Question

In the nuclear decay given below $^A_Z X \rightarrow \,{}^A _{Z+1} Y \rightarrow \,{}^{A-A}_{Z-1}B * \, \rightarrow \,{}^{A-A}_{Z-1}B,$ the particles emitted in the sequence are

• A. $\gamma,\beta,\alpha$
• B. $\beta,\gamma,\alpha$
• C. $\alpha,\beta,\gamma$
• D. $\beta,\alpha,\gamma$

Answer 1

Answer: (D)

${ }_{Z}^{A} X \xrightarrow{\beta^{-}}{ }_{Z+1}^{A} Y \xrightarrow{\alpha}{ }_{Z-1}^{A-4} B^{*} \xrightarrow{\gamma}{ }_{Z-1}^{A-4} B$
First X decays by $\beta$ emission emitting $\overline{v}$, antineutrino simultaneously.$ Y$ emits $\alpha$ resulting in the excited level of $B$ which in tum emits a $\gamma$ ray.
$\therefore \beta,\alpha,\gamma$ is the answer.