By the succesive disintegration of ₉₂U²³⁸, the final product obtained is ₈₂Pb²⁰⁶, then

⇦

⇨

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:

1. The nuclei of which one of the following pairs of nuclei are isotones?
2. A source of unknown frequency gives 4 beats/s, when sounded with a source harmonic
3. A circular coil of radius 10cm and 100 turns carries a current 1 A.
4. Wave nature of light is verified by
5. What is the value of incidence L for which the current is maximum in a series LCR

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