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

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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

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Topics: Radioactivity (83)
Subject: Physics (2479)

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