anonymous
  • anonymous
A sample of a radioactive element has a mass of 80g. How much parent and daughter materials are in the sample after two half-lives?
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@green_1
anonymous
  • anonymous
@whpalmer4
green_1
  • green_1
it would be 20g

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anonymous
  • anonymous
Thank you !! :)
whpalmer4
  • whpalmer4
After 1 half-life, 1/2 of the parent has changed to be the daughter. After another half-life, 1/2 of the parent remaining after the first half-life has also changed to be the daughter. t = 0 100% parent 0% daughter t = 1 50% parent 50% daughter t = 2 25% parent 75% daughter t = 3 12.5% parent 87.5% daughter t = 4 6.25% parent 93.75% daughter and so on (I'm using a unit of 1 half-life for t)
anonymous
  • anonymous
Thank you. :)
whpalmer4
  • whpalmer4
You can write a formula for this: \[P(t) = P_0(\frac{1}{2})^{(t/t_{\text{half}})}\]where \(P_0\) is the starting amount of the parent \(P(t)\) is the amount of the parent remaining at time \(t\) \(t_{\text{half}}\) is the half-life of the parent If you try out some values: \[P(0) = P_0(\frac{1}{2})^{0/t_{\text{half}}} = P_0(\frac{1}{2})^0 = P_0*1 = P_0\]At time \(t = 0\), we have all of the original parent after 1 half-life (\(t = t_{\text{half}}\)) we have: \[P(t_{\text{half}}) = P_0(\frac{1}{2})^{(t_{\text{half}}/t_{\text{half}})} = P_0(\frac{1}{2})^1 = \frac{P_0}{2}\]so as expected, 1 half-life means 1/2 of the parent is gone after 2 half-lives: \[P(2t_{\text{half}}) = P_0(\frac{1}{2})^{(2t_{\text{half}}/t_{\text{half}})} = P_0(\frac{1}{2})^2 = \frac{P_0}{4}\] only 1/4 of the parent remains after 2 half-lives All as we expect. We can use that formula to find the amount remaining at any time, not just multiples of the half-life: \[P(6.64t_{\text{half}}) = P_0(\frac{1}{2})^{(6.64t_{\text{half}}/t_{\text{half}})} = P_0(\frac{1}{2})^{6.64} \approx \frac{P_0}{100}=0.01P_0\]after 6.64 half-lives, we have about 1% of the original remaining. For the specific problem at hand, \(P_0 = 80\text{ grams}\) so \[P(2\text{ half-lives} = (80\text{ grams})(\frac{1}{2})^{2/1} = \frac{80\text{ grams}}{4} = 20\text{ grams}\] If we have \(80\) grams of parent material at the start, and only \(20\) grams remains, that means the daughter product is \(80 \text{ grams} - 20 \text{ grams} = 60 \text{ grams}\) (This assumes that the daughter product is not ALSO decaying into something else in that same time, although the sum total of all the daughter products plus the parent would still add up to the starting amount)
anonymous
  • anonymous
Thank you. :)

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