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Question: In the given reaction \(zX^{A} \rightarrow_{z + 1} ⥂ Y^{A} \rightarrow_{z - 1} ⥂ K^{A - 4} \rightarr...

In the given reaction zXA→z+1⥂YA→z−1⥂KA−4→z−1⥂KA−4zX^{A} \rightarrow_{z + 1} ⥂ Y^{A} \rightarrow_{z - 1} ⥂ K^{A - 4} \rightarrow_{z - 1} ⥂ K^{A - 4}

Radioactive radiations are emitted in the sequence.

A

2ε0hce2\frac{2\varepsilon_{0}hc}{e^{2}}

B

β, 6muα, 6muγ\beta,\mspace{6mu}\alpha,\mspace{6mu}\gamma

C

γ, 6muα, 6muβ\gamma,\mspace{6mu}\alpha,\mspace{6mu}\beta

D

β, 6muγ, 6muα\beta,\mspace{6mu}\gamma,\mspace{6mu}\alpha

Answer

β, 6muα, 6muγ\beta,\mspace{6mu}\alpha,\mspace{6mu}\gamma

Explanation

Solution

ZXA→−1β0Z+1YA→2He4(α)ZX^{A}{\overset{\quad - 1\beta^{0}\quad}{\rightarrow}}_{Z + 1}Y^{A}\overset{\quad 2He^{4}(\alpha)\quad}{\rightarrow}

Z−1KA−4→0γ0Z−1KA−4Z - 1K^{A - 4}{\overset{\quad 0\gamma^{0}\quad}{\rightarrow}}_{Z - 1}K^{A - 4}