Question

When ${}_{90}{\text{Th}}^{228}$ gets converted into ${}_{83}{\text{Bi}}^{212}$, the number of $\alpha$ and $\beta$ particles emitted will respectively be-

• A. $4\alpha ;1\beta$
• B. $8\alpha ;7\beta$
• C. $4\alpha ;4\beta$
• D. $4\alpha ;7\beta$

Answer 1

The correct option is A $4\alpha ;1\beta$

For Alpha decay process,

${}_{90}{\text{Th}}^{228}{\underset{}{\overset{}{\to }}}_{83}{\text{Bi}}^{212}+N_{\alpha }{(}_2{\text{He}}^4)+N_{\beta }{(}_{-1}{\text{e}}^0)$

Comparing the mass number,

$\left(i\right)228=212+N_{\alpha }\left(4\right)+N_{\beta }\left(0\right)$

$228-212=4N_{\alpha }$

$\therefore N_{\alpha }=\frac{16}{4}=4$

$\left(ii\right)$ Comparing the atomic number,

$90=83+N_{\alpha }\left(2\right)+N_{\beta }(-1)$

$90-83=(4\times 2)+(-1)N_{\beta }$

$7-8=-N_{\beta }$

$\therefore N_{\beta }=1$

Hence, $4\alpha$-particles and $1\beta$-particle is emitted.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.