Question

Decrease in atomic number is observed in
A.$\alpha$- emission
B.$\beta$ - emission
C.Positron
D.Electron capture

Answer 1

As we know that the unstable nuclear will decompose at the rate of spontaneous or as in common terminology we call it as a decay into the most stable configuration. Since it will do it only in specific ways by with the help of emitting certain particles as well as a certain form being in electromagnetic radiation. The most common emissions are: alpha $\left( \alpha \right)$ particle; beta $\left( \beta \right)$ particle; gamma $\left( \gamma \right)$

Complete step-by-step answer: First of all we need to understand that instability of the nucleus is due to imbalance in the number of protons and neutrons in the atomic nucleus. This causes radioactive disintegration leading to emission of particles.
An Alpha decay refers to emission of an alpha particle (that is $_{2}^{4}He$ nucleus). Helium $\left( _{2}^{4}He \right)$ contains two protons and two neutrons. Thus, after its emissions, the mass number and the atomic number of the emitting nucleus reduces by 4 and 2 respectively. Also, Beta $\left( \beta \right)$ decay is another type of radioactive decay which involves emission of $\beta$ - particles (fast energetic electron and positron).
Now, if we talk about electron capture, basically it is also a type of $\beta$ - emission in which an electron from the inner orbital is captured by the nucleus.
So, we can summarize in the following nuclear reaction, there occurs a decrease in atomic number $\left( Z \right)$ .
$\alpha$ - Emission : Alpha decay or α-decay refers to any decay where the atomic nucleus of a particular element releases $_{2}^{4}He$ as well as transforms into an atom of completely different element. This decay leads to a decrease in the mass number and atomic number, due to release of a helium atom.
$_{Z}{{X}^{A}}{{\to }_{Z}}H{{e}^{4}}{{+}_{Z-2}}{{Y}^{A-4}}$
Positron: Positron emission as it is known as beta plus decay (β+ decay)which is a subtype of radioactive decay called beta decay it is in which a proton inside a radionuclide nucleus is converted into a neutron while releasing a positron and an electron neutrino.
$_{Z}{{X}^{A}}{{\to }_{+1}}{{e}^{0}}{{+}_{Z-1}}{{Y}^{A}}$
Electron Capture: Electron capture is a comparatively minor decay mode caused by the weak force. The best-known example is of potassium 40:11% of the nuclei of that isotope of potassium present in our body decay by electron capture. The electron's capture triggers emission of an invisible neutrino by nucleus.
$_{Z}{{X}^{A}}{{+}_{-1}}{{e}^{0}}{{\to }_{Z-1}}{{Y}^{A}}$
($\beta$ - Emission): In beta decay, radioactive isotope emits an electron as well as positron. This decay occurs by following the radioactive laws, just as an alpha decay does. An example of beta decay is
$_{Z}{{X}^{A}}{{\to }_{-1}}{{e}^{0}}{{+}_{Z+1}}{{Y}^{A}}$

Thus, the above example clearly shows that the option (A), (C) and (D) are correct i.e. $\alpha$- emission; Positron, Emission and Electron capture respectively.

Note: Note that we might be in a little bit of confusion that the Beta emission. Although Beta particles refer to fast energetic electrons and positrons, generally, we consider electron: emission as $\beta$ - emission; similarly whenever we find any further questions in future regarding $\beta$ - emission, referring should be done as electron emission only.