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

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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?

 

[Nate810]

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