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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+// Scilab code Exa4.8.1: To calculate the energy liberated during fusion reaction: Page 194 (2011)
+// 5*H(1,2)= He(2,3)+He(2,4)+H(1,2)+2*n(0,1)+25MeV is the given reaction
+N = 6.023e+026/2*10; // Number of atoms in 10Kg of H-2, atoms
+E = 25/5*1.6e-013; // Energy liberate during fusion of 1 atom of H-2, J
+E_l = E*N; // Energy liberate during fusion of 10 Kg of H-2, J
+printf("\n Energy liberated during fusion of 10 Kg of H-2 = %4.2e J", E_l)
+// Result
+// Energy liberated during fusion of 10 Kg of H-2 = 2.41e+015 J \ No newline at end of file