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UNIVERSITY OF BARCELONA GABRIEL MARIÑO BRITO MASTER IN NANOSCIENCE AND NANOTECHNOLOGY MODELLING AND SIMULATION Practice #1 “Potential energy curve of Diatomic Curve” 1.1 RHF Calculation of the HCl molecule at a fixed geometry 1.1.1 Draw a schematic showing how the molecule is oriented in relation to the Cartesian axes 1.1.2 Nuclear energy repulsion How could you calculate this last value? Check that this data is coherent with your input The Nuclear Energy Repulsion could be calculated by Coulomb Laws for the interactions of two point charges [1], so in this practice we should take care to transform the units of the output file that are in hartress to electronvolts. F = k ! ! ! ! ! ! [1] Calculations Distance 1,3 Å K 9,9875x10 9 Nm 2 /C 2 . Energy Force (Cl – H) 144,711eV/Å (output) Nuclear Energy 6,913 Hartrees 6,92 Hartrees (Output ) Comparing the value of the output data, we can observe that the work for the calculated value is the same as the output value. 1.1.3 How many iterations have been required in this HFLCAO calculation? After 7 RHFLCAO interactions that had perform to do the calculation of the RHF energy. 1.1.4 Electric dipole moment: Taking into account the orientation of the molecule on the Cartesian axes, check that this moment is a vector directed from the atom with a net negative charge to the one with positive charge The coordinates of the vector form in the HCl molecule are: DX DY DZ 0,00000 0,0000 1,527104 Input Data Internuclear Distance 1,3 Å K 9.,875x10 9 Nm2/C2. Atomic Charge (Cl) 17 atoms Atomic Charge (H) 1 atoms

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  • UNIVERSITY OF BARCELONA GABRIEL MARIO BRITO

    MASTER IN NANOSCIENCE AND NANOTECHNOLOGY MODELLING AND SIMULATION

    Practice #1 Potential energy curve of Diatomic Curve 1.1 RHF Calculation of the HCl molecule at a fixed geometry 1.1.1 Draw a schematic showing how the molecule is oriented in relation to the Cartesian axes 1.1.2 Nuclear energy repulsion - How could you calculate this last value? Check that this data is coherent

    with your input The Nuclear Energy Repulsion could be calculated by Coulomb Laws for the interactions of two point charges [1], so in this practice we should take care to transform the units of the output file that are in hartress to electronvolts. F = k !!!!!! [1]

    Calculations Distance 1,3 K 9,9875x109 Nm2/C2. Energy Force (Cl H) 144,711eV/ (output) Nuclear Energy 6,913 Hartrees 6,92 Hartrees (Output ) Comparing the value of the output data, we can observe that the work for the calculated value is the same as the output value.

    1.1.3 How many iterations have been required in this HF-LCAO calculation? After 7 RHF-LCAO interactions that had perform to do the calculation of the RHF energy. 1.1.4 Electric dipole moment: Taking into account the orientation of the molecule on the Cartesian axes,

    check that this moment is a vector directed from the atom with a net negative charge to the one with positive charge The coordinates of the vector form in the HCl molecule are:

    DX DY DZ 0,00000 0,0000 1,527104

    Input Data Internuclear Distance 1,3 K 9.,875x109 Nm2/C2. Atomic Charge (Cl) 17 atoms Atomic Charge (H) 1 atoms

  • UNIVERSITY OF BARCELONA GABRIEL MARIO BRITO

    MASTER IN NANOSCIENCE AND NANOTECHNOLOGY MODELLING AND SIMULATION

    1.2 Potential energy curve for HCL 1.2.1 Build up a table of values of the distance R in Angstroms and U (TOTAL ENERGY) and Vnuc

    (NUCLEAR ENERGY REPULSION) energies. Obtain the electronic energy values (Eel) for each distance, subtracting the Vnuc column to the U column. Represent graphically, the total energy as a function of distance, together with its two components: Eel, Vnu. Identify the three curves. To what limit must tend the internuclear repulsion for R-> 0 and R-> ?

