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Calculate the Fermi energy and Fermi speed for sodium, which has a density of \(0.97 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) and has one conduction electron per atom.

Short Answer

Expert verified
The Fermi energy is approximately 2.02 eV and the Fermi speed is about \(1.08 \times 10^6\) m/s.

Step by step solution

01

Understand the Fermi Energy Formula

The Fermi energy \(E_F\) is given by the formula \(E_F = \frac{h^2}{8m} \left( \frac{3n}{\pi} \right)^{2/3}\), where \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{ J s}\), \(m\) is the electron mass \(9.109 \times 10^{-31} \text{ kg}\), and \(n\) is the electron number density.
02

Find the Number Density

To find \(n\), the electron number density, use \(n = \frac{N_A \times \text{density}}{M}\), where \(N_A = 6.022 \times 10^{23} \text{ mol}^{-1}\) is Avogadro's number, and \(M = 22.99 \text{ g/mol} = 22.99 \text{ kg/mol}\) is the molar mass of sodium. Substituting, \(n = \frac{6.022 \times 10^{23} \times 0.97 \times 10^3}{22.99} \approx 2.53 \times 10^{28} \text{ electrons/m}^3\).
03

Calculate the Fermi Energy

Using the formula for \(E_F\), substitute the known values: \(E_F = \frac{(6.626 \times 10^{-34})^2}{8 \times 9.109 \times 10^{-31}} \left( \frac{3 \times 2.53 \times 10^{28}}{\pi} \right)^{2/3} \approx 3.24 \times 10^{-19} \text{ J}\).
04

Convert Fermi Energy to Electronvolts

Convert \(E_F\) from joules to electronvolts, knowing that 1 eV = \(1.602 \times 10^{-19} \text{ J}\). Thus, \(E_F \approx \frac{3.24 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 2.02 \text{ eV}\).
05

Calculate Fermi Speed

The Fermi speed \(v_F\) is given by \(v_F = \sqrt{\frac{2E_F}{m}}\). Substituting for \(E_F = 3.24 \times 10^{-19} \text{ J}\) and \(m = 9.109 \times 10^{-31} \text{ kg}\), we find \(v_F = \sqrt{\frac{2 \times 3.24 \times 10^{-19}}{9.109 \times 10^{-31}}} \approx 1.08 \times 10^6 \text{ m/s}\).

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Key Concepts

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

Electron Number Density
The concept of electron number density is crucial when dealing with the properties of metals like sodium. It tells us how many conduction electrons are present in a unit volume of the material. This value is usually calculated using the formula:
  • \( n = \frac{N_A \times \text{density}}{M} \)
where:
  • \( n \) is the electron number density in electrons per cubic meter.
  • \( N_A \) is Avogadro's number, approximately \( 6.022 \times 10^{23} \text{mol}^{-1} \).
  • The density of the metal is given in kilograms per cubic meter.
  • \( M \) is the molar mass, which for sodium is \( 22.99 \text{ kg/mol} \).
This formula effectively links the atomic properties of the element (sodium in this case) with its macroscopic electron behavior. By knowing the density and molar mass of sodium, we can determine that sodium has roughly \( 2.53 \times 10^{28} \) electrons per cubic meter available for conduction. This high electron density is part of what makes metals good conductors.
Fermi Speed
Fermi speed is a concept that stems from the Fermi energy, which describes the maximum energy that electrons in a material can have at absolute zero temperature. The formula to find the Fermi speed \( v_F \) is:
  • \( v_F = \sqrt{\frac{2E_F}{m}} \)
where:
  • \( E_F \) is the Fermi energy, which was calculated for sodium as approximately \( 3.24 \times 10^{-19} \text{ J} \).
  • \( m \) is the electron mass, \( 9.109 \times 10^{-31} \text{ kg} \).
Given these values, we can determine that the Fermi speed for sodium is approximately \( 1.08 \times 10^6 \text{ m/s} \). This speed represents the velocity of conduction electrons in a metal at absolute zero, allowing a deep understanding of electron behavior in solid-state physics. At such high speeds, electrons can traverse a vast number of atomic distances in a very short time, which is critical in understanding how electricity moves through a wire.
Sodium
Sodium is not just an essential element for life, but also a significant metal in physics due to its simple electronic structure. It has only one conduction electron per atom, making it an excellent candidate for studying basic principles of electron behavior in metals. Its low density, \( 0.97 \times 10^{3} \text{ kg/m}^3 \), and accessible atomic properties simplify calculations, as we've explored with electron number density and Fermi speed.In its metallic form, sodium's single conduction electron contributes to its high electron number density, leading to effective conductivity. Physically, sodium is soft and highly reactive, especially with water, producing hydrogen gas and heat which can ignite the hydrogen. This makes sodium interesting yet challenging to handle in certain experiments. Its reactivity is a practical reminder of the significant role electron configuration and density play, both in theoretical physics and in everyday chemical reactions.

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Most popular questions from this chapter

Do we need to consider quantum effects for everyday rotating objects? Estimate the differences between rotational energy levels for a spinning baton compared to the energy of the baton. Assume the baton consists of a uniform \(32-\mathrm{cm}\) -long bar with a mass of \(260 \mathrm{~g}\) and two small end masses, each of mass \(380 \mathrm{~g},\) and that it rotates at \(1.6 \mathrm{rev} / \mathrm{s}\)

Calculate the longest-wavelength photon that can cause an electron in silicon \(\left(E_{\mathrm{g}}=1.14 \mathrm{eV}\right)\) to jump from the valence band to the conduction band.

The neutrons in a neutron star (Chapter 44) can be treated as a Fermi gas with neutrons in place of the electrons in our model of an electron gas. Determine the Fermi energy for a neutron star of radius \(12 \mathrm{~km}\) and mass 2.5 times that of our Sun. Assume that the star is made entirely of neutrons and is of uniform density.

The band gap of silicon is \(1.14 \mathrm{eV}\). \((a)\) For what range of wavelengths will silicon be transparent? (See Example \(40-14 .)\) In what region of the electromagnetic spectrum does this transparent range begin? (b) If window glass is transparent for all visible wavelengths, what is the minimum possible band gap value for glass (assume \(\lambda=450 \mathrm{nm}\) to \(750 \mathrm{nm}\) )? [Hint: If the photon has less energy than the band gap, the photon will pass through the solid without being absorbed.

(1I) For a one-dimensional potential well of width \(\ell,\) start with \(E=n^{2} \frac{h^{2}}{8 m \ell^{2}}, \quad n=1,2,3, \cdots\) and show that the number of states per unit energy interval for an electron gas is given by \(g_{\ell}(E)=\sqrt{\frac{8 m \ell^{2}}{h^{2} E}}\) Remember that there can be two electrons (spin up and spin down) for each value of \(n .[\) Hint. Write the quantum number \(n\) in terms of \(E .\) Then \(g_{t}(E)=2 d n / d E\) where \(d n\) is the number of energy levels between \(E\) and \(E+d E . ]\)

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