# Decrease in the atomic number is observed during which of the following:

a. Alpha emission

b. Beta minus emission

c. Positron emission

d. Electron capture

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**Hint:**Answer the given question with the help of theories of alpha $\left( \alpha \right)$, beta $\left( \beta \right)$ and gamma $\left( \gamma \right)$ decays. In the $\beta $ decay, the nucleus emits an electron and gets transformed into a nucleus.

**Formula used:**

${}_Z^AX$

Where,

$Z$ is the atomic number and $A$ is the mass number

**Complete step by step answer:**

An interesting property of certain nuclei is their ability to transform on their own from one value of atomic number or mass number. Such unstable nuclei undergo some sort of decay process, until stability is reached. In certain decay processes no change of atomic and mass numbers occurs.

To answer the given question, let us write the decay equation for each option separately.

Consider $X$ as a nuclear particle with $Z$ as the atomic number and $A$ as the mass number. That is:

$ \Rightarrow {}_Z^AX$

The given term is common for all the options.

Consider Option (A)- Alpha emission:

In alpha emission the alpha particle has 4 mass numbers and 2 atomic numbers $\left( {_{2}^{4}{\alpha} } \right)$. After the alpha decay the particle will be like \[_{Z - 2}^{A - 4}Y\]. That is,

$ \Rightarrow {}_Z^AX \to _{Z - 2}^{A - 4}Y$

From this we can understand that the atomic number of the particle is decreased.

Consider option (B)- Beta minus emission:

In beta emission, the beta particle has zero mass number and $ - 1$ atomic number $\left( {{}_{ - 1}^0\beta } \right)$. After the beta decay the particle will be like \[_{Z + 1}^AY\]. That is,

$ \Rightarrow {}_Z^AX \to _{Z + 1}^AY$

In beta minus emission the particle emits an electron and gets transformed into the nucleus with the atomic number $Z + 1$. This means that the atomic number is increased and not decreased.

Consider option(C) - Positron emission:

In positron emission, the positron particles have o mass number and have the positive charge as the atomic number $\left( _{+1}^{0}{\beta}\right)$. Therefore, the decay equation will be like:

$ \Rightarrow {}_Z^AX \to _{Z - 1}^{A + 0}Y$.

Positron emission is a type of the beta decay process. In which the positron is emitted instead of the electron. In the given decay equation, it is clear that the atomic number is decreased.

Consider option (D) - Electron capture:

Electron capture is similar to the positron emission. That is the mass number is 0 and has the negative charge as the atomic number $\left(_{-1}^{0}{\beta} \right)$. The decay equation will be like:

$ \Rightarrow {}_Z^AX \to _{Z - 1}^{A + 0}Y$

Another type of the beta decay process is Electron capture. From the equation it is clear that atomic number is reduced.

Therefore, from the given explanations it is clear that the alpha emission, positron emission and the electron capture process the atomic number gets reduced after the decay process. In the beta decay process the atomic number will get increased.

**Hence, the correct answers are option (A), (C) and (D).**

**Note:**One of the interesting applications of the decay process is the technique of radioactive dating. The estimation of the age of any object that was alive once, using the natural radioactivity of $_6^{14}C$ can be done.