The half-life of 198 Au is 2.7 days. Find the number of active nuclei present in a sample containing 1.414×1010 nuclei after a time that is half of its half-life.
A
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B
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C
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D
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Solution
The correct option is B
Okay , let's do this!!! First piece of information is that the half life is 2.7 days I know t1/2=ln2λ So this means λ=ln2t1/2 Moving on Next piece of info is N0=1.414×1010 And the question is asking Number of active nuclei present after a time that is half of half-life i.e. 12t1/2 Looks scary. Let's put the formula and see what all we know in it. N=N0e−λt N=1.414×1010xe−ln2t1/2×t1/22N=1.414×1010xe−ln22 N=1.414×1010×2−12 (That's how we do in log. remember) N=1010 So the number of active nuclei left after 1×1010t=12t1/2