Nuclear Physics and RadioactivityMedium
Question
Two nuclei of mass number 3 combine with another nucleus of mass number 4 to yield a nucleus of mass number 10. If the binding energy per nucleon for the mass number 3, 4 and 10 are $5.6MeV,7.4MeV$ and 6.1 MeV , respectively, then in the process, $\Delta{Mc}^{2} =$ $\_\_\_\_$ MeV .
Options
A.6.9
B.7.9
C.2.2
D.4.3
Solution
Sol. [Key Concept: Compute $\sum(BE\ per\ nucleon \times \ A)$ for the reactants minus the product; that difference is $\Delta\ {Mc}^{2}$.]
$$\Delta{mc}^{2} = \lbrack 2 \times 3 \times 5.6 + 4 \times (7.4) - 6.1 \times 10\rbrack$$
$$= 2.2MeV$$
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