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Melting metal and casting pipe is a process that is inherently dirty and dangerous. The largest US manufacture, with plants in 10 States is McWane. Over.

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Presentation on theme: "Melting metal and casting pipe is a process that is inherently dirty and dangerous. The largest US manufacture, with plants in 10 States is McWane. Over."— Presentation transcript:

1 Melting metal and casting pipe is a process that is inherently dirty and dangerous. The largest US manufacture, with plants in 10 States is McWane. Over the past seven years nine McWane workers have been killed and at least 4,600 have been injured on the job. Real Metallurgy LIFE IN FOUNDRIES

2 Physical Metallurgy 5th Lecture MS&E 410 D.Ast

3 The density of states Assignment: Read Prof. Bakers lecture notes - Lecture 5. We learned in Lecture 2

4 Why the density of states D(E), in 3-D, varies with E 1. The density of states in k space is constant. The volume in k space occupied by one state (two electrons ) is v state = 8  3 /V entire crystal This simply comes from the condition that a wave must fit in a box with length L x,L,L z which requires = L, L/2, L/3, …. As k=2  / k= (2  /L) k= 2 (2  /L) k = 3 (2  /L) ………. V = L x L y L z

5 Because the density of states is constant in k space that we love k space ! Volume  k 3 Surface  k 2 Therefore N(E+dE) - N(E)  k 2 dk dN(E)  k 2 dk E  k 2 ; dE  2k dk ; dk  dE\/E dN(E)  E/ E dE dN/dE  E

6 Predicting for Alkali (bcc)

7 The fall of D(E) in the Alkali metals 1. What is falling of are empty states. 2. If the first B.Z. would be a sphere, D(E) would come down vertical. But the first B.Z. is not. Once you “expand the Fermi sphere to the point where it touches a zone face” further filling finds less and less states (relative to a sphere) The Fermi sphere in Alkali metals is entirely within the first B.Z.

8 bcc fcc 1st Zone 2nd Zone 3rd Zone Filling in 2 electrons we will penetrate through the 1st Zone

9 The free electron model for fcc The Fermi surface for one electron does not touch a B.Z. face. For two electrons per atom it penetrates past the hexagonal face (L point) but not the square face

10 The Noble Metals Cu, Ag, Au, have 1 s electron floating over a closed d shell and therefore are similar, electronically, to the Alkali metals. The Fermi surface, again, is a sphere. However, it has necks that connect to the next cells. The necks come from the underlying d electrons. The area of the neck is 1/51 of the area of a cross section of the sphere From Ashcroft and Mermin’s book

11 In the d electron case, you get more insight using a “chemical” model that is based on local atomic function. Local atomic functions are : s, p, d, f They look very different from a free electron wave 1. They are localized 2. They have directional lobes 3. They interact with nearest neighbors via overlapping.. electron clouds The transition metals

12 The key concept We can represent the interactions with the nearest neighbor, called overlap integral, with “hopping back and forth” This simple concept is incredibly powerful !

13 The electron hopes from the central atom (our starting point) to the nearest neighbor and back to the starting point. How often that happen is controlled by the overlap integral. Since the probability of 2 hopes (one forward one back) is  2 the sum of all “2 atom hops” is 6  2 In a hexagonal network there are 6 (six) such hops. This sum is the second momentum of the electron distribution

14 You may have learned “moments” in semiconductors processing as the Pearson 4 description of implant profiles (See auxiliary slides) 1. The first momentum is where the zero point of the energy scale of a band is. The electron hops on itself 2. The second momentum is a measure of the width of the band 3. The third momentum is a measure of the skew 4. There is forth momentum, called kurtosis, which do not care for here* * Kurtosis is a measure of whether the data are peaked or flat relative to a normal distribution. It is important in the implantation of the Boron implant profile into Si.

15 The width of the band is proportional to the square root of the second momentum. Thus the width of the band is proportional to z, the coordination number, and the square root of  the overlap integral. This is very cute, but unless we can calculate  it does not help us much. But we know that the d electrons have small orbits and therefore the second overlap integral must be small

16 In case you have forgotten the hydrogen atom : The d orbits are small in the transition metals relative to the s electrons in the shell above the d’s ! Therefore, their overlap is not very large.

