Question

In the radioactive decay process of Uranium the initial nuclide is \[{}_{92}{U^{238}}\] and final nuclide is \[{}_{82}P{b^{206}}\] when uranium nucleus decays to Lead, then the number of \[\alpha \]-particles and \[\beta \]-particles emitted will respectively be:
A) \[\left( {\text{A}} \right){\text{ }}8,6\]
B) \[\left( {\text{B}} \right){\text{ }}8,4\]
C) \[\left( {\text{C}} \right){\text{ 6}},8\]
D) \[\left( {\text{D}} \right){\text{ 4}},8\]

Answer 1

Hint:An element that emits radiations such as alpha rays, beta rays, and gamma rays are called radioactive elements. It decays on radiating radiations. This phenomenon is called radioactivity. There are two types of processes in radioactivity. They are nuclear fusion and nuclear fission. When the two or smaller nuclei combine to form a bigger nucleus, it is called nuclear fusion. When the single bigger nucleus bursts into two or smaller nuclei are called nuclear fission.

Complete step by step answer:
From the given question, the process can be simplified as
\[{}_{92}{U^{238}} \to {}_{82}P{b^{206}} + x({}_2H{e^4}) + y({}_{ - 1}{e^0}).....\left( 1 \right)\]
From the equation, we can say that the given equation is nuclear fission. The nuclear fission can happen in \[3\] ways by emitting alpha, beta, and gamma rays.
The alpha particles are simply the helium atoms. The beta rays are simply electrons which is represented as \[{}_{ - 1}{e^0}\]
To find the value of \[x\] and \[y\] in the equation by equating the atomic and mass numbers on both sides.
Considering the mass numbers first,
\[238 = 206 + 4x\]
Taking \[x\] terms as LHS and integers as RHS we get,
\[ \rightarrow 4x = 238 - 206\]
On subtracting we get,
\[ \rightarrow 4x = 32\]
Let us divided \[4\] on both side we get,
\[ \rightarrow x = \dfrac{{32}}{4}\]
\[\therefore x = 8\]
Therefore \[8\] alpha particles are released in the given reaction. Hence the equation \[\left( 1 \right)\] becomes,
\[{}_{92}{U^{238}} \to {}_{82}P{b^{206}} + 8({}_2H{e^4}) + y({}_{ - 1}{e^0})\]
To find the number of beta particles, we are equating the atomic numbers. Therefore,
\[92 = 82 + 16 + ( - y)\]
Let us add the terms we get,
\[ \rightarrow 92 = 98 - y\]
Taking \[y\] terms as LHS and integers as RHS we get,
\[ \rightarrow y = 98 - 92\]
On subtracting we get
\[\therefore y = 6\]
Hence 6 beta particles are released in this reaction. And the equation (1) becomes,
\[{}_{92}{U^{238}} \to {}_{82}P{b^{206}} + 8({}_2H{e^4}) + 6({}_{ - 1}{e^0})\]

There are 8 alpha particles and 6 beta particles are released. Hence option A is correct.

Note:The phenomenon of emission of radiation by the elements is called radioactivity. The elements emit radiation are called radioactive elements. It is due to the unstable nucleus. The radioactive elements radiate \[\alpha \]-rays, \[\beta \]-rays and \[\gamma \]-rays. In the process of decaying, one element changes into another element as the atomic number of the element changes.