| ⇦ |
| ⇨ |
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: - Two identical long conducting wires AOB and COD are placed at right angle
- Radius of orbit of satellite of earth is R. Its kinetic energy is proportional to
- Two liquid drops having diameters of 1 cm and 1.5 cm. The ratio of excess
- Colours appear on a thin soap film and soap bubbles due to the phenomenon of
- In the phenomenon of electric discharge through gases at low pressure,
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two identical long conducting wires AOB and COD are placed at right angle
- Radius of orbit of satellite of earth is R. Its kinetic energy is proportional to
- Two liquid drops having diameters of 1 cm and 1.5 cm. The ratio of excess
- Colours appear on a thin soap film and soap bubbles due to the phenomenon of
- In the phenomenon of electric discharge through gases at low pressure,
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