17 The d band has 5 states in which we can put 10 electrons The energy of the d electrons is To calculate D(E) is hard work but we can make good guesses from the 1st,2nd,3rd,4th moment Those moments can be gotten by just counting “hops” To call the integral E bond makes strictly speaking no sense, as we are not dealing with bonds. It’s a misnomer to talk about “bond energy” when talking about bands, but the term stuck. What it really is the sum of the energies of the electron present in a band

18 Let us just assume that the d band is rectangle ! We do know that the d band is narrow but we do not have any number. But we know that D(E) must go with the inverse of the width, because D(E). E is the total number of states in the d band which is 5 (not to be confused with the maximum number of electrons which is 10. And not with the number of states are occupied, as not every state is filled)

19 For convenience, we set energy at the center of the band to zero and do the integral The factor 2 takes care of the spin. If 6 electrons are in the band, it will look like this

20 The energy the d electrons gain when “they go off the isolated atom” and start “interacting with their nearest neighbors” reaches a maximum when Which occurs for N=5. This predicts that a transition element with a half filled d shell has the highest cohesive energy. The solution is

21 Our ultra primitive model ! Experiment Friedel, in France, was the first one to use this model

22 HW Check the upper integration limit on the bond integral by inspecting the case of a) no electron is in the d band b) the d band has 5 electrons c) it is full. 2. Evaluate the bond integral step by step by inserting the upper and lower limit. Does your result agree with mine ?

23 Those who took Lara Estroff’s and VanDover course should recall this diagram A three hop. Central atom to first nearest neighbor, from their to an adjacent nearest neighbor, back to origin The third moment tells us if the band is skewed or not. If it is zero, there is no skew.

24 There are no three hops in a square lattice (2). Thus there is no skew. There are four hops, which tells us if the density of states is unimodal or bimodal. Read Sutton if interested

25 The moment theory can explain the change in structure in the transition metals as you move to higher Z. But you need to evaluate 4th and 5th moments. This tells you that the structure is controlled by second and third nearest neighbors. No surprise, remember the HW to count the first and second nearest neighbors in bcc and fcc

26 Alloy phases If the size of the atoms is not too different (< 15%) the principle component of alloying is to change the average number of free electrons per atom The underlying principle is how you fill the bands. The common wisdom is that what you want to avoid is “hitting a zone boundary”, as there is a jump in energy. See nearly free electron model, lecture 4, should you have forgotten. The old theory (which you can still find) however, is much to simple.

27 Starting simple  brass = Cu (50 at %) + Zn (50 at %) Bcc fcc hcp Obviously, there is no relation to the “parent phases”. The Cu fcc phase is called  -phase Well… Cu as one electron per atom, Zn has 2 thus the average electron concentration is now 1.5. In general, the more valence electron the element added to Cu has, the lower the existence range of the alpha phase. This suggests an electronic effect :-)

28 The phase diagram of Cu plus a) Zn (valence 2) b) Ga (valence 3) c) Ge (valence 4) HW Make a plot of the maximum existence width of the  -phase against the average electron per atom 2. Make a plot of the “center” of the  phase versus the average electron concentration per atom

29 HW What is the typical Zn content of cartridge brass ? 2. It’s main application is the fabrication of cartridges. Cartridge casings are made by a series of deep drawing, each one followed by an anneal. 2.a Which of the phase present (  to  do you expect to have the most ductility ? 3.a Within such a phase, what composition would you choose and why (hint include in your analysis the price of Cu/lb and of Zn/lb) Yellow brass, red brass, admiralty brass, gilding brass, bronze, free-cutting braze, naval braze…. Mankind loves those Cu-Zn alloys !

30 The classic explanation of the Cu-Zn and related phase diagrams has been based on the idea that the more electron you add, the more “you blow up the Fermi sphere”. Once you hit a zone boundary, you are in trouble. The crystal can avoid this by switching to an other crystal structure with a larger first Brillouin zone. The zone theory become untenable when it was established that the Fermi surfaces touches the zone face in all three noble metals. Without ill effects, they are happy to be fcc. Today’s explanation are based on pseudopotential theory. But I can give a physical picture that comes closer to than the simple model of the past.

31 1. When you disturb the electron gas (for example by putting a a single Ca ++ impurity into the electrons sea around Na + ions, the electron gas responds by doing a density fluctuation. n is the number of unit cells. These oscillations are known as Friedel Oscillations (FO) and occur even in one dimensional electron systems (known as Luttinger liquids)

32 2. The wavelength of this oscillation is determined by 2k F ; NOT k F (k F is the wave vector at the Fermi energy E F. I.e. radius of Fermi sphere in free electron model) 3. In general, the wavelength of this oscillation, in no way, matches the periodicity of the crystal lattice. Or, using math language, 2k f is incommensurate with the reciprocal lattice. 4. When 2k f matches the periodicity, the electron gas can minimize its energy (somewhat - this is perturbation theory) 5. What regulates the Hume-Rothery phase rules, therefore is a) perturbation theory b) the vector 2k f c) reciprocal lattice 6. For those interested, I put a paper on the subject into the 410 folder (ftp.ccmr.cornell.edu/tmp/410)

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34 Auxiliary Slides

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36 From the book of Charles Mischke, Richard Budynas

37 Massalski

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