Question:
Binding energy per nucleon of deuteron and Helium nuclei are 1.1 MeV and 7 MeV respectively. If a single Helium nucleus was formed by adding two Deuterons, the energy released is
Binding energy per nucleon of deuteron and Helium nuclei are 1.1 MeV and 7 MeV respectively. If a single Helium nucleus was formed by adding two Deuterons, the energy released is
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Always convert "per nucleon" values to total binding energy by multiplying by the mass number before subtracting!
Updated On: Jun 3, 2026
- 23.6 MeV
- 32.4 MeV
- 28.6 MeV
- 13.6 MeV
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The energy released ($Q$-value) in a nuclear reaction is equal to the difference between the total binding energy of the products and the total binding energy of the reactants: $Q = \text{BE}_{\text{products}} - \text{BE}_{\text{reactants}}$.
Step 2: Meaning
Total binding energy is calculated by multiplying the binding energy per nucleon by the total number of nucleons (mass number $A$) present in that nucleus.
Step 3: Analysis
The reaction is: $_{1}\text{H}^{2} + {}_{1}\text{H}^{2} \rightarrow {}_{2}\text{He}^{4}$. * For one Deuteron ($A=2$): $\text{BE} = 2 \times 1.1 = 2.2\text{ MeV}$. For two Deuterons, total reactant $\text{BE} = 2 \times 2.2 = 4.4\text{ MeV}$. * For one Helium nucleus ($A=4$): product $\text{BE} = 4 \times 7 = 28\text{ MeV}$. The energy released is: $$\Delta E = 28\text{ MeV} - 4.4\text{ MeV} = 23.6\text{ MeV}$$
Step 4: Conclusion
The energy released during the fusion process is 23.6 MeV.
Final Answer: (A)
The energy released ($Q$-value) in a nuclear reaction is equal to the difference between the total binding energy of the products and the total binding energy of the reactants: $Q = \text{BE}_{\text{products}} - \text{BE}_{\text{reactants}}$.
Step 2: Meaning
Total binding energy is calculated by multiplying the binding energy per nucleon by the total number of nucleons (mass number $A$) present in that nucleus.
Step 3: Analysis
The reaction is: $_{1}\text{H}^{2} + {}_{1}\text{H}^{2} \rightarrow {}_{2}\text{He}^{4}$. * For one Deuteron ($A=2$): $\text{BE} = 2 \times 1.1 = 2.2\text{ MeV}$. For two Deuterons, total reactant $\text{BE} = 2 \times 2.2 = 4.4\text{ MeV}$. * For one Helium nucleus ($A=4$): product $\text{BE} = 4 \times 7 = 28\text{ MeV}$. The energy released is: $$\Delta E = 28\text{ MeV} - 4.4\text{ MeV} = 23.6\text{ MeV}$$
Step 4: Conclusion
The energy released during the fusion process is 23.6 MeV.
Final Answer: (A)
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