| ⇦ | 
| ⇨ |
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
Related Questions: - Which of the following is not a thermodynamical coordinate?
- Two closed organ pipes when sounded simultaneously give 4 beats/second.
- On heating a ferromagnetic substance above curie temperature
- Two coherent sources of intensity ratio α interfere. In interference pattern,
- A transistor is operated in common-emitter configuration at Vᵥ = 2 V such that a change
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which of the following is not a thermodynamical coordinate?
- Two closed organ pipes when sounded simultaneously give 4 beats/second.
- On heating a ferromagnetic substance above curie temperature
- Two coherent sources of intensity ratio α interfere. In interference pattern,
- A transistor is operated in common-emitter configuration at Vᵥ = 2 V such that a change
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