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

Hint: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.