Physics - Modern Physics Question with Solution | TestHub

Let the number of $\alpha$ particles emitted be $x$ and the number of $\beta$ particles emitted be $y$.

The nuclear reaction is given by:

${}_{88}{\mathrm{Ra}}^{226}\to {}_{82}{\mathrm{Pb}}^{206}+x({}_2{\mathrm{He}}^4)+y({}_{-1}{\mathrm{e}}^0)$

Equating the mass numbers on both sides:

$226=206+4x$

$4x=20\Rightarrow x=5$

Equating the atomic numbers on both sides:

$88=82+2x-y$

Substituting $x=5$:

$88=82+10-y$

$y=92-88=4$

The number of $\alpha$ and $\beta$ particles emitted are $5$ and $4$ respectively.

Answer: 5,4