The binding energy per nucleon of deuterium and helium atom is 1.1 MeV and 7.0 MeV. If two deuterium nuclei fu

Question

The binding energy per nucleon of deuterium and helium atom is 1.1 MeV and 7.0 MeV. If two deuterium nuclei fuse to form a helium atom, the energy released is:

Accepted Answer

1. **Identify Nucleons:** 
   - Deuterium (${}_{1}^{2}	ext{H}$) has a mass number $A=2$ (1 proton, 1 neutron).
   - Helium (${}_{2}^{4}	ext{He}$) has a mass number $A=4$ (2 protons, 2 neutrons).
2. **Calculate Initial Binding Energy (Reactants):**
   - Binding Energy per nucleon for Deuterium = $1.1 	ext{ MeV}$.
   - Total Binding Energy for one Deuterium nucleus = $2 	imes 1.1 = 2.2 	ext{ MeV}$.
   - Since two Deuterium nuclei fuse, total initial Binding Energy = $2 	imes 2.2 = 4.4 	ext{ MeV}$.
3. **Calculate Final Binding Energy (Products):**
   - Binding Energy per nucleon for Helium = $7.0 	ext{ MeV}$.
   - Total Binding Energy for the Helium nucleus = $4 	imes 7.0 = 28.0 	ext{ MeV}$.
4. **Calculate Energy Released ($Q$):**
   - The energy released in a nuclear reaction is the difference between the total binding energy of the products and the reactants .
   - $Q = (BE)_{	ext{products}} - (BE)_{	ext{reactants}}$
   - $Q = 28.0 	ext{ MeV} - 4.4 	ext{ MeV} = 23.6 	ext{ MeV}$.

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