Question: What is the most likely mode of decay for each?

(a)${}^{\mathbf{15}}\mathbf{\text{C}}$

(b) ${}^{\mathbf{120}}\mathbf{\text{Xe}}$

(c)${}^{\mathbf{224}}\mathbf{\text{Th}}$

The representation of an atom using its mass number and atomic number is called a nuclide.

The mass number( of an atom is represented with A.

Z is the atomic number (number of protons)

N: the total number of neutrons (A - Z)

Beta decay: nuclides with a high N/Z ratio (above the stability zone) will beta decay. When a neutron decays into a proton, N lowers, Z grows, and the N/Z ratio decreases.

Positron emission/electron capture: nuclides with a low N/Z ratio (below the stability zone) will emit positrons or collect electrons. As a result of the proton being converted to a neutron N rises, Z falls, and the N/Z ratio rises.

Alpha decay: nuclides with Z>83 (too heavy) will undergo alpha decay, so both N and Z values will decrease by 2 (and become lighter).

${}^{15}\text{C}$

C has an atomic number of 6.

Z=6 protons are present.

The neutron number is

$\begin{array}{rcl}N& =& A-Z\\ & =& 15-6\\ & =& 9\end{array}$

The proton-to-neutron ratio is:

$\begin{array}{rcl}\frac{N}{Z}& =& \frac{9}{6}\\ & =& 1.5:1\end{array}$

Because this nucleus's proton to neutron ratio exceeds the stability zone, it will experience beta decay.

${}^{120}\text{Xe}$

Xe has an atomic number of 54.

Z = 54 is the number of protons.

The number of neutrons is calculated as:

$\begin{array}{rcl}\text{N}& =& \text{A}-\text{Z}\\ & =& 120-54\\ & =& 66\end{array}$

$\frac{\text{X}}{\text{Z}}=\frac{66}{54}$

$=1.22:1$ is the proton to neutron ratio.

Because the proton to neutron ratio of this nucleus is lower than the zone of stability, positron emission or electron capture will occur.

The has an atomic number of 90.

The atomic number makes it too heavy to be stable. Therefore, it will decay alpha particles.

The required graph is:

Band of stability