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Q. Atomic mass number of an element is $232$ and its atomic number is $90$. The end product of this radioactive element is an isotope of lead (atomic mass $208$ and atomic number $82)$. The number of $\alpha$-and $\beta$-particles emitted are

Nuclei

Solution:

Number of $\alpha$-particles emitted $=\frac{232-208}{4}=6$
Decrease in charge number due to $\alpha$-emission $=12$.
But actual decrease in charge number $=90-82=8$.
Clearly, four $\beta$-particles are emitted.