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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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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If light emitted from sodium bulb is passed through sodium vapour,
- A particle of mass M is suited at the centre of a spherical shell of same mass and radius
- For a wave propagating in a medium, identify the property that is independent
- Two radioactive materials X₁ and X₂ have decay constants 5λ and λ respectively
- Maximum velocity of the photoelectrons emitted by a metal surface is 1.2×10⁶ ms⁻¹.
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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