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Question : 8
Total: 26
A nucleus with mass number A = 240 B E ∕ A = 7.6 MeV A = 120 B E ∕ A = 8.5 MeV
OR
Calculate the energy in fusion reaction :
1 2 H + 1 2 H → 2 3 He + n 1 2 H = 2.23 MeV
and of2 3 He = 7.73 MeV
OR
Calculate the energy in fusion reaction :
and of
Solution:
Calculation of energy released
Binding energy of nucleus with mass number 240 ,
E b n = 240 × 7.6 MeV
Binding energy of two fragments
= 2 × 120 × 8.5 MeV
Energy released = 240 ( 8.5 − 7.6 ) MeV
= 240 × 0.9
= 216 MeV
OR
Calculation of Energy in the fusion Reaction
Total Binding energy of Initial System i.e.
1 2 H + 1 2 H = ( 2.23 + 2.23 ) MeV
= 4.46 MeV
Binding energy of Final System i.e.2 3 He
= 7.73 MeV
Hence energy released
= 7.73 MeV − 4.46 MeV
= 3.27 MeV
Binding energy of nucleus with mass number 240 ,
Binding energy of two fragments
OR
Calculation of Energy in the fusion Reaction
Total Binding energy of Initial System i.e.
Binding energy of Final System i.e.
Hence energy released
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