| ⇦ | 
| ⇨ |
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: - A nucleus at rest splits into two nuclear parts having radii in the ratio 1:2.
- When aluminium is added as an impurity to silicon, the resulting material is
- Find the maximum velocity with which a train can be moved on a circular track
- The amount of heat energy required to raise the temperature of 1 g of Helium
- If radius of earth’s orbit is made 1/4, the duration of a year will become
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A nucleus at rest splits into two nuclear parts having radii in the ratio 1:2.
- When aluminium is added as an impurity to silicon, the resulting material is
- Find the maximum velocity with which a train can be moved on a circular track
- The amount of heat energy required to raise the temperature of 1 g of Helium
- If radius of earth’s orbit is made 1/4, the duration of a year will become
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