In the nuclear decay given below : ZAX→Z+1AY→Z−1A−4B→z−1A−4B The particles emitted in the sequence are :

zAX→βz+1AY→αz−1A−4B→γz−1A−4Bβ,α,γ

In the nuclear decay given below : ZAX→Z+1AY→Z−1A−4B→z−1A−4B The particles emitted in the sequence are :

zAX→βz+1AY→αz−1A−4B→γz−1A−4Bβ,α,γ

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In the nuclear decay given below : $_{Z}^{A} X \to_{Z + 1}^{A} Y \to_{Z - 1}^{A - 4} B \to_{z - 1}^{A - 4} B$ The particles emitted in the sequence are :

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$β, γ, α$

$β, α, γ$

$α, β, γ$

$γ, β, α$

answer is D.

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$_{z}^{A} X {\overset{β}{\to}}_{z + 1}^{A} Y {\overset{α}{\to}}_{z - 1}^{A - 4} B {\overset{γ}{\to}}_{z - 1}^{A - 4} B$

$β, α, γ$

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In the nuclear decay given below : ZAX→Z+1AY→Z−1A−4B→z−1A−4B The particles emitted in the sequence are :

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In the nuclear decay given below : $_{Z}^{A} X \to_{Z + 1}^{A} Y \to_{Z - 1}^{A - 4} B \to_{z - 1}^{A - 4} B$ The particles emitted in the sequence are :