| ⇦ | 
| ⇨ |
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: - The magnifying power of a convex lens of focal length 10 cm, when the image
- If the radioactive dacay constant of radium is 1.07 x 10⁻⁴ year⁻¹, then its half-life
- A current I flows along the length of an infinitely long, straight, thin walled pipe.
- The electric field associated with an e.m. wave in vacuum is given by
- When a mass m is attached to a spring, it normally extends by 0.2 m.
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The magnifying power of a convex lens of focal length 10 cm, when the image
- If the radioactive dacay constant of radium is 1.07 x 10⁻⁴ year⁻¹, then its half-life
- A current I flows along the length of an infinitely long, straight, thin walled pipe.
- The electric field associated with an e.m. wave in vacuum is given by
- When a mass m is attached to a spring, it normally extends by 0.2 m.
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