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(C) The emission of an $\alpha$-particle $(_{2}He^{4})$ from the parent nucleus $_{92}X^{238}$ results in a daughter nucleus with a mass number reduce

Vedclass · Accepted Answer

$144$

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(C) The emission of an $\alpha$-particle $(_{2}He^{4})$ from the parent nucleus $_{92}X^{238}$ results in a daughter nucleus with a mass number reduced by $4$ and an atomic number reduced by $2$.$_{92}X^{238} ightarrow _{90}Y^{234} + _{2}He^{4}$The number of neutrons in the daughter nucleus $_{90}Y^{234}$ is calculated as:$	ext{Number of neutrons} = 	ext{Mass number} - 	ext{Atomic number}$$	ext{Number of neutrons} = 234 - 90 = 144$.

After the emission of an $\alpha$-particle from the atom $_{92}X^{238}$,the number of neutrons in the resulting atom will be:

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