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JEE Advanced Chemistry Chemical Kinetics 2020 JEE Advanced 2020 (Paper 1)

JEE Advanced Chemistry Question (2020) — Solution

Question

U 92 238 is known to undergo radioactive decay to form Pb 82 206 by emitting alpha and bet a particles. A rock initially contained 68 × 10 - 6 g of U 92 238 . If the number of alpha particles that it would emit during its radioactive decay of U 92 238  to Pb 82 206 in three half-lives is Z × 10 18 , then what is the value of Z ?

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

92 238 U → 82 206 Pb + 2 4 He + 6   − 1         0 β 0 . 286 × 10 - 6 mole 0 . 286 × 10 - 6 × N a = 0 . 286 × 6 . 02 × 10 17 = 1 . 72 × 10 17 After 3 half lives N = No ( 2 ) 3 = 1 . 72 × 10 17 8 = 0 . 215 × 10 17 So, no. of molecule of uranium decayed = [ 1 . 72 - 0 . 215 ] × 10 17 = 1 . 5 × 10 17 - So No. of α particle produced = 8 × 1 . 5 × 10 17 = 12 × 10 17 = 1 . 2 × 10 18  

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