SOLUTION: The half-life of Palladium-100 is 4 days. After 20 days a sample of Palladium-100 has been reduced to a mass of 5 mg. What was the initial mass (in mg) of the sample? What is th

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Question 1136353: The half-life of Palladium-100 is 4 days. After 20 days a sample of Palladium-100 has been reduced to a mass of 5 mg.
What was the initial mass (in mg) of the sample?
What is the mass 4 weeks after the start?
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
y=pe%5E%28-kx%29
-
1%2F2=1%2Ae%5E%28-k%2A4%29
ln%281%2F2%29=ln%28e%5E%28-4k%29%29
-4k=ln%281%2F2%29
k=%281%2F4%29ln%282%29
k=%281%2F4%290.69315
k=0.17329
-
highlight_green%28y=pe%5E%28-0.17329x%29%29model


5 grams after 20 days:
5=pe%5E%28-0.17329%2A20%29
5=p%2Ae%5E%28-3.4658%29
5%2F0.01325=p
highlight%28p=160%29mg

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Scientists like to write exponential decay functions using base e; and sometimes there are good reasons for that. But with half life problems, it seems to me far easier to write the exponential decay function as

A = I(1./2)^t

where t is the number of half lives.

In this problem, the half life is 4 days, and the stated period is 20 days. That is an integer number of half lives; the solution is simple.

20 days if 5 half lives; the original mass has been reduced by a factor of 2^5=32. If 5g are left, the original amount was 5*32 = 160g.

Note the other tutor, using e and natural logarithms, ended up with the wrong answer by transposing a couple of digits near the end of the calculations.

The amount remaining after 4 weeks = 28 days = 7 half lives is the original 160g, reduced by a factor of 2^7 = 128 (or the amount remaining after 20 days, reduced by another factor of 2^2=4 for the 2 additional half lives). That amount is 160/128 = 5/4 = 1.25g.