Question

In the nuclear reaction, ${}_{92}\mathrm{U}^{238}\to {}_{82}\mathrm{Pb}^{206}$the number of $\alpha$ and $\beta$ particles emitted is:

• A. 7$\alpha$, 5$\beta$
• B. 6$\alpha$, 4$\beta$
• C. 4$\alpha$, 3$\beta$
• D. 8$\alpha$, 6$\beta$

Answer 1

The correct option is D

8$\alpha$, 6$\beta$

Step 1:
The $\mathrm{\alpha }$-particles are represented as ${}_2\mathrm{He}^4$ and $\mathrm{\beta }$-particles as ${}_{-1}\mathrm{e}^0$.

Consider the number of $\mathrm{\alpha }$-particles be $\mathrm{X}$ and the number of $\mathrm{\beta }$-particle be $\mathrm{Y}$.

${}_{92}\mathrm{U}^{238}\to {}_{82}\mathrm{Pb}^{206}+\mathrm{X}{}_2\mathrm{He}^4+\mathrm{Y}{}_{-1}\mathrm{e}^0$

Step 2:

Equate the mass number of both sides of the reaction.

$\begin{array}{rcl}238& =& 206+4X+0Y\\ 4X& =& 238-206\\ 4X& =& 32\\ X& =& 8\end{array}$

Here, the number of $\mathrm{\alpha }$-particles is 8.

Equate the atomic mass of both sides of the reaction.

$\begin{array}{rcl}92& =& 82+2\times 8-1Y\\ Y& =& 82+16-92\\ Y& =& 6\end{array}$

Here, the number of $\beta$-particles is 6.

Final answer:

There are 8 $\mathrm{\alpha }$-particles and 6 $\beta$-particles. Thus, option (D) is correct.