24.21 P - Chemistry: Molecular Nature Of Matter And Change

Question

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

(a) ${}_{\mathbf{47}}{}^{\mathbf{111}}\mathbf{\text{Ag}}$

(b)  ${}_{\mathbf{17}}{}^{\mathbf{41}}\mathbf{\text{Cl}}$

(c)  ${}_{\mathbf{44}}{}^{\mathbf{110}}\mathbf{\text{Ru}}$

Step-by-Step Solution

Answer

The most likely mode of decay for each given is:

a) Beta decay.

b) Beta decay.

c) Beta decay.

Step 1: Definition of beta decay.

Beta decay occurs in a nucleus with too many protons, or neutrons occurs when one proton or neutron changes into the other.

Step 2: Subpart (a)

The alphabet A represents the mass number (sum of the number of protons and number of neutrons).

Atomic number Z (number of protons)

Number of neutrons= (A-Z)

 

Beta decay occurs in nuclides with a high N/Z ratio (above the zone of stability). As a neutron decays into a proton, N  decreases, Z rises, and the N/Z ratio falls.

Positron emission/electron capture: nuclides having a low  N/Z  ratio (below the stability zone) will emit positrons or capture electrons. Because the proton is converted to a neutron, N increases, Z falls, and the N/Z ratio rises.

Alpha decay: Nuclides with   (too heavy) suffer alpha decay, resulting in a 2 percent decline in both N and Z values (and become lighter).

 

 ${}_{\mathbf{47}}{}^{\mathbf{111}}\mathbf{\text{Ag}}$

The number of protons is Z = 47.

The number of neutrons is:

 $\begin{array}{rcl}N& =& A-Z\\ & =& 111-47\\ & =& 64\end{array}$

Proton to neutron ratio is: 

 $\begin{array}{rcl}\frac{\text{Z}}{\text{Z}}& =& \frac{\text{6i}}{47}\\ & =& 1.36:1\end{array}$

 

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

Step 3: Subpart (b)

${}_{17}{}^{41}\text{Cl}$

The number of protons is Z = 17.
 The number of neutrons is:

 $\begin{array}{rcl}N& =& A-Z\\ & =& 41-17\\ & =& 24\end{array}$

Proton to nucleon quantitative relation is: 

$\begin{array}{rcl}\frac{\text{K}}{\text{Z}}& =& \frac{24}{17}\\ & =& 1.41:1\end{array}$

Proton to nucleon quantitative relation of  this nuclei is over the zone of stability. 

Therefore it exhibits beta decay.

Step 4: Subpart (c)

${}_{44}{}^{110}\text{Ru}$

The number of protons is Z = 44.

The number of neutrons is:

 $\begin{array}{rcl}N& =& A-Z\\ & =& 110-44\\ & =& 66\end{array}$

Proton to nucleon quantitative relation is:

 $\begin{array}{rcl}\frac{N}{\text{Z}}& =& \frac{66}{44}\\ & =& 1.5:1\end{array}$

Proton to nucleon quantitative relation of this nuclei is over the zone of stability. 

Therefore it will exhibit beta decay.