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Chemical Kinetics and Nuclear Chemistry question

2022 · 29 Jul · Shift 2 · Q19
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Chemical Kinetics and Nuclear Chemistry question

2022 · 29 Jul · Shift 2 · Q19

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
Assuming 1 μg1 \,\mu \mathrm{g}1μg of trace radioactive element X with a half life of 30 years is absorbed by a growing tree. The amount of X remaining in the tree after 100 years is ‾\underline{\hspace{2cm}}​× 10−1μg\times\, 10^{-1} \mu \mathrm{g}×10−1μg. [Given : ln 10 = 2.303; log 2 = 0.30]
Numerical answer
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Correct answer: 1

  1. Use the radioactive decay formula

For radioactive decay, N=N0(12)t/t1/2N = N_0\left(\frac{1}{2}\right)^{t/t_{1/2}}N=N0​(21​)t/t1/2​ where:

  • N0=1 μgN_0 = 1\,\mu\text{g}N0​=1μg
  • t1/2=30t_{1/2} = 30t1/2​=30 years
  • t=100t = 100t=100 years

So, N=1×(12)100/30N = 1\times \left(\frac{1}{2}\right)^{100/30}N=1×(21​)100/30

  1. Simplify the exponent

10030=103\frac{100}{30} = \frac{10}{3}30100​=310​ Thus, N=(12)10/3N = \left(\frac{1}{2}\right)^{10/3}N=(21​)10/3

  1. Evaluate using logarithms

Taking common log, log⁡N=103log⁡(12)\log N = \frac{10}{3}\log\left(\frac{1}{2}\right)logN=310​log(21​) Since log⁡(12)=−log⁡2=−0.30\log\left(\frac{1}{2}\right) = -\log 2 = -0.30log(21​)=−log2=−0.30 we get log⁡N=103(−0.30)=−1.0\log N = \frac{10}{3}(-0.30) = -1.0logN=310​(−0.30)=−1.0

Hence, N=10−1 μgN = 10^{-1}\,\mu\text{g}N=10−1μg

  1. Match with the required format

The question asks: ‾×10−1 μg\underline{\hspace{2cm}}\times 10^{-1}\,\mu\text{g}​×10−1μg Since the amount is exactly 1×10−1 μg1\times 10^{-1}\,\mu\text{g}1×10−1μg

Therefore, the required integer is: 1\boxed{1}1​

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