MSE 561, Atomic Modeling in Material Science Assignment 1

Transcription

1 Depatment of Mateial Science and Engineeing, Univesity of Pennsylvania MSE 561, Atomic Modeling in Mateial Science Assignment 1 Yang Lu 1. Analytical Solution The close-packed two-dimensional stuctue is shown below, whee neighbos ae coloed based on distances to the cental atom (the blue one). 1, 2 and 3 ae fist, second and thid neaest neighbo spacing espectively Let 1 =, it s easy to find out that 2 = 3, 3 = 2. We only conside the inteactions between the cental atom (epesented by i=1) and its neighbos, because all the atoms ae equivalent. And the total enegy can be witten as, Ep = N Φ(1i ) 2 i 1 Eq.(1) whee N is the total numbe of atoms and Φ(1i) is detemined by Lennad-Jones potential. Now let s conside the following cases, i) only the fist neaest neighbos (ed ones) ae included in Eq.(1) In this case: < cut = 7.50Å, 3 > cut = 7.50Å; 4.33Å< <7.50Å must be satisfied. Fistly, if < 0 = 7.0Å, Eq.(1) can be witten as, Ep = 1 N 6 4 ε [( )12 ( )6 ] = 12N ε [( )12 ( )6 ] 2 Eq.(2)

2 It s easy to find out min = case. 6 2 = 3.82Å (<4.33Å). So when = 4.33Å, Emin = Nε in this Secondly, if > 0 = 7.0Å, Eq.(1) can be witten as, = 3N[A( cut ) 3 + B( cut Eq.(3) Then = 7.0Å, Emin = Nε in this case. ii) the fist and second neaest neighbos ae both included in Eq.(1) In this case: 3 < cut = 7.50Å, 2 > cut = 7.50Å; 3.75Å< <4.33Å must be satisfied. Let s fistly conside 3 < 0 = 7.00Å, < 4.04Å. Then Eq.(1) can be witten as, = 12Nε[( )12 + ( 3 )12 ( )6 ( 3 )6 ] Eq.(4) Take deivative of Ep and it s easy to get min = = 3.80Å (consistent with ou assumption) and coesponding Emin = Nε. Secondly, if 3 > 0 = 7.00Å, > 4.04Å. Then Eq.(1) can be witten as, = 12Nε[( )12 ( )6 ]+ 3N[A( 3 cut ) 3 + B( 3 cut Eq.(5) Let dep/d = 0, It tuns out that it s not min but max that obtained fom Eq.(6). So when = 4.04Å, Emin = Nε in this case. 24ε [ 2( )12 + ( )6 ]+ 3 3A( 3 cut ) B( 3 cut ) = 0 iii) The fist, second and thid neaest neighbos ae all included in Eq.(1) This means 2 < cut = 7.50Å, < 3.75Å. Fistly of all, if 2 < 0 = 7.00Å, < 3.50Å, Eq.(1) can be witten as, Eq.(6) = 12Nε[( )12 + ( 3 )12 + ( 2 )12 ( )6 ( 3 )6 ( 2 )6 ] Eq.(7) Let dep/d = 0, then min = = 3.79Å (>3.50Å). So when = 3.50Å, Emin = Nε in this case. Secondly, if 2 > 0 = 7.00Å, 3.50Å < < 3.75Å ( 3 > 0 is impossible hee), Eq.(1) will become, = 12Nε[( )12 + ( 3 )12 ( )6 ( 3 )6 ]+ 3N[A(2 cut ) 3 + B(2 cut Eq.(8) Let dep/d = 0, 2 12ε [ 2( )12 + ( )6 2 3( 3 )12 + 3( 3 )6 ]+ 3A(2 cut ) 2 + 2B(2 cut ) = 0 Eq.(9)

