Q7 P - Chemistry: Molecular Nature Of Matter And Change

Question

Why might it be difficult to use only a nuclide's $\mathit{N}\mathbf{/}\mathit{Z}$ ratio to predict whether it will decay by ${\mathit{\beta }}^{\mathbf{+}}$emission or by ${\mathbf{\text{e}}}^{\mathbf{-}}$capture? What another factor is important?

 

Step-by-Step Solution

Answer

For lighter stable nuclei, neutron to proton ratio $(\text{N}/\text{Z})$ is l:1.

As the mass of stable nuclei increases, neutron to proton ratio ( $\text{N}/\text{Z}$ ) increases as well. So, in heavy stable nuclei,  N to Z ratio is around .$1.5:1$Nucleus is more stable when  N and  Z numbers are even, than when  N and  Z numbers are odd.

 

Step 1: Given information

Let  A be mass number

Z be atomic number (number of protons)

 N be number of neutrons $(\text{N}=\text{A}-\text{Z})$

 

Step 2:Neutron to proton ratio

 

The ratio of neutrons to protons in a stable nucleus is thus around $\mathbf{1}:\mathbf{1}$for small nuclei ($\mathbf{Z}<\mathbf{20}$). The ratio increases slowly with atomic number up to about $\mathbf{1}.\mathbf{58}$at high Z

 

Step 3: Ratio

- For lighter stable nuclei, neutron to proton ratio (N/Z) is 1:1.- As the mass of stable nuclei increases, neutron to proton ratio ( $\text{N}/\text{Z}$ ) increases as well. So, in heavy stable nuclei, N to Z  ratio is around $1.5:1$- Nucleus is more stable when N and Z numbers are even, than when  N and Z numbers are odd