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Q.

In the radioactive decay: zxA→z+1YA→z−1ZA−4→Z−1Z*A−4The sequence of emission is:

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a

β,γ,α

b

β,α,γ

c

γ,α,β

d

α,β,γ

answer is B.

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Detailed Solution

zxA→z+1YA+−1e0; Z+1YA→z−1ZA−4+2He4;z−1ZA−4→Z−1ZA−4→Z−1ZA−4*+γ
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