| ⇦ |
| ⇨ |
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: - When the temperature increases, the viscocity of
- The shortest wavelength in Lyman series is 91.2 nm. The largest wavelength
- which of the following can not be emitted by radioactive substances during their decay?
- The ratio of magnetic dipole moment of an electron of charge e and mass m
- A machine gun fires 10 bullets/s,each of mass 10 g,the speed of each bullet
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- When the temperature increases, the viscocity of
- The shortest wavelength in Lyman series is 91.2 nm. The largest wavelength
- which of the following can not be emitted by radioactive substances during their decay?
- The ratio of magnetic dipole moment of an electron of charge e and mass m
- A machine gun fires 10 bullets/s,each of mass 10 g,the speed of each bullet
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