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In the nuclear decay given below:
$^A_ZX \rightarrow ^{A}_{Z+1}Y \rightarrow ^{A-4}_{Z-1}B^* \rightarrow ^{A-4}_{Z-1}B$
the particles emitted in the sequence are
$^A_ZX \rightarrow ^{A}_{Z+1}Y \rightarrow ^{A-4}_{Z-1}B^* \rightarrow ^{A-4}_{Z-1}B$
the particles emitted in the sequence are
A
$\alpha$, $\beta$, $\gamma$
B
$\beta$, $\alpha$, $\gamma$
C
$\gamma$, $\beta$, $\alpha$
D
$\beta$, $\gamma$, $\alpha$
Detailed Solution
Step 1: $^A_ZX \rightarrow ^A_{Z+1}Y$. The mass number is unchanged and the atomic number increases by 1, so a $\beta^-$ particle is emitted.
Step 2: $^A_{Z+1}Y \rightarrow ^{A-4}_{Z-1}B^*$. The mass number decreases by 4 and the atomic number decreases by 2, so an $\alpha$ particle is emitted.
Step 3: $^{A-4}_{Z-1}B^* \rightarrow ^{A-4}_{Z-1}B$. Neither A nor Z changes; the excited nucleus only drops to its ground state, emitting a $\gamma$ photon.
Hence the sequence is $\beta$, $\alpha$, $\gamma$.
Step 2: $^A_{Z+1}Y \rightarrow ^{A-4}_{Z-1}B^*$. The mass number decreases by 4 and the atomic number decreases by 2, so an $\alpha$ particle is emitted.
Step 3: $^{A-4}_{Z-1}B^* \rightarrow ^{A-4}_{Z-1}B$. Neither A nor Z changes; the excited nucleus only drops to its ground state, emitting a $\gamma$ photon.
Hence the sequence is $\beta$, $\alpha$, $\gamma$.