Find the $Q$-value and the kinetic energy of the emitted $\alpha$-particle in the $\alpha$-decay of $(a) \; ^{226}_{88} Ra$ and $(b) \; ^{220}_{86} Rn$.
Given:
$m(^{226}_{88} Ra) = 226.02540 \; u$
$m(^{222}_{86} Rn) = 222.01750 \; u$
$m(^{220}_{86} Rn) = 220.01137 \; u$
$m(^{216}_{84} Po) = 216.00189 \; u$
$m(^{4}_{2} He) = 4.002603 \; u$

$(a)$ The $\alpha$-decay of $^{226}_{88} Ra$ is given by: $^{226}_{88} Ra \rightarrow ^{222}_{86} Rn + ^{4}_{2} He$.
The $Q$-value is calculated as: $Q = [m(^{226}_{88} Ra) - (m(^{222}_{86} Rn) + m(^{4}_{2} He))] \times 931.5 \; MeV/u$.
$Q = [226.02540 - (222.01750 + 4.002603)] \times 931.5 \; MeV = 0.005297 \times 931.5 \approx 4.934 \; MeV$.
The kinetic energy of the $\alpha$-particle is $K_{\alpha} = (\frac{A-4}{A}) \times Q = (\frac{222}{226}) \times 4.934 \approx 4.847 \; MeV$.
$(b)$ The $\alpha$-decay of $^{220}_{86} Rn$ is given by: $^{220}_{86} Rn \rightarrow ^{216}_{84} Po + ^{4}_{2} He$.
The $Q$-value is: $Q = [m(^{220}_{86} Rn) - (m(^{216}_{84} Po) + m(^{4}_{2} He))] \times 931.5 \; MeV/u$.
$Q = [220.01137 - (216.00189 + 4.002603)] \times 931.5 \; MeV = 0.006877 \times 931.5 \approx 6.406 \; MeV$.
The kinetic energy of the $\alpha$-particle is $K_{\alpha} = (\frac{216}{220}) \times 6.406 \approx 6.289 \; MeV$.