Q5 / 150 · MHT CET 2008 · Chemistry
In successive emission of $\alpha$ -and $\beta$ -particles, the number of $\alpha$ -and $\beta$ -particles should be emitted for the conversion of ${ }_{92} \mathrm{U}^{238}$ to ${ }_{82} \mathrm{~Pb}^{206}$ are
- A$7 \alpha, 5 \beta$
- B$6 \alpha, 4 \beta$
- C$4 \alpha, 3 \beta$
- Correct answerD$8 \alpha, 6 \beta$
Solution
${ }_{92} \mathrm{U}^{238} \longrightarrow{ }_{82} \mathrm{~Pb}^{206}+m_{2} \mathrm{He}^{4}+n_{-1} e^{0}$
On comparing both sides
$$
\begin{aligned}
238 &=206+4 m \\
\Rightarrow \quad m &=8 \\
92 &=82+2 m-n \\
2 m-n &=10 \\
\Rightarrow \quad n &=6
\end{aligned}
$$
$\therefore \alpha$ -particles emitted, $m=8$ $\beta$ -particles emitted, $n=6$
Hence, the correct answer is option: 'D'.