Question

The nucleus ${}_{84}{}^{}\mathrm{X}^{202}$ emits an $\mathrm{\alpha }$-particle and forms the nucleus $Y$. Represents this change in the form of an equation.

Answer 1

Step 1:- Given data

$$\begin{gathered} MassnumberofnucleusX \\ =202 \\ andatomicnumberofnucleusX \\ =84 \end{gathered}$$

When a radioactive nucleus ${}_{84}{}^{}\mathrm{X}^{202}$emits an $\mathrm{\alpha }$-particle $\left({}_2{}^4\mathrm{He}\right)$ the mass number decreases by $4units$ and atomic number decreases by $2units.$

Step 2:- Finding mass number and atomic number of nucleus $Y$

$$\begin{gathered} MassnumberofnucleusY \\ =202-4 \\ =198 \\ andatomicnumberofnucleusX \\ =84-2 \\ =82 \end{gathered}$$

Step 3:- write the nuclear equation

Nuclear equation is represented as,

${}_{84}{}^{}\mathrm{X}^{202}\to {}_{82}{}^{}\mathrm{Y}^{198}+{}_2{}^4\mathrm{He}+\mathrm{energy}$

Hence, formation of nucleus $\mathrm{Y}$ of atomic number $82$ and mass number is $198.$