Question

For the following fusion reaction, \[ {}^1_1\text{H} + {}^1_1\text{H} \longrightarrow {}^4_2\text{He} + 2\beta^+ + 2\nu + \gamma \] The Q-value (energy of the reaction) in MeV (rounded off to one decimal place) is ________. (Given: Mass of ${}^1$H nucleus = 1.007825 u; Mass of ${}^4$He nucleus = 4.002604 u)

Hint:

Fusion reactions release large energy due to high mass-to-energy conversion, as per $E = \Delta m c^2$.

Answer 1

Correct Answer: 26.7

Step 1: Calculate mass defect.
Mass of reactants $= 4 \times 1.007825 = 4.0313$ u
Mass of products $= 4.002604$ u
Mass defect, $\Delta m = 4.0313 - 4.0026 = 0.0287$ u
Step 2: Convert mass defect to energy.
\[ 1 \, \text{u} = 931.5 \, \text{MeV} \] \[ Q = \Delta m \times 931.5 = 0.0287 \times 931.5 = 26.7 \, \text{MeV} \] Step 3: Conclusion.
Hence, Q-value = 26.7 MeV.