| ⇦ | 
| ⇨ |
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: - When 4 A current flows for 2 min in an electroplating experiment,
- Two particles are oscillating along the same line with the same frequency
- An electron in a circular orbit of radius 0.05 nm performs 10¹⁶ rev/s. The magnetic
- For having large magnification power of a compound microscope
- A nucleus at rest splits into two nuclear parts having radii in the ratio 1:2.
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- When 4 A current flows for 2 min in an electroplating experiment,
- Two particles are oscillating along the same line with the same frequency
- An electron in a circular orbit of radius 0.05 nm performs 10¹⁶ rev/s. The magnetic
- For having large magnification power of a compound microscope
- A nucleus at rest splits into two nuclear parts having radii in the ratio 1:2.
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