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Calculate the energy released in MeV in the following nuclear reaction : " "_(92)^(238)U to" "_(90)^(234)Th + " "_(2)^(4)He + Q mass of " "_(92)^(238)U = 238.05079 u, mass of " "_(90)^(234)Th = 234.043630 u, mass of " "_(2)^(4)He = 4.00260 u, 1 u=931.5 MeV/c^(2). |
| Answer» Solution :ENERGY released in the reaction `Q= [m (" "_(92)^(238)U) -m (" "_(90)^(234)Th) -m (" "_(2)^(4)He)]c^(2) =(238.05079 - 234.04363-4.00260) u xx c^(2)= (0.00456 u). c^(2) = 0.00456 xx 931.5 MeV= 4.25 MeV`. | |