    Intramolecular Distance U (eV) Vnuc (eV) Eel (eV)

    0,6 -453,64015860 14,99335540 -468,63351400 0,7 -454,26315750 12,85144748 -467,11460500 0,8 -454,64843290 11,24501655 -465,89344940 0,9 -454,88298610 9,99557026 -464,87855630 1,0 -455,02045240 8,99601324 -464,01646560 1,1 -455,09548250 8,17819385 -463,27367630 1,2 -455,13052760 7,49667770 -462,62720530 1,3 -455,14014990 6,92001018 -462,06016010 1,4 -455,13380300 6,42572374 -461,55952680 1,6 -455,09562540 5,62250827 -460,71813370 1,8 -455,04374990 4,99778513 -460,04153500 2,0 -454,99038070 4,49800662 -459,48838730 3,0 -454,81224670 2,99867108 -457,81091780 4,0 -454,75589720 2,24900331 -457,00490050

    -455,500 -455,000 -454,500 -454,000 -453,500 0 0,5 1 1,5 2 2,5 3 3,5 4

    Total E

    nergy (eV)

    Intramolecular Distance

    Total Energy (eV)

  • UNIVERSITY OF BARCELONA GABRIEL MARIO BRITO

    MASTER IN NANOSCIENCE AND NANOTECHNOLOGY MODELLING AND SIMULATION

    According to the Coulomb Laws the limit of the internuclear repulsion for R-> 0 should tends to infinity and for R-> should tends to zero. 1.3 Equilibrium Internuclear Distance and Dissociation Energy 1.3.1 How many HF calculations were necessary to find the equilibrium geometry in our case? After 3 HF calculations the equilibrium is reached for the geometry. 1.3.2 Write down the equilibrium distance obtained by the calculation and compare it with the

    experimental value: 1.275 . How would explain the difference? We can determinate a difference between the calculation (1.3024) and the experimental values (1.275) about of the less state of energy reached, because the set point performed some calculations using previous results. 1.3.3 Using the U(4) value as an approximation to U() (we can consider the H-Cl bond almost broken

    for 4 ), estimate the molecules dissociation energy as De = U() - U(Re) and compare this value with the experimental one: De = 4,62 eV.

    0,0 2,0 4,0 6,0 8,0 10,0

    12,0 14,0 16,0

    0 0,5 1 1,5 2 2,5 3 3,5 4 Nuclear Ene

    rgy repu

    lsin (eV)

    Intermolecular Distance

    Nuclear Energy repulsin (eV)

    -470,000 -468,000 -466,000 -464,000 -462,000 -460,000 -458,000 -456,000 0 0,5 1 1,5 2 2,5 3 3,5 4

    Electron

    ic Ene

    rgy (eV)

    Intermolecular Distance

    Electronic Energy (eV)

  • UNIVERSITY OF BARCELONA GABRIEL MARIO BRITO

    MASTER IN NANOSCIENCE AND NANOTECHNOLOGY MODELLING AND SIMULATION

    The U(4) value is 10.45 eV obtained by multiplying it for 27.2 the estimate of the molecules dissociation energy that has been determined on 0.3842 Hartrees. relatively equal to the experimental value of 4.62 eV. 1.3.4 Use of Configuration Interaction method: Plot UCI(R) and compare it with UHF(R). Comment the

    differences.

    Intramolecular Distance U (eV) Vnuc (eV) Eel (eV) 0,6 -453,64579720 14,99335540 -468,63915260 0,7 -454,26891920 12,85144748 -467,12036670 0,8 -454,65438950 11,24501655 -465,89940600 0,9 -454,88958510 9,99557026 -464,88515540 1,0 -455,02829160 8,99601324 -464,02430480 1,1 -455,10526820 8,17819385 -463,28346200 1,2 -455,14306670 7,49667770 -462,63974440 1,3 -455,15634000 6,92001018 -462,07635020 1,4 -455,15462560 6,42572374 -461,58034930 1,6 -455,12900380 5,62250827 -460,75151210 1,8 -455,09462980 4,99778513 -460,09241490 2,0 -455,06392890 4,49800662 -459,56193560 3,0 -455,01375070 2,99867108 -458,01242180 4,0 -455,01239780 2,24900331 -457,26140110

    -455,500 -455,000 -454,500 -454,000 -453,500 0 0,5 1 1,5 2 2,5 3 3,5 4

    Total E

    nergy (eV)

    Intramolecular Distance

    Total Energy (eV)

  • UNIVERSITY OF BARCELONA GABRIEL MARIO BRITO

    MASTER IN NANOSCIENCE AND NANOTECHNOLOGY MODELLING AND SIMULATION

    The shape of the curves follows relatively the same path, changing a little bit in the values of the Electronic Energy and the Total Energy, excepting for the nuclear Energy repulsion values that are the same. 1.3.5 Optimize the geometry of the HCl molecule using the CI method. Recalculate the dissociation energy

    and compare the final theoretical values (Re and De) with the previously mentioned experimental values.

    Recalculation of the dissociation energy gave 3.93 eV value, quite similar to the experimental value (4.62eV) 1.4 Effect of Basis set 1.4.1 The CI and HF methods are both variational: they converge to the best numerical solution as more

    basis functions are included in the description of the molecular orbitals. Recalculate the potential energy curve for the CI and obtain new estimations of Re and De using the CI method and the 6-31G(d,p) basis.

