| ⇦ | 
| ⇨ |
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: - Heat produced in a resistance according to which law?
- The magnetic susceptibility is negative for
- The mass of a lift is 2000kg. When the tension in the supporting cable is 28000N, then its acceleration is
- Two Carnot engine A and B are operated in series. The engine A receives
- All components of the electromagnetic spectrum in vacuum have the same
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Heat produced in a resistance according to which law?
- The magnetic susceptibility is negative for
- The mass of a lift is 2000kg. When the tension in the supporting cable is 28000N, then its acceleration is
- Two Carnot engine A and B are operated in series. The engine A receives
- All components of the electromagnetic spectrum in vacuum have the same
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply