Question

What is $Q$-value of a nuclear reaction? Find (i) nature, (ii) $Q$-value, and (iii) mass defect of nuclear reaction: \[ 2({}^1H^2) \; \longrightarrow \; {}^4He^2 + g \] if the binding energy per nucleon of deuterium and helium are $1.25 \, \text{MeV}$ and $7.2 \, \text{MeV}$ respectively.

Hint:

To calculate $Q$-value, always use binding energy differences. A positive $Q$ means the reaction releases energy.

Answer 1

Step 1: Definition of $Q$-value.
The $Q$-value of a nuclear reaction is the difference between the total binding energy of products and reactants: \[ Q = (B.E._{products} - B.E._{reactants}) \] If $Q>0$, the reaction is exothermic; if $Q<0$, it is endothermic.
Step 2: Binding energy of reactants.
Deuterium (${}^2H$) has 2 nucleons and B.E. per nucleon $= 1.25 \, \text{MeV}$. \[ B.E._{D} = 2 \times 1.25 = 2.5 \, \text{MeV} \] Since two deuterium nuclei are involved: \[ B.E._{reactants} = 2 \times 2.5 = 5.0 \, \text{MeV} \]
Step 3: Binding energy of product helium.
Helium-4 has 4 nucleons and B.E. per nucleon $= 7.2 \, \text{MeV}$. \[ B.E._{He} = 4 \times 7.2 = 28.8 \, \text{MeV} \] So, \[ B.E._{products} = 28.8 \, \text{MeV} \]
Step 4: Calculate $Q$-value.
\[ Q = B.E._{products} - B.E._{reactants} = 28.8 - 5.0 = 23.8 \, \text{MeV} \] (approx $23.6 \, \text{MeV}$ considering significant figures).
Step 5: Nature of reaction.
Since $Q>0$, the reaction is exothermic (energy releasing).
Step 6: Mass defect.
Energy released $Q = \Delta m c^2$. In atomic mass units ($1u = 931 \, \text{MeV}$), \[ \Delta m = \frac{Q}{931} = \frac{23.8}{931} \approx 0.025 \, u \]
Step 7: Conclusion.
- The reaction is exothermic.
- $Q$-value $\approx 23.6 \, \text{MeV}$.
- Mass defect $\Delta m \approx 0.025 u$.