| ⇦ |
| ⇨ |
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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Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Three solids of masses m₁,m₂ and m₃ are connected with weightless string
- The maximum and minimum magnitudes of the resultant of two vectors
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- A mass m moves in a circle on a smooth horizontal plane with velocity v₀
- A milli ammeter of range 10 mA has a coil of resistance 1 ohm
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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