| ⇦ | 
| ⇨ |
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 body A experiences perfectly elastic collision with a stationary body B
- A lead bullet of 10 gm travelling at 300 m/s strikes against a block
- A body under the action of a force F⃗ = 6i⃗–8j⃗+10k⃗, acquires an accelerationof 1 m/s². The mass of this body must be
- A solid sphere, a hollow sphere and a ring are released from top of an inclined
- Ohm’s law is not applicable except to
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A body A experiences perfectly elastic collision with a stationary body B
- A lead bullet of 10 gm travelling at 300 m/s strikes against a block
- A body under the action of a force F⃗ = 6i⃗–8j⃗+10k⃗, acquires an accelerationof 1 m/s². The mass of this body must be
- A solid sphere, a hollow sphere and a ring are released from top of an inclined
- Ohm’s law is not applicable except to
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