Question

${}^{{\mathbf{234}}}{\mathbf{U}}\;$has 92 protons and 234 nucleons total in its nucleus. It decays by emitting an alpha particle. After the decay it becomes.
(A) ${}^{232}U$
(B) ${}^{232}Pa$
(C) ${}^{230}Th$
(D) ${}^{230}Ra$

Answer 1

We know that an alpha particle is equivalent to an helium nucleus ($H{e^{2 + }}$). When an alpha particle releases from a nucleus it will reduce $2$ electrons, $2$ protons and $2$ neutrons. So we have to check the options with an element less in $2$ electrons, $2$ protons and $2$ neutrons than with${}^{{\mathbf{234}}}{\mathbf{U}}\;$.

Complete step by step answer:
The reaction should be of the form.
$U{}_{92}^{234}\xrightarrow{a}Th{}_{90}^{230} + \alpha$.
The mass number of thorium is $230$ and its atomic number, $Z{\text{ }}is90.$
The mass number of thorium is $226$ and its atomic number is $88.$
As at one instant, only one particle can be emitted, unless it is given two successive emissions of particle, Ra cannot be obtained.
Uranium$- 234$ produces thorium$- 230$ by alpha decay.
An article is a helium nucleus.
It contains $2$ protons and $2$ neutrons, for a mass number of $4$.
The correct answer is $\left( C \right)$.

Note: The sum of the subscripts $\left( {atom{\text{ }}each{\text{ }}or{\text{ }}charges} \right)$ is the same on each side of the equation. Also, the same on each superscript (mass is the same on each side of this equation.