Question 21.SE.3: Predict the mode of decay of (a) carbon-14, (b) xenon-118.

Analyze We are asked to predict the modes of decay of two nuclei.

Plan To do this, we must locate the respective nuclei in Figure 21.1 and determine their positions with respect to the belt of stability in order to predict the most likely mode of decay.

Solve

(a) Carbon is element 6. Thus, carbon-14 has 6 protons and 14 – 6 = 8 neutrons, giving it a neutron-to-proton ratio of 1.25. Elements with Z < 20 normally have stable nuclei that
contain approximately equal numbers of neutrons and protons (n/p = 1). Thus, carbon-14 is located above the belt of stability, and we expect it to decay by emitting a beta particle to decrease the n/p ratio:

{}_6^{14}C \longrightarrow{}_7^{14}N +{}_{-1}^0 e

This is indeed the mode of decay observed for carbon-14, a reaction that lowers the n/p ratio from 1.25 to 1.0.

(b) Xenon is element 54. Thus, xenon-118 has 54 protons and 118 – 54 = 64 neutrons, giving it an n/p ratio of 1.18. According to Figure 21.1, stable nuclei in this region of the belt  of stability have higher neutron-to-proton ratios than xenon-118. The nucleus can increase this ratio by either positron emission or electron capture:

\begin{aligned}{}_{54}^{118}Xe & \longrightarrow{}_{53}^{118}I +{}_{+1}^0 e \\{}_{54}^{118}Xe +{}_{-1}^0 e & \longrightarrow{}_{53}^{118}I\end{aligned}

In this case, both modes of decay are observed.

Comment Keep in mind that our guidelines do not always work. For example, thorium-233, which we might expect to undergo alpha decay, actually undergoes beta emission. Furthermore, a few radioactive nuclei lie within the belt of stability. Both {}_{60}^{146}Nd and {}_{60}^{148}Nd, for example, are stable and lie in the belt of stability. {}_{60}^{147}Nd, however, which lies between them, is radioactive.