A uranium nucleus (atomic number 92, mass number 238) emits an $\alpha $-particle and the resultant nucleus emits $\beta $-particle. What are the atomic number and mass numbers of the final nucleus:

Hint: Uranium is a radioactive element which is very heavy i.e. it has a huge atomic mass number and is unstable and therefore it emits alpha and beta particles to become more stable and forms an element which is stable.

Complete step by step answer:
It is given in the problem that the uranium nucleus has atomic number 92 and mass number 238 and it emits an $\alpha $-particle and the resultant nucleus emits $\beta $-particle and we need to find the atomic number and mass number of the resulting nucleus.
As it is given that there is only one alpha particle emitted and only one beta particle emitted then the reaction will be equal to,
${}_{92}^{238}U \to {}_2^4\alpha + {}_{ - 1}^0\beta + {}_a^bX$
Now the atomic number of the newly formed nucleus is equal to,
The atomic number of the alpha particle is 2 and the charge on the beta particle is -1. Therefore we will subtract the atomic number of the alpha particle and the charge of the beta particle with the atomic number of the Uranium.
$ \Rightarrow a = 92 - 2 - \left( { - 1} \right)$
$ \Rightarrow a = 92 - 2 + 1$
$ \Rightarrow a = 91$.
The atomic mass of the alpha particle is equal to 4 a.m.u and the beta particle does not have mass therefore we subtract the mass of the alpha particle with the Uranium.
The atomic mass of the newly formed nucleus is equal to,
$ \Rightarrow b = 238 - 4$
$ \Rightarrow b = 234$
So the atomic number of the newly formed particle is equal to $a = 91$ and the atomic mass of the new nucleus is equal to$b = 234$.
The newly formed nucleus can be represented as,
${}_{91}^{234}X$

Note:The alpha particle has two protons and two neutrons which is why its atomic number becomes 2 and mass number becomes 4 whereas beta will only increase the atomic number and we do not have any change in mass number.