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₉₂U²³⁵ undergoes successive disintegrations with the end product of ₈₂P²⁰³. The number of α and β-particles emitted are
Options
(a) α=8,β=6
(b) α=3,β=3
(c) α=6,β=4
(d) α=6,β=0
Correct Answer:
α=8,β=6
Explanation:
₉₂U²³⁵ → end product ₈₂P²°³ α and β emitted.
ΔA = 235 – 203 = 32
Therefore, 8 alpha particles are emitted. The charge should be 92 – 16 = 76.
But as the final charge is 82, six β⁻ particles had been emitted to make up the final atomic number Z = 82.
.·. 8 alpha particles and six β⁻ have been emitted.
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Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The masses of three wires of copper are in the ratio 1:3:5 and lengths
- The radius of curvature of a convave mirror is 24 cm and the imager is magnified
- The horizontal range and maximum height attained by a projectile are R and H,respectively
- Two small spherical shells A and B are given positive charges of 9C and 4C
- A point charge q is situated at a distance r on axis from one end of a thin
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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