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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 temperature of a metal block is increased from 27⁰C to 84⁰C. The rate of radiated energy
- A charge Q is enclosed by a Gaussian spherical surface of radius R
- Maximum velocity of the photoelectrons emitted by a metal surface is 1.2×10⁶ ms⁻¹.
- To observed diffraction, the size of the obstacle
- In Moseley’s law √ν=a(z-b), the values of the screening constant for K-series
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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