Question

A uranium-238 nucleus, $_{92}^{238}U$, undergoes decays to form uranium-234, $_{92}^{234}U$. Which series of decays and give this result?

Answer 1

If there is a decrease in the mass number by 4 and a decrease in the atomic number by 2 in the nucleus of an atom then, there is alpha-decay ($\alpha -decay$). If there is an increase in the atomic number in the nucleus by 1 then, it is negative beta-decay (${{\beta }^{-}}-decay$). If there is a decrease in the number of atomic numbers in the nucleus by 1 then, it is positive beta-decay (${{\beta }^{+}}-decay$).

Complete answer:
Radioactive decay means there is an emission of particles from the nucleus of the element which will convert it into a new nucleus. So, the given nucleus in the question is uranium-238 nucleus, $_{92}^{238}U$, and it undergoes radioactive decays to form uranium-234, $_{92}^{234}U$.
There is only a decrease in the mass number by 4 units.
So, in this decay, two types of decay will take place, i.e., alpha-decay and negative beta-decay.
If there is a decrease in the mass number by 4 and a decrease in the atomic number by 2 in the nucleus of an atom then, there is alpha-decay ($\alpha -decay$). If there is an increase in the atomic number in the nucleus by 1 then, it is negative beta-decay (${{\beta }^{-}}-decay$).
First, uranium-238 nucleus, $_{92}^{238}U$ will undergo alpha-decay. The reaction is given below:
$_{92}^{238}U\to _{90}^{234}U+_{2}^{4}He$
Now, the formed nucleus is $_{90}^{234}U$, this will undergo negative beta-decay two times to increase the atomic number by 2. The reaction is given below:
$_{90}^{234}U\to _{91}^{234}U+_{-1}^{0}e$
$_{9}^{234}U\to _{92}^{234}U+_{-1}^{0}e$

Note:
There is another type of radioactive decay other than alpha-decay and beta-decay, which is gamma-decay in which high energy gamma particle is emitted and after the gamma-decay, there is no change in the atomic and mass number.