| ⇦ |
| ⇨ |
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 projectile is thrown in the upward direction making an angle of 60°with the horizontal
- A gramophone record is revolving with an angular velocity ?. A coin is placed at
- One-fourth length of a spring of force constant K is cut away. The force constant
- The ratio of the numerical value of the average velocity and average speed
- The magnetic moment of a diamagnetic atom is
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A projectile is thrown in the upward direction making an angle of 60°with the horizontal
- A gramophone record is revolving with an angular velocity ?. A coin is placed at
- One-fourth length of a spring of force constant K is cut away. The force constant
- The ratio of the numerical value of the average velocity and average speed
- The magnetic moment of a diamagnetic atom is
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