4.2 × 10^14 J will be the energy released by the fusion of a 1.25 -kg mixture of deuterium and tritium, which produces helium.
Emitted energy in the reaction will be
E = (1876.136 + 2809.45) (3728.42 +939.565)
E =017.6 Mev
Now total atomic bar weight A = (2+3) = 5 grams
total no of atoms = 2Na
for 1-25 kg
= (2Na/5)X1250
= (2x6.023×10^23/5)*1250
= 3.0115×10^26
In Joules released energy in fusion a 1.25kg mixture of denterium and trilium will be
Q = 2.65 × 10^33 × 1-6 × 10^-19 J
= 4.2 × 10^14 J
the energy released by the fusion deuterium and tritium is 4.2 × 10^14 J.
the complete question is:
The atomic masses of deuterium and tritium are 2.014102 u and 3.016049 u respectively. Calculate the energy in joules released by the fusion of a 1.25 -kg mixture of deuterium and tritium, which produces helium. There are equal numbers of deuterium and tritium nuclei in the mixture
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