Question

Consider the nuclear reaction: \[ {}^{2}_{1}H + {}^{2}_{1}H \rightarrow {}^{4}_{2}He \] Binding energy per nucleon of \( {}^{2}_{1}H \) and \( {}^{4}_{2}He \) are \(1.1\,\text{MeV}\) and \(7\,\text{MeV}\) respectively. Find the energy released in the nuclear reaction.

• A. \(23.6\,\text{MeV}\)
• B. \(24.2\,\text{MeV}\)
• C. \(5.9\,\text{MeV}\)
• D. \(3\,\text{MeV}\)

Answer 1

The Correct Option is B

Concept: The energy released in a nuclear reaction equals the increase in total binding energy. \[ \text{Energy released} = \text{Total BE of products} - \text{Total BE of reactants} \] Binding energy of a nucleus is \[ \text{Total BE} = (\text{Binding energy per nucleon}) \times (\text{Number of nucleons}) \] Step 1: {\color{red}Calculate total binding energy of reactants.} Each deuterium nucleus \( {}^{2}_{1}H \) has \[ \text{BE per nucleon} = 1.1\,\text{MeV} \] Number of nucleons \(=2\) \[ \text{BE of one } {}^{2}_{1}H = 2 \times 1.1 = 2.2\,\text{MeV} \] Since there are two deuterium nuclei: \[ \text{Total BE of reactants} = 2 \times 2.2 = 4.4\,\text{MeV} \] Step 2: {\color{red}Calculate total binding energy of product.} For \( {}^{4}_{2}He \): \[ \text{BE per nucleon} = 7\,\text{MeV} \] Number of nucleons \(=4\) \[ \text{Total BE of } {}^{4}_{2}He = 4 \times 7 = 28\,\text{MeV} \] Step 3: {\color{red}Find energy released in the reaction.} \[ E = 28 - 4.4 \] \[ E = 23.6\,\text{MeV} \] Thus, the energy released in the reaction is \[ \boxed{23.6\,\text{MeV}} \]