In the nuclear decay given below

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Question

${}_Z^AX\xrightarrow{{{\beta ^ - }}}{}_{Z + 1}^AY\,\xrightarrow{\alpha }{}_{Z - 1}^{A - 4}{B^*}\xrightarrow{\gamma }{}_{Z - 1}^{A - 4}B$

First $X$

Vedclass · Accepted Answer

$\beta ,\;\alpha ,\;\gamma $

Vedclass · Answer

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

In the nuclear decay given below

$_z{X^A}{ \to _{z + 1}}{Y^A}{ \to _{z - 1}}{K^{A - 4}}{ \to _{z - 1}}{K^{A - 4}}$

the particles emitted in the sequence are

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In the nuclear decay given below $_z{X^A}{ \to _{z + 1}}{Y^A}{ \to _{z - 1}}{K^{A - 4}}{ \to _{z - 1}}{K^{A - 4}}$ the particles emitted in the sequence are