Question

the alpha decay of a radioactive nuclide (x) emits a he - 4 nucleus and produces an isotope of $_ {92}^{235}u$. what is x? $_ {91}^{235}pa$ $_ {93}^{235}np$ $_ {92}^{236}u$ $_ {94}^{239}pu$

Explanation:

Step1: Recall alpha - decay equation

In alpha - decay, the general equation is $^{A}_{Z}X
ightarrow^{A - 4}_{Z - 2}Y+^{4}_{2}He$. Here, the daughter nucleus is $^{235}_{92}U$ and the alpha - particle is $^{4}_{2}He$.

Step2: Calculate mass number of parent

The mass number $A$ of the parent nucleus $X$ is $A=235 + 4=239$ (since mass number is conserved in nuclear reactions).

Step3: Calculate atomic number of parent

The atomic number $Z$ of the parent nucleus $X$ is $Z=92+2 = 94$ (since atomic number is conserved in nuclear reactions).

Step4: Identify the element

The element with atomic number $Z = 94$ is Plutonium ($Pu$). So the parent nucleus $X$ is $^{239}_{94}Pu$.

Answer:

$^{239}_{94}Pu$