Question

The atomic mass number of an element is $232$ and its atomic number $90$. The end products of this radioactive element are an isotope of Lead (Atomic mass $208$ and atomic number $82$). The number of $\mathrm{\alpha }-$ and $\mathrm{\beta }-$particles emitted is:

Answer 1

Step 1: Given data

The mass number of the radioactive element$=232$

The atomic number of the radioactive element$=90$

The atomic number of the end product$=82$

The atomic mass of the end product$=208$

Step 2: Calculating the number of alpha and beta particles

From the given data writing an equation, ${}_{90}\mathrm{Th}^{232}\to {}_{82}\mathrm{Pb}^{208}+{\mathrm{n}}_1\mathrm{\alpha }+{\mathrm{n}}_2{\mathrm{\beta }}^{-}$

Applying mass balance, we know the mass of an alpha particle$=4$

$232=208+4{\mathrm{n}}_1$ where ${\mathrm{n}}_1$ is the number of alpha particles

${\mathrm{n}}_1=\frac{232-208}{4}=6$

Now balancing atomic numbers, since an alpha particle is very similar to the nucleus of a Helium atom that consists of two neutrons and two protons, its atomic number is $2$

$90=82+2{\mathrm{n}}_1-{\mathrm{n}}_2$ where ${\mathrm{n}}_2$ is the number of beta particles

Now substituting the value of ${\mathrm{n}}_1$ we get,

$90=82+2\times 6-{\mathrm{n}}_2$

${\mathrm{n}}_2=4$

Therefore, the number of alpha and beta particles emitted are $6$and $4$