used to modify the electrical properties of semiconductors. By adding impurities into an intrinsic semiconductor, its electrical behavior can be tailored. For germanium, doping with arsenic introduces additional energy levels in the band gap. These are called donor levels, situated just below the conduction band. In this scenario:
  • Doped arsenic atoms donate extra electrons to the conduction band.
  • This increases the number of charge carriers available for conduction.
  • It effectively shifts the Fermi level closer to the conduction band.
Doping is vital for creating n-type and p-type semiconductors, essential for crafting components like transistors and diodes.
Electron Probability Occupancy
**Electron probability occupancy** describes the likelihood of an electron occupying a particular energy state within a semiconductor. This is fundamentally governed by the Fermi-Dirac distribution.In the given exercise, the probability of an electron occupying a state at the bottom of the conduction band is calculated as \(4.4 \times 10^{-4}\) at a temperature of 300 K. This probability assists in finding the Fermi level's position as part of the semiconductor's energy structure.Key points include:
  • At higher energies, this probability decreases exponentially.
  • Temperature plays a significant role in occupancy probabilities.
  • These probabilities provide insights into electrical conductivity behaviors.
From practical applications in technology to fundamental physics research, understanding electron occupancy is essential for engaging with semiconductor materials.

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Most popular questions from this chapter

Hydrogen is found in two naturally occurring isotopes; normal hydrogen (containing a single proton in its nucleus) and deuterium (having a proton and a neutron). Assuming that both molecules are the same size and that the proton and neutron have the same mass (which is almost the case), find the ratio of (a) the energy of any given rotational state in a diatomic hydrogen molecule to the energy of the same state in a diatomic deuterium molecule and (b) the energy of any given vibrational state in hydrogen to the same state in deuterium (assuming that the force constant is the same for both molecules). Why is it physically reasonable that the force constant would be the same for hydrogen and deuterium molecules?

(a) Consider the hydrogen molecule \(\left(\mathrm{H}_{2}\right)\) to be a simple harmonic oscillator with an equilibrium spacing of 0.074 \(\mathrm{nm}\) , and estimate the vibrational energy-level spacing for \(\mathrm{H}_{2}\) . The mass of a hydrogen atom is \(1.67 \times 10^{-27} \mathrm{kg}\) . (Hint: Estimate the force constant by equating the change in Coulomb repulsion of the protons, when the atoms move slightly closer together than \(r_{0},\) to the "spring" force. That is, assume that the chemical binding force remains approximately constant as \(r\) is decreased slightly from \(r_{0} .\) ) (b) Use the results of part (a) to calculate the vibrational energy-level spacing for the deuterium molecule, \(\mathrm{D}_{2}\) . Assume that the spring constant is the same for \(\mathrm{D}_{2}\) as for \(\mathrm{H}_{2}\) . The mass of a deuterium atom is \(3.34 \times 10^{-27} \mathrm{kg}\) .

42.27. For a solid metal having a Fermi energy of 8.500 eV, what is the probability, at room temperature, that a state having an energy of 8.520 eV is occupied by an electron?

An Ionic Bond. (a) Calculate the electric potential energy for \(a K^{+}\) ion and \(a B r^{-}\) ion separated by a distance of \(0.29 \mathrm{nm},\) the equilibrium separation in the KBr molecule. Treat the ions as point charges. (b) The ionization energy of the potassium atom is 4.3 \(\mathrm{eV}\) . Atomic bromine has an electron affinity of 3.5 \(\mathrm{eV}\) . Use these data and the results of part (a) to estimate the binding energy of the KBr molecule. Do you expect the actual binding energy to be higher or lower than your estimate? Explain your reasoning.

The hydrogen iodide (HI) molecule has equilibrium separation 0.160 \(\mathrm{nm}\) and vibrational frequency \(6.93 \times 10^{13} \mathrm{Hz}\) . The mass of a hydrogen atom is \(1.67 \times 10^{-27} \mathrm{kg}\) , and the mass of an iodine atom is \(2.11 \times 10^{-25} \mathrm{kg}\) . (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 vibration-rotation transitions: (i) \(n=1\) , \(l=1 \rightarrow n=0, \quad l=0 ; \quad(\text { ii) } n=1, \quad l=2 \rightarrow n=0, \quad l=1 ;\) (iii) \(n=2, l=2 \rightarrow n=1, l=3\)

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