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Problem set 3 and 4 for physics 463, winter 2008, which covers topics on molecular solids and ionic crystals, including lennard-jones potential, cohesive energy, nearest neighbor separation, madelung constant, lattice energy, phonon density of states, and debye approximation. The problems require calculations and comparisons with experimental values.
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Physics 463, Winter 2008 Problem Set 3 and 4, due Friday, February 22, by 5pm. page 1 of 2
U (r) = 4
σ r
( (^) σ
r
When considering the cohesive energy in a solid, we must sum up this interaction energy over all of the atoms. For a monoatomic fcc structure in which each atom has 12 nearest neighbors, the corresponding cohesive energy for a molecular solid with N atoms (neglecting the KE)
( (^) σ
R
( (^) σ
R
where R is the nearest neighbor distance and the 1/2 ensures that we count each atom only once.
(a) What is the origin of the attractive and repulsive terms in the LJ potential? (b) Find the nearest neighbor separation in terms of the parameter σ (hint, this occurs when the energy is minimized). (c) for Ne, = 0. 0031 eV and σ = 2. 74 A (this is obtained by measurements in the gas phase). For˚ soild Ne what is the theoretical cohesive energy per atom? what is the theoretical nearest neighbor separation? The experimental values are 3.13 ˚A and 0.02 eV respectively. The slight departure is due to quantum mechanical zero point motion of the Ne atoms. Do you expect the agreement to be better or worse as one moves down the periodic table (Ar, Kr, Xe)?
Uij = λe−rij^ /ρ^ ±
q^2 rij
where we use cgs units (as in Kittel), q is the charge of the ions, rij is the distance between the two ions, and we choose (+) or (-) depending on whether we are considering like ions (repulsive) or unlike ions (attractive) and λ and ρ are parameters to be extracted from the experimental data.
(a) Typically ρ is on order of 10% of the nearest neighbor separation. Thus, show that the exponential term can be neglected past nearest neighbors such that the total cohesive energy may be written for N pairs of positive and negative ions as
zλe−R/ρ^ −
αq^2 R
where z is the number of nearest neighbors, and R is the nearest neighbor separation negative ions, and α is the Madelung constant: α =
j
pij
where rij = pij R. (b) The Madelung constant depends only on the structure. For the sodium chloride structure it is 1.747565, for the cesium chloride structure it is 1.762675 and for the zinc-blende structure it is 1.6381. Calculate the lattice energy of KCl in the the zinc-blende, sodium chloride and zinc-blende structure. Compare with the experimental value of the lattice energy which is 7.17 eV per molecular unit. (See figure 3.10 in Kittel for a plot of the energy per KCl pair showing the Madelung and
Physics 463, Winter 2008 Problem Set 3 and 4, due Friday, February 22, by 5pm. page 2 of 2
repulsive contributions).
The experimentally determined values of R, ρ and zλ for KCl are:
R ρ zλ KCl 3.147 ˚A 0.326 ˚A 2. 05 × 10 −^8 erg
(a) Calculate the dispersion relationship for the two phonon branches in terms of the quantities given above assuming M 1 > M 2. Show that there is an energy gap between the acoustic and optical phonon branches at the edge of the Brillouin zone. How does the gap depend on the difference between the masses? Sketch the dispersion relation. (b) Show that in the limit that M 1 = M 2 that the gap closes and the dispersion relation is equivalent to that of a monoatomic crystal with lattice spacing a/ 2.
(a) Derive the general form of the lattice contribution to the specific heat:
CV = 9N kB
0
x^4 ex (ex^ − 1)^2
x.
where ΘD = ¯hωD/kB , is the Debye Temperature. You may assume that all three phonon branches have the same speed of sound v. (b) Show that at sufficiently low temperatures the specific heat is proportional to T 3 and that at very high temperatures the specific heat approaches the classical limit 3 N kB. hint: You may find the following integral useful: ∫ (^) ∞
0
x^3 ex^ − 1
x. =
n=
x^3 e−nxx. = 6
n=
n^4
π^4 15
(c) Derive an expression with the thermal energy density u = U/N for a crystal in the low temperature limit. Compare your result to the Stefan-Boltzmann law for black body radiation
uEM =
π^2 15
(kB T )^4 (¯hc)^3 and comment on the significance of the similarities and differences.