| ⇦ | 
| ⇨ |
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 small oil drop of mass 10⁻⁶kg is hanging in at rest between two plates seperated
- The wavelength λₑ of an electron and λₚ of a photon are of same energy E are related by
- A particle free to move along X-axis has potential energy given as U(X) =k(1-e⁻ˣ²)
- If the wavelength of light in vacuum be, the wavelength in a medium of refractive index
- Two particles A and B, move with constant velocities V₁ and V₂. At the initial
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A small oil drop of mass 10⁻⁶kg is hanging in at rest between two plates seperated
- The wavelength λₑ of an electron and λₚ of a photon are of same energy E are related by
- A particle free to move along X-axis has potential energy given as U(X) =k(1-e⁻ˣ²)
- If the wavelength of light in vacuum be, the wavelength in a medium of refractive index
- Two particles A and B, move with constant velocities V₁ and V₂. At the initial
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