13.26 Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes:

\(\begin{aligned} &{ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{82}^{209} \mathrm{~Pb}+{ }_{6}^{14} \mathrm{C} \\ &{ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{86}^{219} \mathrm{Rn}+{ }_{2}^{4} \mathrm{He} \end{aligned}\)

Calculate the Q-values for these decays and determine that both are energetically allowed.

Hint: The Q-value is given by \(E=\Delta mc^2\)​.
Step 1: Find the Q-value for first reaction.
For the emission of C614, the nuclear reaction:R88223aP82209b+C614
We know that:
Mass of R88223a, m1=223.01850 u
Mass of P82209b, m2=208.98107 u
Mass ofC614, m3=14.00324 u
Hence, the Q-value of the reaction is given as:
Q=(m1-m2-m3) c2=(223.01850-208.98107-14.00324)c2=(0.03419 c2) u
But 1 u = 931.5 MeV/c2
Q=0.03419×931.5=31.848 MeV
Hence, the Q-value of the nuclear reaction is 31.848 MeV. Since the value is positive, the reaction is energetically allowed.


Step 2: Find the Q-value for the second reaction.

For the emission of \({ }_{2}^{4} \mathrm{He}\), the nuclear reaction is: \({ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{86}^{219} \mathrm{Rn}+{ }_{2}^{4} \mathrm{He}\)
We know that:
Mass of \({ }_{88}^{223} R a, m_{1}=223.01850~u\)
Mass of \({ }_{86}^{219} R n, m_{2}=219.00948~u\)
Mass of \({ }_{2}^{4} \mathrm{He}, m_{3}=4.00260~u\)
Q-value of this nuclear reaction is given as:
Q=(m1-m2-m3) c2=(223.01850-219.00948-4.00260)C2=(0.00642 c2)u==0.00642×931.5=5.98 MeV
Hence, the Q value of the second nuclear reaction is 5.98 MeV. Since the value is positive, the reaction is energetcally allowed.