Question

A certain stable nuclide after absorbing a neutron, emits $\beta -$ particle and the new nuclide formed splits spontaneously into two $\alpha -$ particles. The nuclide is-

• A. ${}_3{\text{Li}}^6$
• B. ${}_4{\text{Be}}^6$
• C. ${}_3{\text{Li}}^7$
• D. ${}_2{\text{He}}^4$

Answer 1

The correct option is C ${}_3{\text{Li}}^7$

Initially the stable nuclide absorbs a neutron,

${}_Z{\text{X}}^A{+}_0{\text{n}}^1{\to }_Z{\text{Y}}^{A+1}......\left(1\right)$

And the nuclide emits $\beta -$ particle and the new nuclide splits spontaneously into two $\alpha -$ particles,`

${}_Z{\text{Y}}^{A+1}{\to }_{-1}{\text{e}}^0+2{\text{He}}_2^4......\left(2\right)$

Comparing mass number in the second equation,

$A+1=(2\times 4)=8$

$A=8-1=7$

Comparing atomic number in the second equation,

$Z=-1+(2\times 2)=3$

$\therefore {}_z{\text{X}}^A{=}_3{\text{Li}}^7$

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