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A nucleus ᴢXᴬ emits an α- particle with velocity v. The recoil speed of the daughter nucleus is
Options
(a) A-4/4v
(b) 4v/A-4
(c) v
(d) v/4
Correct Answer:
4v/A-4
Explanation:
According to the question, zXᴬ → z-₂ Yᴬ⁻⁴ + ₂He⁴
Let the recoil speed of daughter nucleus is ʋ’. By the law of conservation of momentum, we have Initial momentum = Final momentum
0 = (A – 4)ʋ’ + 4ʋ
(A – 4) ʋ’ = -4ʋ ⇒ ʋ’ = 4ʋ / (A – 4)
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