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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.
Related Questions: - The barrier potential of a p-n junction depends on
- The wavelength of the matter wave is independent of
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- A bomb of mass 16 kg at rest explodes into two pieces of masses of 4 kg and 12 kg.
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The barrier potential of a p-n junction depends on
- The wavelength of the matter wave is independent of
- On heating a ferromagnetic substance above curie temperature
- Zenor breakdown will occur if
- A bomb of mass 16 kg at rest explodes into two pieces of masses of 4 kg and 12 kg.
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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