Question

Complete the following nuclear reaction for $\alpha$ and $\beta$ decay:
$\left( i \right){}_{92}^{238}U\to ?+{}_{2}^{4}He+Q$
$\left( ii \right){}_{11}^{22}Na\to {}_{10}^{22}Ne+?+v$

Answer 1

As a first step, you could define a nuclear reaction. After that you could define what alpha and beta decay reactions are. After that you could try to understand what all particles are released in these reactions. Thereby, you will be able to easily identify the missing things in the given equations.

Complete step-by-step solution:
In the question, we are given two nuclear reactions for $\alpha$ and $\beta$ decay.
As a first step let us recall what a nuclear reaction is.
Nuclear reaction is basically defined as the process when two nuclei or maybe an external subatomic particle and a nucleus collide with each other to produce one or more than one new nuclides. The compulsory condition for nuclear reaction is the transformation of at least one nuclide to another.
Now, what is an alpha decay?
It can be defined as a radioactive decay where an atomic nucleus emits an alpha particle and thus transform into a new atomic nucleus and the newly formed atom will now have its mass number reduced by 4 and atomic number reduced by 2 when compared to its parent atom.
(i) Here ${}_{92}^{238}U$ produces a new atom which would be${}_{92-2}^{238-4}X$. That is,${}_{90}^{234}X$
The newly formed atom would be thorium and the reaction can be written as,
$\left( i \right){}_{92}^{238}U\to {}_{90}^{234}Th+{}_{2}^{4}He+Q$
Now, what is beta decay?
Here the original nuclide is transformed into an isobar of that nuclide by emitting a beta particle from an atomic nucleus.
(ii) In the second reaction we see that the beta particle is missing. So, we could rewrite the equation as,
$\left( ii \right){}_{11}^{22}Na\to {}_{10}^{22}Ne+{{\beta }^{+}}+v$

Note: We have easily arrived at these conclusions with help of the hint given in the question itself that we are being given alpha and beta decay reactions. Also, you may have now realized that an alpha particle is nothing but a helium atom. Helium is that element that has 4 as its mass number and 2 as its atomic number.