Question

The energy of the reaction
$$Li^7+p\longrightarrow 2 _2He^4$$
if the binding energy per nucleon in $$Li^7$$ and $$He^4$$ nuclei are $$5.60\ MeV$$ and $$7.06\ MeV$$, respectively, is :

Answer 1

Correct option is D. $$17.28\ MeV$$
Given:

Binding energy per nucleon in $$Li^7$$ = $$5.6 \ MeV$$

Binding energy for $$Li^7$$ = $$ 7 \times 5.6 = 39.20 \ Mev$$

Similarly. binding energy for $$He^4$$ = $$4 \times 7.06 = 28.24 \ MeV$$

Since, we have $$2He^4$$ $$\Rightarrow$$ B.E for $$2He^4$$ = $$ 2 \times 28.24 = 56.48 \ MeV$$

Energy of the reaction = B.E of $$2He^4$$ - B.E of $$Li^7$$ = $$(56.48 - 39.20) \ MeV = 17.28 \ MeV$$