A) \[\frac{4v}{A-4}\]
B) \[\frac{4v}{A+4}\]
C) \[\frac{2v}{A+4}\]
D) \[\frac{2v}{A-4}\]
Correct Answer: A
Solution :
[a] The emission of an \[\alpha \]-particle from the atom of element reduces its atomic number by 2 and mass number by 4. Hence, the radioactive emission is as follows; \[_{Z}{{X}^{A}}{{\xrightarrow{\alpha -particle}}_{Z-2}}{{Y}^{A-4}}+{}_{Z}H{{e}^{4}}(\alpha \text{-}particle)\]Also from law of conservation of momentum, \[m\times 0={{m}_{y}}{{v}_{y}}+{{m}_{\alpha }}{{v}_{\alpha }}\] \[=(A-4){{v}_{y}}+4v\] \[\Rightarrow {{v}_{y}}=\frac{4v}{A-4}\]
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