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Q.
A radioactive reaction is ${ }_{92} U ^{238} \rightarrow{ }_{82} Pb ^{206} .$ How many $\alpha$ and $\beta$ particles are emitted?
Nuclei
Solution:
${ }_{92} U ^{238} \rightarrow{ }_{82} Pb ^{206}$
Let the number of $\alpha$ and $\beta$ particles emitted be $x$ and $y$ respectively. Then
${ }_{92} U ^{238} \rightarrow{ }_{82} Pb ^{206}+x_{2} He ^{4}+y_{-1} e^{0}$
Equating the mass and atomic numbers on both sides, we get
$206+4 x=238$
$ \Rightarrow x=\frac{238-206}{4}=8$
$82 + 2x - y = 92$
$\Rightarrow y = 82 + 2 \times 8 - 92 = 6$
$\therefore $ The number of $\alpha$ and $\beta$ particles emitted are $8$ and $6$ respectively