3 It tuns out that min = 3.39Å (<3.50Å). So when = 3.50Å, Emin = Nε in this case. In all of the thee cases, total enegy becomes lage when moe atoms lie in the "tail" egion. In ode to make all the neighbos befoe the "tail" egion, will become smalle and smalle as moe and moe neighbos ae included. This means min will always lie out of equied ange (min > ), afte case iii). Theefoe, thee s no need to poceed this pocess futhe to include the foth and othe neaest neighbos. In conclusion, total enegy is minimized as Nε (-3.203kJ/mol) when the fist and second neaest neighbos ae included and neaest neighbo spacing equal to 3.80Å. 2. Numeical Solution Besides the analytical solution in the fist section, we could use a pogam to decide the fom of Eq.(1) with a given and then output the total enegy Ep. Afte calculating total enegies using all the possible neaest neighbo spacings, we shall diectly find out Emin and the coesponding min. Fist of all, let s decide the ange of in the pogam. Accoding to case i), has to be smalle than 7.5Å to include at least the fist neighbos. Due to eason descibed in the conclusion of section 1, we don t need to conside neighbos fathe than the thid neaest ones. 4 = 7, so has to be lage than 2.835Å. An aay R(4666) will be ceated to accommodate all the spacings fom 2.835Å to 7.5Å (each step is 0.001Å). So the main pogam can be coded as, PROGRAM MAIN IMPLICIT NONE INTEGER I REAL*8 R(4666), EP OPEN(1, FILE='OUT') DO I=1, 4666 FORTRAN CODES ARE DELETED IN THIS COPY R(I)= *(I-1) CONTACT ME IF YOU NEED MORE INFO. CALL F(R(I), EP) WRITE (1, *) R(I), EP ENDDO CLOSE(1) 3

4 END The suboutine F is used to detemine the analytical fom of Eq.(1) based on the discussion in section 1: if 7.000Å < 7.500Å, Eq.(3) will be used; if 4.330Å < 7.000Å, Eq.(2) will be used; if 4.042Å < 4.330Å, Eq.(5) will be used; if 3.750Å < 4.042Å, Eq.(4) will be used; if 3.500Å < 3.750Å, Eq.(8) will be used; if 2.835Å < 3.500Å, Eq.(7) will be used. So F can be coded as, SUBROUTINE F(R, EP) IMPLICIT NONE REAL*8 R, EP REAL*8 NA, e, A, B NA = 6.02D23 e = *1.6D-19 A = D-3*1.6D-19 B = D-3*1.6D-19 IF (R.LT ) THEN EP=12.0*NA*e*((3.405/R)**12.0+(3.405/R/2.0)**12.0+(3.405/R/ FORTRAN CODES ARE DELETED IN THIS COPY +SQRT(3.0))**12.0-(3.405/R)**6.0-(3.405/R/2.0)**6.0-(3.405/R/ CONTACT ME IF YOU NEED MORE INFO. +SQRT(3.0))**6.0) ELSEIF (R.LT ) THEN EP=12.0*NA*e*((3.405/R)**12.0+(3.405/R/SQRT(3.0))** (3.405/R)**6.0-(3.405/R/SQRT(3.0))**6.0)+3.0*NA*(A*(2.0*R )**3.0+B*(2.0*R-7.500)**2.0) ELSEIF (R.LT ) THEN EP=12.0*NA*e*((3.405/R)**12.0+(3.405/R/SQRT(3.0))** (3.405/R)**6.0-(3.405/R/SQRT(3.0))**6.0) ELSEIF (R.LT ) THEN EP=12.0*NA*e*((3.405/R)**12.0-(3.405/R)**6.0)+3.0*NA*(A* 4

5 +(SQRT(3.0)*R-7.500)**3.0+B*(SQRT(3.0)*R-7.500)**2.0) ELSEIF (R.LT ) THEN EP=12.0*NA*e*((3.405/R)**12.0-(3.405/R)**6.0) ELSE FORTRAN CODES ARE DELETED IN THIS COPY CONTACT ME IF YOU NEED MORE INFO. EP=3.0*NA*(A*(R-7.500)**3.0+B*(R-7.500)**2.0) ENDIF END Finally, the enegy dependence on neaest neighbo spacing is shown below, the value of minimum enegy (-3.203kJ/mol) and its position (=3.80Å) ae exactly the same as what we get in analytical solution. 5