This document contains homework problems related to physics concepts like crystal structure, band structure, diffraction, and ionic bonding energies. It includes 4 problems:
1) Identifying the element of a metallic sample based on x-ray diffraction data by matching Bragg angles to allowed crystal structures. The element is determined to be nickel.
2) Calculating the zero-point kinetic energy of particles confined in a linear potential well based on the de Broglie wavelength formula.
3) Analyzing the interatomic potential and compressibility of a 1D linear ionic crystal chain, expressing the work required to compress the chain in terms of material properties.
4) Computing the Madelung energies of barium oxide
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This paper is devoted to homogenization of partial differential operators to use in special structure that is a plate allied to an elastic foundation when it is situated through the basic loads (especially with the harmonic forces) with a Non-axially deformation of the cantilever. Furthermore, it contains the Equations of motion that they can be derived from degenerate Non-linear elliptic ones. Through the mentioned processes, there exists many excess works related to computing the bounded conditions for this special application form of study (when the deformation phenomenon has occurred). At the end of the article whole results of the study on a circular plate are debated and new ways assigned to them are discussed. Afterwards all the processes are formulised with the collection of contracting sequences and expanding sequences integrable functions that are intrinsically joints with the characteristic functions to expanding the behaviour of an elastic foundation. Thenceforth all the resultant functions are sets and compared with the other ones (without the loads). Sample pictures and analysis of the study were employed with the ANSYS software to obtain the better observations and conclusions.
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This paper is devoted to homogenization of partial differential operators to use in special structure that is a plate allied to an elastic foundation when it is situated through the basic loads (especially with the harmonic forces) with a Non-axially deformation of the cantilever. Furthermore, it contains the Equations of motion that they can be derived from degenerate Non-linear elliptic ones. Through the mentioned processes, there exists many excess works related to computing the bounded conditions for this special application form of study (when the deformation phenomenon has occurred). At the end of the article whole results of the study on a circular plate are debated and new ways assigned to them are discussed. Afterwards all the processes are formulised with the collection of contracting sequences and expanding sequences integrable functions that are intrinsically joints with the characteristic functions to expanding the behaviour of an elastic foundation. Thenceforth all the resultant functions are sets and compared with the other ones (without the loads). Sample pictures and analysis of the study were employed with the ANSYS software to obtain the better observations and conclusions.
Question answers Optical Fiber Communications 4th Edition by Keiserblackdance1
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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).
€
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.