24.22 P - Chemistry: Molecular Nature Of Matter And Change

Question

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

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

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

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

Step-by-Step Solution

Answer

The most likely mode of decay for each given is:

a) Beta decay.

b) Position emission or electron capture.

c) Alpha decay.

Step 1: Definition of nuclide

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

Step 2: Subpart (a)

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.

Step 3: Subpart (b)

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

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.

Step 4: Subpart (c)

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