Question

The binding energy per nucleon of $_{3}^{7}Li$ and $_{2}^{4}He$ nuclei are $5.06MeV$ and $7.06MeV$, respectively. In the nuclear reaction $_{3}^{7}Li+_{1}^{1}H\to_{2}^{4}He+_{2}^{4}He+Q$ the value of energy $Q$ released is
A). $19.6MeV$
B). $-2.4MeV$
C). $8.4MeV$
D). $17.3MeV$

Answer 1

Binding energy per nucleon of Helium and Lithium are given. During the nuclear reaction, some energy is released. We need to find the amount of energy released. The energy released in a nuclear reaction is the difference between the binding energy of reactants and the binding energy of the products. By finding the binding energy of each reactant and products in the reaction, we can find out the amount of released energy.

Formula used:

~B.{{E}_{_{1}^{1}H}}~$$ **Complete step-by-step solution:** The energy released in a nuclear reaction is the difference between the binding energy of reactants and the binding energy of the products. Here, $$Q~=~B.{{E}_{_{2}^{4}He}}+B.{{E}_{_{2}^{4}He}}-~B.{{E}_{_{3}^{7}Li}} - B.{{E}_{_{1}^{1}H}} $$ Where, $$B.{{E}_{_{2}^{4}He}}$$ is the binding energy Helium $$B.{{E}_{_{3}^{7}Li}}$$ is the binding energy of Lithium $$B.{{E}_{_{1}^{1}H}}$$ is the binding energy of hydrogen Given, The binding energy per nucleon of $$_{3}^{7}Li=5.06MeV$$ The binding energy per nucleon of $$_{2}^{4}He=7.06MeV$$ Take binding energy of $$_{1}^{1}H$$ as zero, $$B.{{E}_{_{1}^{1}H}}=0$$ Now, $$B.{{E}_{_{3}^{7}Li}}$$, binding energy of $$_{3}^{7}Li=7\times 5.06=39.2MeV$$ $$B.{{E}_{_{2}^{4}He}}$$, binding energy of $$_{2}^{4}He=4\times 7.06=28.24MeV$$ Then, Energy released, $$\text{Q =2}\times \left( \text{Binding energy of }_{2}^{4}He \right)\text{ -Binding energy of }_{\text{3}}^{\text{7}}\text{Li}$$ $$Q=2\times 28.24-7\times 5.06=17.3MeV$$ **So the answer is option D.** **Note:** If the total mass of the resultant particles is less than the mass of the initial reactants, energy is released in a nuclear reaction. Since the nucleons at the surface of a nucleus interact with fewer nucleons the binding energy per nucleon is reduced. It is based on an average of overall nucleons in the nucleus. For a nucleus to be stable, its binding energy should be larger.