Under certain circumstances,a nucleus can decay by emitting a particle more massive than an $\alpha$-particle. Consider the following decay processes:
$_{88}^{223} Ra \rightarrow _{82}^{209} Pb + _{6}^{14} C$
$_{88}^{223} Ra \rightarrow _{86}^{219} Rn + _{2}^{4} He$
Calculate the $Q$-values for these decays and determine that both are energetically allowed.

(N/A) For the $_{6}^{14} C$ emission reaction:
$^{223}_{88} Ra \rightarrow _{82}^{209} Pb + _{6}^{14} C$
Given masses: $m(^{223}_{88} Ra) = 223.01850 \, u$,$m(_{82}^{209} Pb) = 208.98107 \, u$,$m(_{6}^{14} C) = 14.00324 \, u$.
The $Q$-value is $Q = (m_{initial} - m_{final}) c^2$.
$Q = (223.01850 - 208.98107 - 14.00324) \, u \times c^2 = 0.03419 \, u \times c^2$.
Using $1 \, u = 931.5 \, MeV/c^2$,we get $Q = 0.03419 \times 931.5 \, MeV = 31.848 \, MeV$.
Since $Q > 0$,the reaction is energetically allowed.
For the $_{2}^{4} He$ emission reaction:
$^{223}_{88} Ra \rightarrow _{86}^{219} Rn + _{2}^{4} He$
Given masses: $m(^{223}_{88} Ra) = 223.01850 \, u$,$m(_{86}^{219} Rn) = 219.00948 \, u$,$m(_{2}^{4} He) = 4.00260 \, u$.
The $Q$-value is $Q = (223.01850 - 219.00948 - 4.00260) \, u \times c^2 = 0.00642 \, u \times c^2$.
$Q = 0.00642 \times 931.5 \, MeV = 5.98 \, MeV$.
Since $Q > 0$,the reaction is energetically allowed.