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प्रश्न
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.
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उत्तर
The binding energy of the nucleus of mass number 240, B1=7.6×240=1824 MeV
The binding energy of each product nucleus, B2=8.5×120=1020 MeV
Then, the energy released as the nuclues breaks is given by
E=2B2−B1=2×1020−1824=216 MeV
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