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Chapter 41: Problem 19

The Fermi energy for silver is \(5.5 \mathrm{eV}\). At \(T=0^{\circ} \mathrm{C}\), what are the probabilities that states with the following energies are occupied: (a) \(4.4 \mathrm{eV},(\mathrm{b}) 5.4 \mathrm{eV},(\mathrm{c}) 5.5 \mathrm{eV},(\mathrm{d}) 5.6 \mathrm{eV}\), and \((\mathrm{e}) 6.4 \mathrm{eV} ?(\mathrm{f})\) At what temperature is the probability \(0.16\) that a state with energy \(E=5.6 \mathrm{eV}\) is occupied?

Short Answer

Expert verified
At 0 K: (a) 1, (b) 1, (c) 0.5, (d) 0, (e) 0. At T 鈮� 699 K, P(5.6 eV) = 0.16.

Step by step solution

01

Understand the Fermi-Dirac distribution

The Fermi-Dirac distribution function gives the probability that a given energy state is occupied by a fermion (like an electron) at a temperature \( T \). It 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 energy, \( k \) is Boltzmann's constant (approximately \( 8.617 \times 10^{-5} \) eV/K), and \( T \) is the temperature in Kelvin.
02

Calculate probabilities at T=0 K

At absolute zero (\( T = 0 \) K), the Fermi-Dirac distribution simplifies. If \( E < E_F \), \( f(E) = 1 \). If \( E > E_F \), \( f(E) = 0 \). For \( E = E_F \), \( f(E) = 0.5 \). Thus, for the given energy levels:a) \( f(4.4 \, \mathrm{eV}) = 1 \) because \( 4.4 \, \mathrm{eV} < 5.5 \, \mathrm{eV} \).b) \( f(5.4 \, \mathrm{eV}) = 1 \) because \( 5.4 \, \mathrm{eV} < 5.5 \, \mathrm{eV} \).c) \( f(5.5 \, \mathrm{eV}) = 0.5 \) because \( 5.5 \, \mathrm{eV} = 5.5 \, \mathrm{eV} \).d) \( f(5.6 \, \mathrm{eV}) = 0 \) because \( 5.6 \, \mathrm{eV} > 5.5 \, \mathrm{eV} \). e) \( f(6.4 \, \mathrm{eV}) = 0 \) because \( 6.4 \, \mathrm{eV} > 5.5 \, \mathrm{eV} \).
03

Determine temperature for occupation probability

We want to find the temperature \( T \) such that the probability of occupancy \( f(E) = 0.16 \) for \( E = 5.6 \, \mathrm{eV} \). Rearrange the Fermi-Dirac distribution for \( T \):\[0.16 = \frac{1}{e^{(5.6 - 5.5)/(kT)} + 1}\]Solving this for \( T \) requires:\[e^{(5.6 - 5.5)/(kT)} = \frac{1}{0.16} - 1 = 5.25\]Taking the natural log:\[\frac{0.1}{kT} = \ln(5.25)\]Now solve for \( T \):\[T = \frac{0.1}{k \ln(5.25)}\]where \( k = 8.617 \times 10^{-5} \mathrm{eV/K} \). Solve the equation:\[T \approx \frac{0.1}{8.617 \times 10^{-5} \times 1.657} \approx 699 \mathrm{K}\]

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Key Concepts

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

Fermi energy
Fermi energy is a fundamental concept when it comes to understanding the behavior of electrons in metals at different temperatures. It represents the highest energy level that an electron can occupy at absolute zero temperature. In essence, it marks the borderline between occupied and unoccupied electron states at zero Kelvin.

In metals, the Fermi energy defines key properties related to electrical conduction and thermal capacity. When we mention the Fermi energy for a material like silver, which is given as 5.5 eV, it implies that at absolute zero, electrons fill up all available energy states up to this energy value. No states above this energy level are occupied at this zero-point.

Understanding Fermi energy helps explain why materials behave significantly differently at various temperatures. It sets the baseline for how electrons are distributed among the energy states of a material. At higher temperatures, however, electrons gain enough thermal energy to move to higher energy levels beyond the Fermi energy.
probability at absolute zero
The probability of electrons occupying certain energy states at absolute zero temperature is simple yet crucial. At absolute zero (

aMoreover, this probability distribution is pivotal in explaining how materials conduct electricity at very low temperatures.
Boltzmann's constant
Boltzmann's constant (

Some useful points related to Boltzmann's constant include:
  • It acts as a bridge between macroscopic and microscopic physics, linking temperature and energy.
  • The constant is crucial for predicting how the probabilities of energy state occupations change with temperature.
  • In the Fermi-Dirac distribution, it helps define how readily electrons can jump from one energy state to another as the temperature varies.

    Thus, Boltzmann's constant is indispensable in extending our understanding of temperature-related phenomena to the quantum level.
temperature dependence of electron states
Temperature plays a vital role in influencing the distribution of electrons in the energy states within a material. As temperatures rise from absolute zero, electrons are provided with additional thermal energy. This energy allows some electrons to get excited to higher energy levels, which were initially unoccupied.

The Fermi-Dirac distribution comes into the picture when dealing with non-zero temperatures. It illustrates how the probability of finding electrons in various energy states changes with temperature. Specifically:
  • At higher temperatures, electrons are more likely to be found in states with energies above the Fermi energy.
  • There is a smoother transition between the occupied and unoccupied states as compared to the sharp transition at absolute zero.
  • The distribution predicts that even states with energy slightly above the Fermi level can have a non-zero probability of being occupied as temperature increases.

    This temperature dependence is a cornerstone in understanding electrical conductivity, thermal properties of materials, and many other physical phenomena.

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Most popular questions from this chapter

Pure silicon at room temperature has an electron number density in the conduction band of about \(5 \times 10^{15} \mathrm{~m}^{-3}\) and an equal density of holes in the valence band. Suppose that one of every \(10^{7}\) silicon atoms is replaced by a phosphorus atom. (a) Which type will the doped semiconductor be, \(n\) or \(p ?\) (b) What charge carrier number density will the phosphorus add? (c) What is the ratio of the charge carrier number density (electrons in the conduction band and holes in the valence band) in the doped silicon to that in pure silicon?

A certain metal has \(1.70 \times 10^{28}\) conduction electrons per cubic meter. A sample of that metal has a volume of \(6.00 \times 10^{-6}\) \(\mathrm{m}^{3}\) and a temperature of \(200 \mathrm{~K}\). How many occupied states are in the energy range of \(3.20 \times 10^{-20} \mathrm{~J}\) that is centered on the energy \(4.00 \times 10^{-19} \mathrm{~J} ?\) (Caution: Avoid round-off in the exponential.)

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.

(a) Find the angle \(\theta\) hetween adjacent nearest-neighbor bonds in the silicon lattice. Recall that each silicon atom is bonded to four of its nearest neighbors. The four neighbors form a regular tetrahedron-a pyramid whose sides and base are equilateral triangles. (b) Find the bond length, given that the atoms at the corners of the tetrahedron are 388 pm apart.

The compound gallium arsenide is a commonly used semiconductor, having an energy gap \(E_{g}\) of \(1.43 \mathrm{eV}\). Its crystal structure is like that of silicon, except that half the silicon atoms are replaced by gallium atoms and half by arsenic atoms. Draw a flattened-out sketch of the gallium arsenide lattice, following the pattern of Fig. 41-10a. What is the net charge of the (a) gallium and (b) arsenic ion core? (c) How many electrons per bond are there? (Hint: Consult the periodic table in Appendix G.)
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