    0,0 2,0 4,0 6,0 8,0 10,0

    12,0 14,0 16,0

    0 0,5 1 1,5 2 2,5 3 3,5 4

    Nuclear Ene

    rgy repu

    lsin (eV)

    Intermolecular Distance

    Nuclear Energy repulsin (eV)

    -470,000 -468,000 -466,000 -464,000 -462,000 -460,000 -458,000 -456,000 0 0,5 1 1,5 2 2,5 3 3,5 4

    Electron

    ic Ene

    rgy (eV)

    Intermolecular Distance

    Electronic Energy (eV)

  • UNIVERSITY OF BARCELONA GABRIEL MARIO BRITO

    MASTER IN NANOSCIENCE AND NANOTECHNOLOGY MODELLING AND SIMULATION

    Intramolecular Distance U (eV) Vnuc (eV) Eel (eV)

    0,6 -458,826166 14,9933554 -473,8195214 0,7 -459,3262139 12,85144748 -472,1776614 0,8 -459,7865938 11,24501655 -471,0316103 0,9 -459,9856942 9,995570263 -469,9812645 1,0 -460,0947078 8,996013237 -469,0907211 1,1 -460,1494589 8,178193852 -468,3276527 1,2 -460,1716403 7,496677698 -467,668318 1,3 -460,1746039 6,920010182 -467,0946141 1,4 -460,1665495 6,425723741 -466,5922733 1,6 -460,1356411 5,622508273 -465,7581493 1,8 -460,10054 4,997785132 -465,0983251 2,0 -460,0693179 4,498006619 -464,5673245 3,0 -459,9921514 2,998671079 -462,9908225 4,0 -458,826166 14,9933554 -473,8195214

    -460,400 -460,200 -460,000 -459,800 -459,600 -459,400 -459,200 -459,000 -458,800 -458,600 0 0,5 1 1,5 2 2,5 3 3,5 4

    Total E

    nergy (eV)

    Intramolecular Distance

    Total Energy (eV)

    0,0 2,0 4,0 6,0 8,0 10,0

    12,0 14,0 16,0

    0 0,5 1 1,5 2 2,5 3 3,5 4

    Nuclear Ene

    rgy repu

    lsin (eV)

    Intermolecular Distance

    Nuclear Energy repulsin (eV)

  • UNIVERSITY OF BARCELONA GABRIEL MARIO BRITO

    MASTER IN NANOSCIENCE AND NANOTECHNOLOGY MODELLING AND SIMULATION

    The comparison between the experimental and the new Re obtained (1.269) and the De (5.37eV) by the optimization geometry calculation, are similar. 1.4.2 Fitting the data in a Morse Function The Morse function [2] describes the shape of the total energy function and with the help of the QtiPlot software, we could find the values that can permit to fit the data obtained in 1.4.1 exercise. U r = D! 1 e!(!!!!) ! [2] = -1.9277 Re =1.269 De =5.37eV

    Practice #2 Molecular Dynamics Simulations 2 Unit 1: Carbon nanotubes morphology and gas physisorption

    -476,000 -474,000 -472,000 -470,000 -468,000 -466,000 -464,000 -462,000 -460,000 0 0,5 1 1,5 2 2,5 3 3,5 4

    Electron

    ic Ene

    rgy (eV)

    Intermolecular Distance

    Electronic Energy (eV)

    -5 0 5

    10 15 20 25 30 35 40

    0 1 2 3 4 5

    Total Energy Vs Morse Function

  • UNIVERSITY OF BARCELONA GABRIEL MARIO BRITO

    MASTER IN NANOSCIENCE AND NANOTECHNOLOGY MODELLING AND SIMULATION

    2.1.1 Use the tubebash.sh tool to generate carbon nanotubes with different morphologie

    Carbon nanotubes with different morphologies with the chiral indices (n=10,m=10) for 80 atoms (n=10,m=0) for 180 atoms

    2.1.2 For each CNT in Ex.1.1, generate a nanotube bundle using the tool nanotube_bundle.x

    The carbon nanotubes separation in the bundle

    5 10

  • UNIVERSITY OF BARCELONA GABRIEL MARIO BRITO

    MASTER IN NANOSCIENCE AND NANOTECHNOLOGY MODELLING AND SIMULATION

    2.1.3 Use tubebash.sh to generate a CNT with chirality (10,0) or (10,10) and 6 units. Next, employing the

    tool initconf.x, build up the simulation cell. This consists of a nanotube bundle with hexagonal symmetry with periodic boundary conditions along X and Y, while Z is left unbound. An ensemble of gas H2 molecules is set on top of the bundle (use between 30 and 90 molecules). Visualize the initial configuration of the MD simulation and identify the hexagonal symmetry according to the boundary conditions.

