24.23 P - Chemistry: Molecular Nature Of Matter And Change

Question

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

(a) ${}^{\mathbf{106}}\mathbf{\text{In}}$

(b) ${}^{\mathbf{141}}\mathbf{\text{Eu}}$

(c) ${}^{\mathbf{241}}\mathbf{\text{Am}}$

Step-by-Step Solution

Answer

The most likely mode of decay for each given is:

a) Positron emission or electron capture.

b) Positron emission or electron capture.

c) Alpha decay.

Step 1: Definition of nuclide

Any species of atom that persists for a quantifiable period of time is referred to as a nuclide.

Step 2: Find the most likely mode of decay for part a

a)

A: a large number..

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).

 ${}^{106}\ln$

 In has an atomic number of 49.

 Z = 49 protons are present.

The neutron count is:

 $\begin{array}{rcl}N& =& A-Z\\ & =& 106-49\\ & =& 57\end{array}$

The proton-neutron ratio is:

 $\begin{array}{rcl}\frac{N}{Z}& =& \frac{57}{49}\\ & =& 1.16:1\end{array}$

This nucleus' proton to neutron ratio is less than the stability zone, hence it will emit positrons or collect electrons.

Step 3: Find the most likely mode of decay for Part b

b) 

 ${}^{141}\text{Eu}$

Eu has an atomic number of 63.

Z = 63 protons are present.

The neutron count is:

 $\begin{array}{rcl}N& =& A-Z\\ & =& 141-63\\ & =& 78\end{array}$

The proton-neutron ratio is:

 $\begin{array}{rcl}\frac{N}{Z}& =& \frac{78}{63}\\ & =& 1.24:1\end{array}$

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: Find the most likely mode of decay for part c

c)

${}^{241}\text{Am}$

The atomic number of Am is 95.

Am has an atomic number of 95.

We can observe that Am has an atomic number of ($95>83)$, which makes it too heavy to be stable.

Thus it will decay into alpha particles.