Asked by Peter
You have 20 grams of phosphorus-32 that decays 5% per day. How long will it take for hlaf of the original amount to decay?
Answers
Answered by
DrBob222
This may not be the math process you want. From a chemists standpoint, I would do the following.
ln(No/N) = kt. We must determine k.
So at zero time we start with 20 g. At the end of 1 day we have 20g-5% = 19 g. Plug that into the first order equation of ln(20/19) = k(1 day)
k = 0.0513
Then ln(20/10) = 0.0513(t)
Solve for t. If I didn't goof that is
13.51 days which I would round to 13.5 days.
ln(No/N) = kt. We must determine k.
So at zero time we start with 20 g. At the end of 1 day we have 20g-5% = 19 g. Plug that into the first order equation of ln(20/19) = k(1 day)
k = 0.0513
Then ln(20/10) = 0.0513(t)
Solve for t. If I didn't goof that is
13.51 days which I would round to 13.5 days.
Answered by
Reiny
Using the "continuous" decay formula, I got
.5 = 1(e^(-.05t) )
ln .5 - -.05t lne, but ln e = 1
t = -.05/ln .5 = .072134.. years
= 26.3 days
.5 = 1(e^(-.05t) )
ln .5 - -.05t lne, but ln e = 1
t = -.05/ln .5 = .072134.. years
= 26.3 days
Answered by
DrBob222
I looked up the half life of P-32. It is 14.28 days. I suspect 5% decay in the problem is just a close number.
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