| ⇦ | 
| ⇨ |
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: - Lumen is the unit of
- Three cells, each of e.m.f 1.5 V and internal resistance 1 ohm are connected in parallel
- A body takes 5 minute for cooling from 50⁰C to 40⁰C. Its temperature
- Time period of oscillation of a magnet is 2 sec. When it is remagnetised
- In a thermodynamic system, working substance is ideal gas. Its internal energy
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Lumen is the unit of
- Three cells, each of e.m.f 1.5 V and internal resistance 1 ohm are connected in parallel
- A body takes 5 minute for cooling from 50⁰C to 40⁰C. Its temperature
- Time period of oscillation of a magnet is 2 sec. When it is remagnetised
- In a thermodynamic system, working substance is ideal gas. Its internal energy
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