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JEE Main 2014 (19 Apr, Shift 1) · Law of Radioactive Decay

Law of Radioactive Decay question from JEE Main 2014 (19 Apr, Shift 1)

Q94JEE Main 2014 · 19 APR · SHIFT 1Medium

Law of Radioactive Decay

A piece of wood from a recently cut tree shows 20 decays per minute. A wooden piece of same size placed in a museum (obtained from a tree cut many years back) shows 2 decays per minute. If half life of C14 is 5730 years, then age of the wooden piece placed in the museum is approximately:

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Answer: (c)

  1. AA0=2−t/T⇒t=Tlog⁡(A0/A)log⁡2\frac{A}{A_0} = 2^{-t/T} \Rightarrow t = T \frac{\log(A_0/A)}{\log 2}.
  2. A0A=202=10\frac{A_0}{A} = \frac{20}{2} = 10, so t=5730×10.301=5730×3.32t = 5730 \times \frac{1}{0.301} = 5730 \times 3.32.
  3. t≈19040t \approx 19040 years, so the answer is (c) 19039 years.
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