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?

Short 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-by-step solution

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 highZ

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