Question

A nucleus with mass number A = 240 and BE/A = 7.6 MeV breaks into two fragments each of A = 120 with BE/A = 8.5 MeV. Calculate the released Energy.

Calculate the energy in the following fusion reaction:

H₁² + H₁² → He₂³ + n,
         Binding energies of   H₁² = 2.22 Mev.  and of ₂H³ = 7.72 MeV .
         
BE = Nuclear Binding Energy

Answer 1

B.E of the nucleus mass no. 240,
B1=7.6×240=1824 MeV
B2=8.5×120=1020 MeV
Energy released as nucleus break is given as E=2B2-B1=2×1020-1824=216 MeV

OR

Given= B.E of 2 He, E1 =2.23 MeV
1
B.E of 3He,E2=7.73 MeV
2
∆E=E2-2E1=7.73-(2×2.23)
=3.27 MeV

Answer 2

Explanation:

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