    960 C and 120 H atoms (60 H2) CNT (10,10) 5 of Separation (CNT) 4 of Separation molecules (H2) Hexagonal Periodicity

    2.1.4 Open the FIELD file contained in the same directory. Discuss the different parameters of the

    potential and make the necessary modifications according to the number of C atoms and H2 molecules in your simulation. Notice that no parameters are defined for C - C between and inside the nanotubes. How do you interpret this? Open the CONTROL file, observe and comment the parameters of the simulation

    CONFIG FILE: Contains Positions, Dimensions, boundary conditions, atomic labels, coordinates, velocities and forces. 2nd line ; 0 (coordinates in the file.) 6parallelogram boundary conditions with no periodicity in the z direction

  • UNIVERSITY OF BARCELONA GABRIEL MARIO BRITO

    MASTER IN NANOSCIENCE AND NANOTECHNOLOGY MODELLING AND SIMULATION

    CONTROL FILE Parameters for the simulation (temperature, timestep, time). In This file: Temperature =150K Steps for the calculation (300000)

    FIELD FILE: It is the force field File Nature and type of the force

    2.1.5 Run the MD simulation using DLPOLY.X. Open the OUTPUT file and examine the outcome of the

    simulation.

  • UNIVERSITY OF BARCELONA GABRIEL MARIO BRITO

    MASTER IN NANOSCIENCE AND NANOTECHNOLOGY MODELLING AND SIMULATION

    i. Analyze the evolution of the total energy and its components with time. Analyze the evolution of

    temperature with time.

    Temperature When the steps increases the temperature reach a stable value.

    Energies The Van der Waal forces decreases when the time pass, thus the energy decreases

  • UNIVERSITY OF BARCELONA GABRIEL MARIO BRITO

    MASTER IN NANOSCIENCE AND NANOTECHNOLOGY MODELLING AND SIMULATION

    ii. Visualize the geometry at the end of the simulation. In order to do this use the draw_final.sh shell

    command.

    Molecules of Hydrogen have been desorved, because they are decomposing Geometry at the last step of the simulation. It is possible to see that most of the H2 molecules have been desorbed. In fact, there are only 36 H atoms from the original 120. Even some molecules seems to be decomposed.

    iii. Run the traj.sh tool and use xmgrace to represent the z-trajectories of the gas molecules.

    Computed z-Trajectories obtained in xmgrace

  • UNIVERSITY OF BARCELONA GABRIEL MARIO BRITO

    MASTER IN NANOSCIENCE AND NANOTECHNOLOGY MODELLING AND SIMULATION

    3 MD simulation of the Ar clusters 3.1 Working Directory UNIT2/Exercise. Visualize the initial structure (Ar_cluster500.xyz) using the VMD or

    Avogadro visualization software. Convert the XYZ geometry file into a CONFIG file for DLPOLY using the tool XYZtoCONFIG.x. (select no boundary conditions). Open the CONFIG file and compare with the XYZ file. Ar cluster (VMD.) Config File OUTPUT:

    System Temperature: 59.121 K Total energy: -200.93 units

    Configurational energy: -288.87 units

    3.2 Running the MD calculations at a higher T value System Temperature: 80K Total energy: -158.76 units Configurational energy: -278.80 units

  • UNIVERSITY OF BARCELONA GABRIEL MARIO BRITO

    MASTER IN NANOSCIENCE AND NANOTECHNOLOGY MODELLING AND SIMULATION

    3.3 Running the MD calculations for a range of T values to calculate the caloric curve. This is a useful tool to monitor the occurrence of phase transitions, which imply a jump of E(T) if first-order, or in a derivative of E(T) if second- or higher order. Run the MD simulation for a Temperature range of your choice (close to the initial value) and plot the Configuration Energy (Ecfg) as a function of T. T defined in control file (50 -80 K)

    System Temperature Total Energy Configurational energy 50 59,121 -200,93 -288,87 52 50,966 -231,98 -307,79 54 52,221 -230,47 -308,14 56 55,53 -221,48 -304,07 58 57,79 -217,65 -303,6 60 61,27 -209,39 -300,52 62 63,498 -206,64 -301,09 64 65,48 -204,54 -301,94 66 65,555 -206,91 -304,42 68 3,08E+43 4,58E+43 6,91E+16 70 8,69E+30 1,29E+31 8,26E+10 72 72,316 -15,848 -124,41 74 73,694 -14,105 -123,72 76 76,037 -10,612 -123,71 78 77,96 -7,801 -123,76 80 80,704 -158,76 -278,8

    -1E+43 0 1E+43 2E+43 3E+43 4E+43 5E+43

    50 55 60 65 70 75 80

    Total Energy