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By the succesive disintegration of ₉₂U²³⁸, the final product obtained is ₈₂Pb²⁰⁶, then how many number of α and β-particles are emitted?
Options
(a) 6 and 8
(b) 8 and 6
(c) 12 and 6
(d) 8 and 12
Correct Answer:
8 and 6
Explanation:
The number of α-particles, n₁ = Change in mass number / 4
n₁ = [(238 – 206) / 4] = 32 / 4 = 8
Now, number of β-particles, n₂ = 82 – (92 – 2 n₁)
= 82 – (92 – 2 × 8) = 82 – 76 = 6
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Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The moment of inertia of a thin uniform rod of mass M and length L about an axis
- The half-life of a radioactive substance is 30 min. The time (in minutes) taken
- The lowest frequency of light that will cause the emission of photoelectrons
- The angle between the two vectors A = 3i+4j+5k and B = 3i+4j-5k will be
- A 50Hz AC signal is applied in a circuit of inductance of (1/π)H and resistance 2100Ω
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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