In \[{\mathbf{\beta }} - \]decay \[{\mathbf{n}}/{\mathbf{p}}\] ratio :
A.Remain unchanged
B.May increase or decrease
C.Increases
D.Decreases

Hint: To answer this question recall the concept of first-order reactions. The first-order decomposition can be defined as the chemical reaction involving only one chemical species, in which the rate of decrease of the concentration of the reactant is directly proportional to its concentration. Radioactive is an example of the first-order decomposition.

Complete step by step answer:
\[{\mathbf{\beta }} - \] decay takes place when the emission occurs, when a proton is formed along with a $\beta $particle from a neutron. In \[{\mathbf{\beta }} - \]emission \[{\mathbf{n}}/{\mathbf{p}}\] ratio decreases. The reaction can be represented as: \[{\;^0}{n_1}{ \to ^1}{p_1}{ + ^{ - 1}}{\beta _0}\]
From the above equation, we get to know that there is a decrease of 1 in total number of neutrons and an increase of 1 in total number of protons. This clearly indicates the fact that in \[{\mathbf{\beta }} - \] emission \[{\mathbf{n}}/{\mathbf{p}}\] ratio decreases.

Hence, the correct option is D.

Note:
The emissions in most of the spontaneous radioactive decay involves alpha $(\alpha )$ particle, the beta $(\beta )$ particle, the gamma-ray, and the neutrino. The alpha particle is the nucleus of doubly charged ${\text{He}}_2^4$. Beta particles can be beta minus beta plus. Beta minus is an electron created in the nucleus during beta decay. Beta plus particle is also known as a positron, is the antiparticle of the electron; when brought together, two such particles will mutually annihilate each other. Gamma decay occurs in the nucleus. In this decay, there is no emission of particles ejected from the nucleus. It is a form of energy radiation decay.