A nuclide of an alkaline earth metal undergoes radioactive decay by emission of three $\alpha $ particles in succession. The group of the periodic table to which the resulting daughter element would belong is
A.Group $14$
B.Group $16$
C.Group $4$
D.Group $6$
Answer
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Hint: The term alpha decay is a type of radioactive decay where the unstable atomic nuclei emits a helium nucleus i.e. an alpha particle and, in this process, it further transforms into another more stable element. There are four nucleons present in the alpha particle i.e. two neutrons and two protons.
Complete step by step answer:
Due to the nuclear instability, an atom’s nucleus exhibits the phenomenon of Radioactivity. In this process energy is lost due to the radiation that is emitted out of the unstable nucleus of an atom.
Now, the term alpha decay is a type of radioactive decay where the unstable atomic nuclei emits a helium nucleus (alpha particle) and in the process, it transforms into another more stable element. The alpha particle consists of four nucleons i.e. two neutrons and two protons. Basically, an alpha particle is identical to the nucleus of a helium atom.
Now, in the given question, the nuclide of an alkaline earth metal undergoes radioactive decay by the emission of three $\alpha $ particles in succession. The alkaline earth metals are present in group 2.
When the decay happens in group $2$ elements, basically it happens in the element of the ${U_{235}}$ series.
Now, the final decay product of uranium series is lead and it belongs to group $14$
Moreover, after $3$ alpha decay, the atomic number decreases by $6$. The parent atom was in the $2nd$ group so the daughter element will belong to the ${14^{th}}$ group since the atomic number decreases by $6$.
Hence, option A is correct.
Note:
Radioactivity is used in domestic smoke detectors, to sterilize medical instruments, to diagnose and treat diseases and to produce electric power. The smoke detector used in the United States is known as Americium-241. The alpha particles are also known as doubly ionized helium nuclei.
Complete step by step answer:
Due to the nuclear instability, an atom’s nucleus exhibits the phenomenon of Radioactivity. In this process energy is lost due to the radiation that is emitted out of the unstable nucleus of an atom.
Now, the term alpha decay is a type of radioactive decay where the unstable atomic nuclei emits a helium nucleus (alpha particle) and in the process, it transforms into another more stable element. The alpha particle consists of four nucleons i.e. two neutrons and two protons. Basically, an alpha particle is identical to the nucleus of a helium atom.
Now, in the given question, the nuclide of an alkaline earth metal undergoes radioactive decay by the emission of three $\alpha $ particles in succession. The alkaline earth metals are present in group 2.
When the decay happens in group $2$ elements, basically it happens in the element of the ${U_{235}}$ series.
Now, the final decay product of uranium series is lead and it belongs to group $14$
Moreover, after $3$ alpha decay, the atomic number decreases by $6$. The parent atom was in the $2nd$ group so the daughter element will belong to the ${14^{th}}$ group since the atomic number decreases by $6$.
Hence, option A is correct.
Note:
Radioactivity is used in domestic smoke detectors, to sterilize medical instruments, to diagnose and treat diseases and to produce electric power. The smoke detector used in the United States is known as Americium-241. The alpha particles are also known as doubly ionized helium nuclei.
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