双中心积分计算氟化氢HF的动能和势能
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双中心积分计算氟化氢HF的动能和势能
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<F 1s | H 1s>,<F 2s | H 1s>,<F 2p0 | H 1s>,氟化氫中有這三種情況是雙中心積分,
列成表格
| n | l | n' | l' | λ | ζ | ζ' | R | Pople John A | 計算值 | 參考值/計算值 | |
| <F 1s | H 1s> | 1 | 0 | 1 | 0 | 0 | 8.7 | 1 | 1.733 | -2.1798 | -2.179823778 | 0.999989092 |
| <F 2s | H 1s> | 2 | 0 | 1 | 0 | 0 | 2.6 | 1 | 1.733 | -4.3051 | -4.305097499 | 1.000000581 |
| <F 2p0 | H 1s> | 2 | 1 | 1 | 0 | 0 | 2.6 | 1 | 1.733 | -2.2822 | -2.282193291 | 1.00000294 |
利用前面得到Tab,VabA和VabB方法得到一個新的方法
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public static double cou( double R ,int zn1, int zn2,double za,double zb, int n1,int L1 ,int m1 ,int n2 ,int L2 ,int m2 ) throws IOException, ParseException, InterruptedException {double d= Tab (R , za , zb , n1, L1 , m1 , n2 , L2 , m2 );//0.00523319906316793double d1= VabA( R , zn1 , za , zb , n1, L1 , m1 , n2 , L2 , m2 );//-2.152403725820715 double d2= VabB( R , zn2 , za , zb , n1, L1 , m1 , n2 , L2 , m2 );//-0.03265325106338999double f=d+d1+d2;return f; }直接代入這個新方法就可以得到這三個積分,
double d=cou( 1.733 ,9,1, 8.7 , 1 ,?? 1,? 0 , 0 , 1 , 0 , 0 ); //-2.1798237778209373
double d=cou( 1.733 ,9,1, 2.6 , 1 ,?? 2,? 0 , 0 , 1 , 0 , 0 ); //-4.305097498533459
double d=cou( 1.733 ,9,1, 2.6 , 1 ,?? 2,? 1 , 0 , 1 , 0 , 0 );//-2.282193290607004
完整的java程序
package greenwood;import java.io.IOException; import java.text.ParseException; import udp.HFoverlapA;public class kin {public static double FACT( double n ) throws IOException, ParseException {double prodt=1.0;for(int a=1 ;a<n+1 ;a++){prodt=prodt*a;}if(n<0){System.out.println( prodt+ " 負數階乘 " );}return prodt;}public static double Tab( double R,double za,double zb, int n1,int L1 ,int m1 ,int n2 ,int L2 ,int m2 ) throws IOException, ParseException, InterruptedException {double f=0;double f1=0;double f2=0;double f3=0;f1= ( -zb*zb/2)*HFoverlapA.Sab ( R,za, zb, n1, L1 , m1 , n2 , L2 );System.out.println( f1 +" f1 Tab " );double d1=( ( 2*n2*zb*zb) /Math.pow ( 2*n2*(2*n2-1) ,0.5 ) );double d2=HFoverlapA.Sab (R, za, zb, n1, L1 , m1 , (n2-1) , L2 );f2=d1*d2;System.out.println( f2 +" f2 Tab " +d1+" "+d2 ); double d5=0;double d6=0;if(n2>1){double d4=2*zb*zb*(L2*(L2+1)-n2*(n2-1) );if(Math.abs(d4)>1e-6){d5=Math.pow ( 2*n2*(2*n2-1)*(2*n2-2)*(2*n2-3) ,0.5);d6= HFoverlapA.Sab (R, za, zb, n1, L1 , m1 , (n2-2) , L2 );f3=(d4/d5)*d6;}System.out.println( f3 +" f3 Tab " +d4+" "+d5+" "+d6 );}f=f1+f2+f3;System.out.println( f+" "+ f1+" "+f2+" "+f3+" f Tab " );return f;}public static double VabA( double R ,int zn, double za,double zb, int n1,int L1 ,int m1 ,int n2 ,int L2 ,int m2 ) throws IOException, ParseException, InterruptedException {double f1=-zn* (2*za/ Math.pow(2*n1*(2*n1-1),0.5 ) )*HFoverlapA .Sab ( R, za, zb , n1-1, L1 , m1 ,n2 ,L2 ); return f1;}public static double VabB( double R ,int zn, double za,double zb, int n1,int L1 ,int m1 ,int n2 ,int L2 ,int m2 ) throws IOException, ParseException, InterruptedException {double f1=-zn* (2*zb/ Math.pow(2*n2*(2*n2-1),0.5 ) )*HFoverlapA .Sab ( R, za, zb , n1, L1 , m1 ,n2-1 ,L2 ); return f1;}public static double cou( double R ,int zn1, int zn2,double za,double zb, int n1,int L1 ,int m1 ,int n2 ,int L2 ,int m2 ) throws IOException, ParseException, InterruptedException {double d= Tab (R , za , zb , n1, L1 , m1 , n2 , L2 , m2 );//0.00523319906316793double d1= VabA( R , zn1 , za , zb , n1, L1 , m1 , n2 , L2 , m2 );//-2.152403725820715 double d2= VabB( R , zn2 , za , zb , n1, L1 , m1 , n2 , L2 , m2 );//-0.03265325106338999double f=d+d1+d2;return f; }public static void res( ) throws IOException, ParseException, InterruptedException {// double d=cou( 1.733 ,9,1, 8.7 , 1 , 1, 0 , 0 , 1 , 0 , 0 ); //-2.1798237778209373// double d=cou( 1.733 ,9,1, 2.6 , 1 , 2, 0 , 0 , 1 , 0 , 0 ); //-4.305097498533459 double d=cou( 1.733 ,9,1, 2.6 , 1 , 2, 1 , 0 , 1 , 0 , 0 );//-2.282193290607004System.out.println( d+" res " );}public static void main(String[] args) throws IOException, ParseException, InterruptedException {res( );}} 《新程序員》:云原生和全面數字化實踐50位技術專家共同創作,文字、視頻、音頻交互閱讀總結
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