Search any question & find its solution
Question:
Answered & Verified by Expert
A certain radioactive material $_{Z} X^{\Lambda}$ starts emitting $\alpha$ and $\beta$ particles successively such that the end product is $\mathrm{Z}-3 \mathrm{Y}^{\mathrm{A}-8}$. The number of $\alpha$ and $\beta$ particles emitted are
Options:
Solution:
2518 Upvotes
Verified Answer
The correct answer is:
2 and 1 respectively
Let thenumber of $\alpha$ and $\beta$ particles emitted be $\mathrm{m}$ and $\mathrm{n}$ respectively. Then $\mathrm{A}-4 \mathrm{~m}=\mathrm{A}-8 \Rightarrow \mathrm{m}=2$
$[\because$ the mass number of a radioactive nuclide decreases by 4 due to emission of one $\alpha$ particle]
Again,
$(Z-2 m)+n=Z-3$
$[\because$ the atomic number decreases by 2 due to emission of $1 \alpha$-particle but increases by 1 due to emission of $1 \beta$-particle]
or $\quad-2 \mathrm{~m}+\mathrm{n}=-3$
or $\quad 2 \mathrm{~m}-\mathrm{n}=3$
or $\quad(2 \times 2)-\mathrm{n}=3 \quad(\because \mathrm{m}=2)$
$\Rightarrow \quad \mathrm{n}=1$
$[\because$ the mass number of a radioactive nuclide decreases by 4 due to emission of one $\alpha$ particle]
Again,
$(Z-2 m)+n=Z-3$
$[\because$ the atomic number decreases by 2 due to emission of $1 \alpha$-particle but increases by 1 due to emission of $1 \beta$-particle]
or $\quad-2 \mathrm{~m}+\mathrm{n}=-3$
or $\quad 2 \mathrm{~m}-\mathrm{n}=3$
or $\quad(2 \times 2)-\mathrm{n}=3 \quad(\because \mathrm{m}=2)$
$\Rightarrow \quad \mathrm{n}=1$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.