Calculate $Q$ value of the following nuclear reaction.
${_{13} A l}^{27} + {_{2} H e}^{4} ⟶ {_{14} S i}^{30} + {_{1} H}^{1} + Q$

The exact mass of ${_{13} A l}^{27}$ is $26.9815 a m u$ , ${_{14} S i}^{30}$ is $29.9738 a m u$ , ${_{2} H e}^{4}$ is $4.0026 a m u$ and ${_{1} H}^{1}$ is $1.0078 a m u$ .

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Solution

Mass of reactants $=$ mass of ${_{13} A l}^{27} +$ mass of ${_{2} H e}^{4}$
Mass of reactants $= 26.9815 + 4.0026 = 30.9841$ amu
Mass of products $=$ mass of ${_{14} S i}^{30} +$ mass of ${_{1} H}^{1}$
Mass of products $= 29.9738 + 1.0078 = 30.9816$ amu
The mass defect $Δ m = 30.9841 - 30.9816 = 0.0025$ amu
Q value $= 0.0025 a m u \times 931.5 M e V / a m u = 2.33 M e V$

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