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Chemical Kinetics and Nuclear Chemistry question

2024 · 29 Jan · Shift 2 · Q27
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Chemical Kinetics and Nuclear Chemistry question

2024 · 29 Jan · Shift 2 · Q27

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
The half-life of radioisotope bromine - 82 is 36 hours. The fraction which remains after one day is ‾×10−2\underline{\hspace{2cm}}\times 10^{-2}​×10−2. (Given antilog 0.2006=1.5870.2006=1.5870.2006=1.587)
Numerical answer
View written solutionFree

Correct answer: 63

  1. Use radioactive decay law in half-life form

For a radioactive substance,

N=N0(12)t/t1/2N = N_0\left(\frac12\right)^{t/t_{1/2}}N=N0​(21​)t/t1/2​

So the fraction remaining after time ttt is

NN0=(12)t/t1/2\frac{N}{N_0}=\left(\frac12\right)^{t/t_{1/2}}N0​N​=(21​)t/t1/2​
  1. Substitute the given values

Half-life of bromine-82:

t1/2=36 ht_{1/2}=36\text{ h}t1/2​=36 h

Time elapsed after one day:

t=24 ht=24\text{ h}t=24 h

Thus,

NN0=(12)24/36=(12)2/3\frac{N}{N_0}=\left(\frac12\right)^{24/36}=\left(\frac12\right)^{2/3}N0​N​=(21​)24/36=(21​)2/3
  1. Evaluate the expression using logarithms
(12)2/3=2−2/3\left(\frac12\right)^{2/3}=2^{-2/3}(21​)2/3=2−2/3

Taking common logarithm,

log⁡(2−2/3)=−23log⁡2\log\left(2^{-2/3}\right)=-\frac23\log 2log(2−2/3)=−32​log2

Using log⁡2≈0.3010\log 2 \approx 0.3010log2≈0.3010,

log⁡(2−2/3)=−23(0.3010)=−0.2007≈−0.2006\log\left(2^{-2/3}\right)=-\frac23(0.3010)=-0.2007\approx -0.2006log(2−2/3)=−32​(0.3010)=−0.2007≈−0.2006

Hence,

NN0=antilog(−0.2006)=1antilog(0.2006)\frac{N}{N_0}=\text{antilog}(-0.2006)=\frac{1}{\text{antilog}(0.2006)}N0​N​=antilog(−0.2006)=antilog(0.2006)1​

Given,

antilog(0.2006)=1.587\text{antilog}(0.2006)=1.587antilog(0.2006)=1.587

So,

NN0=11.587=0.63\frac{N}{N_0}=\frac{1}{1.587}=0.63N0​N​=1.5871​=0.63
  1. Express in the required form
0.63=63×10−20.63 = 63 \times 10^{-2}0.63=63×10−2

So the required integer is

63\boxed{63}63​
  1. Comparison with stored answer

Derived answer = 636363

Stored correct answer = 636363

They match.

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