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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
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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A circular disc of radius 0.2 meter is placed in a uniform magnetic field of induction
- An engine is moving towards a wall with a velocity 50 ms⁻¹ emits a note of 1.2 kHz.
- The electric field associated with an e.m. wave in vacuum is given by
- In potentiometer experiment, a cell of emf 1.25V gives balancing length of 30 cm
- Three capacitors each of capacitance C and of breakdown voltage V are joined in series
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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