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Question

A parent nucleus $$^{m}_{1}p$$ decays into a daughter nucleus $$D$$ through $$\alpha$$ emission in the following way $$^{m}_{1}p\rightarrow D+\alpha$$ The subscript and superscript on the daughter nucleus $$D$$ will be written as

A
$$^{m+4}_{n}D$$
B
$$^{m-4}_{n}D$$
C
$$^{m}_{n}D$$
D
$$^{m-4}_{n-2}D$$
Solution
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Correct option is D. $$^{m-4}_{n-2}D$$

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