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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 current in the coil of inductance 5H decreases at the rate of 2 A/s
- In which process, the P-V indicator diagram is a straight line parallel to volume
- A particle of mass m is driven by a machine that delivers a constant power k watts
- A ball moving with velocity 2 m/s collides head on with another stationary ball
- In an inductor when current changes from 2 A to 18 A in 0.05 sec, the e.m.f. induced
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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