Question

Calculate $Q$ value of the following nuclear reaction.

${{}_{13}Al}^{27}+{{}_{2}He}^4⟶{{}_{14}Si}^{30}+{{}_{1}H}^1+Q$

The exact mass of ${{}_{13}Al}^{27}$ is $26.9815amu$, ${{}_{14}Si}^{30}$ is $29.9738amu$, ${{}_{2}He}^4$ is $4.0026amu$ and ${{}_{1}H}^1$ is $1.0078amu$.

Answer 1

Mass of reactants $=$ mass of ${{}_{13}Al}^{27}+$ mass of ${{}_{2}He}^4$
Mass of reactants $=26.9815+4.0026=30.9841$ amu
Mass of products $=$ mass of ${{}_{14}Si}^{30}+$ mass of ${{}_{1}H}^1$
Mass of products $=29.9738+1.0078=30.9816$ amu
The mass defect $\Delta m=30.9841-30.9816=0.0025$ amu
Q value $=0.0025amu\times 931.5MeV/amu=2.33MeV$