# FormBind

## Contents |

## Abstract

This is tutorial for K-12 students. This involves the analysis of a relevant research problem in nuclear materials - the segregation of chromium (Cr) and helium (He) to grain boundaries in iron (Fe). This project will introduce a student to using spreadsheet tools to extract information about this important reaction in nuclear energy applications. This is part of a program at Pacific Northwest National Laboratory designed to get high school students involved with STEM-related projects.

Author(s): Mark A. Tschopp, Fei Gao (PNNL), Joanna Sun (student, PNNL)

## Background

Here are some questions that may be needed to understand what we are doing in this example.

- How does radiation damage materials? Here is a link to Radiation Material Science that gives a relatively good basic explanation of how radiation affects materials.
- What is a point defect? Here is a link to find out about point defects.
- What is a vacancy? It is a type of point defect. Here is a link to find out about a vacancy defect.
- What is an interstitial atom? This is another type of point defect in a material. Here is a link to find out about interstitial defects. This is not used in the present example.
- How are atoms arranged in a metal? Here is a link to find out about crystal structures. In this example, iron (Fe) is a body-centered cubic structure.
- What is a formation energy? Here is a link to find out about grain boundaries. Grain boundaries join two perfect lattices of different orientations.
- What is segregation? Here is a link to find out about segregation to grain boundaries.
- Why Chromium in Iron? Here is a link to find out about Chromium steels, otherwise known as stainless steels. These are commonly used in the nuclear industry for their corrosion resistance.

## Cohesive energy

What is the cohesive energy?

Let's start with the cohesive energy of a particular type of element: iron. For instance, iron is often in a body-centered cubic structure with 2 atoms/unit cell. Why 2? Think of a cube with atoms at the corners (8 of them) and an atom in the middle. Each of the eight atoms at the corners is shared with 7 other neighboring cubes, so each of the corner atoms counts as 1/8th of an atom. Hence, 8 * 1/8 + 1 = 2/unit cell. The length of the sides of the cube is called the lattice spacing.

Q: So if we had 10 lattice spacings in each direction, how many atoms would we have?

A: 10 x 10 x 10 = 1000 cubes (or unit cells). Since there are 2 atoms per unit cell, this would result in 2000 total atoms.

Let's assume these are iron atoms. Now if we relaxed these atoms in our atomistic codes, we would get a total energy of -8025.972 eV (electron-volts). The cohesive energy is merely the energy of each atom.

Q: What is the cohesive energy of iron?

A: -8025.972 eV / 2000 atoms = -4.012986 eV/atom

Chromium is also a body-centered cubic structure. For 10 x 10 x 10 unit cells of Cr (2000 atoms), the total energy is -7672.56624 eV and the cohesive energy is -3.83628312 eV/atom.

## Formation energy

What is a formation energy?

### Vacancy

Let's start with the formation energy for a vacancy. This is the energy needed to remove an atom and create a vacancy in a perfect single crystal bcc lattice. So, in our previous cell with 2000 atoms, we now would have 1999 atoms. The total energy with the vacancy, Ed (d for defect), is -8020.249014 eV. In order to find out the total energy needed to create this vacancy, we need to subtract the total energy of an equivalent number of atoms in a perfect lattice with no defects. This can be obtained by multiplying the number of atoms by the cohesive energy of these atoms, i.e., Ec (c for cohesive) is -8021.959014 eV. Hence the vacancy formation energy is:

Evf = Ed - Ec

Evf = (-8020.249014) - (1999 * -4.012986)

Evf = (-8020.249014) - (-8021.959014) = 1.71 eV

### Interstitial

The same relation holds for an interstitial atom (an atom that is added to the perfect lattice) except that there would now be 2001 atoms. So the first term would be the total energy with the interstitial and the second term would be the energy given the same number of atoms in a perfect lattice.

Ef = Ed - Ec

Ef = (-8026.464986) - (2001 * -4.012986)

Ef = (-8026.464986) - (-8029.984986) = 3.52 eV

### Chromium atom

There can also be a formation energy for another type of element. Let's consider adding a chromium atom to the iron lattice. Now we have 1999 iron atoms and 1 chromium atom. The total energy for the Fe and Cr combination is -8026.08262 eV.

Q: How do we subtract the equivalent energies of Fe and Cr in perfect lattices, though? (These are the righthand terms above)

A: We can consider each atom in a lattice consisting of only like atoms. For instance, the energy of the 1999 iron atoms can be calculated using the cohesive energy for iron and the 1 chromium atom can be calculated using the cohesive energy for chromium.

Now our calculation for the formation energy is:

Ef,CrV = Ed - Ec

Ef,CrV = (-8026.08262) - (1999 * -4.012986 + 1 * -3.83628312) = -0.2873 eV

### Chromium at Grain Boundary

What if we are trying to calculate a formation for Chromium in a simulation cell with a grain boundary? Let's say that we chose an Fe atom that is far away from the grain boundary and are going to substitute Cr for this Fe atom. Remember the example above where the lefthand and righthand terms have to have the same number of atoms. So, again, let's assume that we have 200 atoms in our simulation cell with a grain boundary and it has a total energy of -790 eV. After substituting the Cr atom, the energy is -789.88940288 eV. Therefore,

Ef,CrV = Ed - Ec

Ef,CrV = (-790.11059712 + 1 * -4.012986) - (-790 + 1 * -3.83628312) = -0.2873 eV

Notice that Ed contains the grain boundary simulation cell with the defect (Chromium), but it also contains the cohesive energy for 1 Fe atom. This means that this energy is associated with 200 Fe atoms (199 + 1) and 1 Cr atom. The Ec term contains the energy for the grain boundary simulation cell without the Cr atom along with the cohesive energy for 1 Cr atom. Therefore, this term is associated with 200 Fe atoms (200 + 0) and 1 Cr atom; again, the number of atoms of each element type is equal for Ed and Ec. Another way to think of this is that a Cr atom is taken from a single crystal Cr lattice and substituted into the grain boundary simulation cell, while the Fe atom removed is inserted into a single crystal Fe lattice.

Now, we'll substitute the Cr for an atom right at the grain boundary. The energy of this configuration (199 Fe, 1 Cr) is -789.38940288 eV.

Ef,CrV = Ed - Ec

Ef,CrV = (-790.61059712 + 1 * -4.012986) - (-790 + 1 * -3.83628312) = -0.7873 eV

Aha! A lower formation energy at the grain boundary site for Cr in this case. This means that it is energetically favorable for Cr to reside at the grain boundary sites.