# Calculate the ratio of the Fermi wavevector to the radius of the largest sphere

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In summary, sodium is a monovalent free-electron metal with a body-centred cubic structure. The ratio of the Fermi wavevector to the radius of the largest sphere in the first Brillouin zone is approximately \sqrt{\frac{2}{\pi}}:1. The Fermi energy of sodium can be calculated using its cubic cell side and is equal to 5.09 × 10^{-19} J. The change in Fermi energy per degree temperature rise can also be calculated using the linear thermal expansion coefficient and is equal to -1.81 × 10^{-23} JK^{-1}.
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## Homework Statement

Sodium is, to a good approximation, a monovalent free-electron metal which has the
body-centred cubic structure.
(a) Calculate the ratio of the Fermi wavevector to the radius of the largest sphere that
can be inscribed in the first Brillouin zone. Remember that the reciprocal lattice of bcc
is fcc, and that the Brillouin zone is the cell formed in reciprocal space by the bisector
planes of the vectors from the origin to every reciprocal lattice point.
(b) If the cubic cell side of Sodium is 0.4225 nm, calculate its Fermi energy.
(c) Calculate the change in the Fermi energy of Sodium per degree temperature rise if
its linear thermal expansion coefficient is $\alpha$= 7 × 10$^{-5}$ K$^{-1}$.
(The linear thermal expansion coefficient is related to the lattice parameter, a, by
$\frac{da}{dT}$=$\alpha$a)

## The Attempt at a Solution

a) Fermi wavevector:
k$_{F}$=$\sqrt{\frac{2}{\pi}}$($\frac{\pi}{a}$)

r=$\frac{\pi}{a}$

so the ratio is $\sqrt{\frac{2}{\pi}}$:1

Is it really as simple as that or have I done something wrong?

b) Fermi energy:E_{F}=\frac{h^{2}}{2m}k_{F}^{2}k_{F}=\sqrt{\frac{2}{\pi}}(\frac{\pi}{a})E_{F}=\frac{h^{2}}{2m}(\sqrt{\frac{2}{\pi}}(\frac{\pi}{a}))^{2}E_{F}=\frac{h^{2}}{2m}\frac{4}{\pi a^{2}}a=0.4225 nmE_{F}=\frac{h^{2}}{2m}\frac{4}{\pi(0.4225nm)^{2}}E_{F}= 5.09 × 10^{-19} Jc) \frac{dE_{F}}{dT}=\frac{\partial E_{F}}{\partial a}\frac{da}{dT}\frac{\partial E_{F}}{\partial a}= -\frac{8h^{2}}{\pi m a^{3}}\frac{da}{dT}= \alpha aso \frac{dE_{F}}{dT}= -\frac{8h^{2} \alpha a^{2}}{\pi m a^{3}}dE_{F}= -\frac{8h^{2} \alpha a^{2}}{\pi m a^{3}} dTdE_{F}= -\frac{8h^{2} \alpha a}{\pi m } dTdE_{F}= -1.81 × 10^{-23} JK^{-1} dT

## What is the formula for calculating the ratio of the Fermi wavevector to the radius of the largest sphere?

The formula for calculating this ratio is given by:
Fermi wavevector/radius of largest sphere = (3/4π)^(1/3) * (number of particles/volume)^(1/3)

## Why is it important to calculate this ratio in physics?

This ratio is important in physics because it helps us understand the behavior of particles in a system. It gives us information about the density and energy of the particles, which is crucial in many areas of physics such as quantum mechanics and condensed matter physics.

## What is the significance of the Fermi wavevector in this ratio?

The Fermi wavevector is a fundamental constant in quantum mechanics that represents the maximum momentum a particle can have in a system. It is closely related to the energy and density of particles in a system, making it an important factor in calculating this ratio.

## How does the ratio change with different numbers of particles and volumes?

The ratio will change as the number of particles and the volume of the system change. As the number of particles increases, the ratio will decrease, indicating a higher density of particles in the system. Similarly, as the volume increases, the ratio will decrease, indicating a lower density of particles.

## Can this ratio be used to predict the behavior of particles in a system?

Yes, this ratio can give us valuable information about the behavior of particles in a system. It can help us understand the energy and density of the particles, which can then be used to make predictions about their behavior and interactions with each other.

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