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// Scilab code Exa6.3.2 : To calculate the kinetic energy of protons and no. of possibile reactions: Page 265 (2011)
V = 5; // Voltage of accelerator, MV
// Declare three cells (for three reactions): Page no. : 133(2011)
R1 = cell(3,2)
R2 = cell(10,2)
// Enter data for first cell (Reaction)
R1(1,1).entries = "p";
R1(1,2).entries = 1;
R1(2,1).entries = 'd';
R1(2,2).entries = 1;
R1(3,1).entries = 'He';
R1(3,2).entries = 2;
E_p = (R1(1,2).entries)*V
E_d = (R1(2,2).entries)*V
E_He = (R1(3,2).entries)*V
// Enter data for second cell (Reaction)
R2(1,1).entries = "p"
R2(1,2).entries = 1
R2(2,1).entries = "N"
R2(2,2).entries = 14
R2(3,1).entries = "O"
R2(3,2).entries = 15
R2(4,1).entries = "y"
R2(4,2).entries = 0
R2(5,1).entries = "d"
R2(5,2).entries = 1
R2(6,1).entries = "n"
R2(6,2).entries = 0
R2(7,1).entries = "He"
R2(7,2).entries = 3
R2(8,1).entries = "C"
R2(8,2).entries = 13
R2(9,1).entries = "He"
R2(9,2).entries = 4
R2(10,1).entries = "C"
R2(10,2).entries = 12
printf("\nProtons energy = -%d MeV \n Deuterons energy = -%d MeV \n Double charged He-3 = -%d MeV", E_p, E_d, E_He)
printf("\n Possible reaction at these energies are")
printf("\n %s + %s(%d) ---> %s(%d)+ %s", R2(1,1).entries,R2(2,1).entries,R2(2,2).entries,R2(3,1).entries,R2(3,2).entries,R2(4,1).entries)
printf("\n %s + %s(%d) ---> %s(%d) + %s ", R2(5,1).entries,R2(2,1).entries,R2(2,2).entries,R2(3,1).entries,R2(3,2).entries,R2(6,1).entries)
printf("\n %s(%d) +%s(%d) ---> %s(%d)+ %s", R2(7,1).entries,R2(7,2).entries,R2(8,1).entries,R2(8,2).entries,R2(3,1).entries,R2(3,2).entries,R2(6,1).entries)
printf("\n %s(%d) + %s(%d) ---> %s(%d) +%s", R2(9,1).entries,R2(9,2).entries,R2(10,1).entries,R2(10,2).entries,R2(3,1).entries,R2(3,2).entries,R2(6,1).entries)
// Result
// Protons energy = -5 MeV
// Deuterons energy = -5 MeV
// Double charged He-3 = -10 MeV
// Possible reaction at these energies are
// p + N(14) ---> O(15)+ y
// d + N(14) ---> O(15) + n
// He(3) +C(13) ---> O(15)+ n
// He(4) + C(12) ---> O(15) +n
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