the alpha decay of a radioactive nuclide (x) emits a he - 4 nucleus and produces an isotope of…

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

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

Answer

Explanation:

Step1: Recall alpha - decay equation

In alpha - decay, the general equation is ${Z}^{A}X\rightarrow{Z - 2}^{A - 4}Y+{2}^{4}He$. Here, $Y$ is ${92}^{235}U$, $A_Y = 235$, $Z_Y=92$, and the alpha - particle is ${2}^{4}He$ with $A{He}=4$ and $Z_{He} = 2$.

Step2: Calculate mass number of $X$

Using the mass - number conservation $A_X=A_Y + A_{He}$. Substitute $A_Y = 235$ and $A_{He}=4$, so $A_X=235 + 4=239$.

Step3: Calculate atomic number of $X$

Using the atomic - number conservation $Z_X=Z_Y+Z_{He}$. Substitute $Z_Y = 92$ and $Z_{He}=2$, so $Z_X=92 + 2=94$.

Answer:

$_{94}^{239}Pu$