| ⇦ | 
| ⇨ |
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: - A magnifying glass of focal length 5 cm is used to view an object by a person
- Fission of nuclei is possible because the binding energy per nucleon in them
- A radioactive nucleus emits a beta particle. The parent and daughter nuclei are
- The magnetic susceptibility is negative for
- A sample of radioactive element has a mass of 10 gram at an instant t=0.
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A magnifying glass of focal length 5 cm is used to view an object by a person
- Fission of nuclei is possible because the binding energy per nucleon in them
- A radioactive nucleus emits a beta particle. The parent and daughter nuclei are
- The magnetic susceptibility is negative for
- A sample of radioactive element has a mass of 10 gram at an instant t=0.
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