Add Activity 4 and 5
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"cells": [
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{
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"cell_type": "markdown",
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"id": "d053c16f-22ca-44ce-a58d-03e73c1a5554",
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"metadata": {
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"editable": true,
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"slideshow": {
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"slide_type": ""
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},
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"tags": []
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},
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"source": [
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"# From single scattering to multiple scattering"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"id": "2f4bc57a-1dbf-4b24-8660-f5866e2303ea",
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"metadata": {
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"editable": true,
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"slideshow": {
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"slide_type": ""
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},
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"tags": []
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},
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"outputs": [],
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"source": []
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3 (ipykernel)",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.11.13"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 5
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}
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{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "ff0fc2d9-b53e-4d29-883b-6d6303d76eb2",
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"metadata": {
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"editable": true,
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"slideshow": {
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"slide_type": ""
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},
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"tags": []
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},
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"source": [
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"# Forward scattering regime"
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]
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},
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{
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"cell_type": "markdown",
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"id": "7f10b898-4fc8-40fd-a477-05e736a5a255",
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"metadata": {
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"editable": true,
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"slideshow": {
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"slide_type": ""
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},
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"tags": []
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},
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"source": [
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"In photoelectron diffraction, it is well known that for high kinetic energy of the photoelectron (typically > 900 eV), the scattering factor has a strong peak in the forward direction. It means that photoelectrons are almost not deviated after a scattering event.\n",
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"\n",
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"Peaks of intentisity are then usually observed for dense atomic directions of the sample. This is the forward scattering approximation.\n",
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"\n",
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"For such high kinetic energy, the matrix inversion algorithm cannot be used since the memory needed for storing the matrix itself would be generally too high. The matrix will contain\n",
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"{math}```(N \\times (L_{max}+1)^2)^2``` elements of complex type with double precision (64 bits)"
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]
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},
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{
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"cell_type": "markdown",
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"id": "524878ef-3b96-47f5-9e9c-dc54f860b1a4",
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"metadata": {
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"editable": true,
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"slideshow": {
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"slide_type": ""
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},
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"tags": []
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},
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"source": [
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"How much memory would you need for this matrix for a hemispherical cluster of copper 15 angstroms thick (1 MFP) where the number of basis functions Lmax = 25 ?"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 11,
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"id": "e6f5b739-bcbf-4a2f-9c51-5e409079392a",
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"metadata": {
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"editable": true,
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"slideshow": {
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"slide_type": ""
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},
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"tags": []
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},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"606 atoms, 2.685 TB\n"
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]
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}
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],
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"source": [
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"import numpy as np\n",
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"\n",
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"# lattice constant of fcc copper\n",
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"a = 3.6\n",
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"# radius of the cluster\n",
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"r = 15\n",
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"# volume of the cluster\n",
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"V = .5 * (4/3) * np.pi * r**3\n",
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"# volume of the cell\n",
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"v = a**3\n",
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"# number of atoms in the unit cell\n",
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"n = 4\n",
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"# number of atoms in the cluster\n",
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"N = int(V/v * n)\n",
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"\n",
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"Lmax = 25\n",
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"M = (N * (Lmax+1)**2 )**2 * 2 * 64 / 8\n",
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"print(f\"{N:d} atoms, {M/1e12:.3f} TB\")"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"id": "c3161733-d4d7-498b-90a8-6fdaf2c4c306",
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"metadata": {
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"editable": true,
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"slideshow": {
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"slide_type": ""
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},
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"tags": []
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},
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"outputs": [],
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"source": []
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3 (ipykernel)",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.11.13"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 5
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}
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@ -5,5 +5,6 @@ format: jb-book
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root: intro
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root: intro
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chapters:
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chapters:
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- file: Activity01/Activity01
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- file: Activity01/Activity01
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- file: Activity02/Activity02
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- file: Activity03/Activity03
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#- file: Activity02/Activity02
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#- file: Activity03/Activity03
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