1. PHYSICS 4340 Homework # 6
Due Mar. 8
Reading: You should finish Ch 3 through page 72 and then begin reading Chapter 7
– band structure.
1. An x-ray (lambda= 1.542 Å) powder photograph is made of a metallic element. The
structure is known to be either FCC or BCC. Diffraction rings are observed at the
following Bragg angles (theta), in degrees: 22.3, 26.0, 38.3, 46.6, 49.4, 61.2,...
What element is it?
Bragg Law is nλ=2dsinθ where d=2π/|G| and |G|=(2π/a)(h2+k2+l2)1/2
So nλ = 2(2π/|G|)sinθ
nλ 2π
h 2 + k 2 + l 2 = sin θ
4π a
⎛ nλ ⎞ 2 2
sin θ = ⎜ ⎟ (h + k 2 + l 2 )
2
⎝ 2a ⎠
We don’t know a, but by matching the sequence of Bragg angles with the allowed values
of h2+k2+l2, we can see if (λ/2a) is a constant (modulo n).
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If FCC, SG=0 unless h,k,l are all even or all odd.
h,k,l,=111,200,220,311,222,400,331 ….
h2+k2+l2=3,4,8,11,12,16 ….
If BCC, SG=0 unless h+k+l is even
h,k,l=110,200,220,310,222,312
h2+k2+l2=2,4,8,10,12,14 …

2. θ 22.3, 26.0, 38.3, 46.6, 49.4, 61.2,...
sin2θ 0.144 0.192 0.384 0.528 0.576 0.768
sin 2 θ
0.048 0.048 0.048 0.048 0.048 0.048
(h 2 + k 2 + l 2 ) fcc
sin 2 θ
0.072 0.048 0.048 0.052 0.048 0.055
€ (h 2 + k 2 + l 2 ) bcc
So the structure if FCC
€
So letting n=0, we see (λ/2a)=0.048 for FCC. There is no nice relation for BCC. So get
a=3.52Å, since λ =1.542 Å
It should be Nickel.
2. Kittel 3.1
L
2k 2 2π
ε= k= where λ = de Broglie wavelength
2M λ
2
 2 ⎛ 2π ⎞  2π 2
λ =2L, ε = ⎜ ⎟ = is the zero point K.E. per particle
€ 2M ⎝ 2L ⎠ 2ML2
3. Kittel 3.5
€ a) Linear ionic crystal
+ - + - + - 2N ions
Nαq 2
Attractive term: U = − where α=2ln2 for the 1D chain [page 71]
R
€

3. NA ⎛ A αq 2 ⎞
Repulsive term: U = n or U(R) = N ⎜ n − ⎟
R ⎝ R R ⎠
∂U ⎛ −nA αq 2 ⎞ nA αq 2
In equilibrium, = 0 = N ⎜ n +1 + 2 ⎟ , i.e. n +1 = 2
€ ∂R ⎝ R R ⎠ R R
A αq 2
= 2
R n +1 € nR €
⎛ Aq 2 αq 2 ⎞ αq 2 1
€ So U(R0 ) = N ⎜ − ⎟ = −N (1 − ) ; linear ionic crystal: α=2ln2, so
⎝ nR0 R0 ⎠ R0 n
2Nq 2 ln2 1
U(R0 ) = (1 − )
R0 n
€ b) Compress crystal so R0 R0(1-δ)
€
write a Taylor series expression for
2
∂U 1 2∂ U
U(R0 − R0δ ) = U(R0 ) − (R0δ ) + (R0δ ) − ...
∂R R0 2 ∂R 2 R0
The linear term vanishes at equilibrium (R=R0)
€
∂ 2U ∂ ⎛ −nA αq 2 ⎞ ⎛ n(n +1)A 2αq 2 ⎞
Now = 0 = N ⎜ n +1 + 2 ⎟ = N ⎜ − 3 ⎟
∂R 2 ∂R ⎝ R R ⎠ ⎝ R
n +2
R ⎠
nA αq 2
Remembering that = (part a)
R n +1 R 2
€
∂ 2U ⎛ αq 2 2αq 2 ⎞ αq 2
= N ⎜(n +1) 3 − 3 ⎟ = N(n −1) 3
∂R 2 R0
€⎝ R R ⎠ R
2
1 2∂ U
Total work to distort= U(R0 − R0δ ) − U(R0 ) = (R0δ )
€ 2 ∂R 2 R0
We want work/unit length. Here the unit length is R0.
€

4. 2
1 2∂ U
(R δ )
2 0 ∂R 2 R0 1
Since the total length is 2NR0, the work/unit length = ≡ Cδ 2
2NR0 2
∂ 2U
∂R 2 αq 2
R0 €
= (n −1);
2NR0 2R0
∂ 2U αq 2 ∂ 2U (n −1)q 2 log2
= (n −1); C = R0 2 2 =
€ ∂R 2 R0
2R0 4 ∂R R0
R0 2
4. Kittel 3.7
€
Divalent ionic crystals
BaO in the NaCl structure. We must calculate the Madelung energies.
αke 2
Ba+O-: U = − ; α = 1.748 (pg. 71); R 0 = 2.76Α (given)
R0
−1.748(14.40eV ⋅ A)
= = −9.12eV (gain)
2.76A
€
2
++ -- αk (2e)
Ba O : U = − = 4⋅ (−9.12) = −36.48eV (gain)
R0
€
To create Ba+ and O- from Ba and O costs 5.19eV and gains us 1.5eV for a total cost of
3.69eV.
€
To create Ba++ and O— from Ba+ and O- costs 9.96eV and gains us (-9.0eV), for a total
cost of 18.96eV. Relative to Ba and O, Ba++ and O— cost us 22.65eV.
So the Ba+ O- solid gains 9.12-3.69eV=5.43eV gain
The Ba++O—solid gains 36.48eV-22.65eV=13.83eV gain
Ba++O—is the stable form.