Asked by Heather
The model for radioactive decay is y=y0e−kt. A radioactive substance has a half-life of 230 years. If 35 grams are present today, in how many years will 14 grams be present?
Answers
Answered by
oobleck
e^(-t/230) = 14/35
-t/230 = ln(14/35)
t = 230 ln(35/14) = 210.747
makes sense, since it will take 230 years to drop by half to 17.5 g
-t/230 = ln(14/35)
t = 230 ln(35/14) = 210.747
makes sense, since it will take 230 years to drop by half to 17.5 g
Answered by
Anonymous
less than half takes longer than a half life
m = M e^-kt
m/M = 1/2 when t = 230
0.5 = e^-230 k
ln 0.5 = -230 k = -0.693
k = 0.003
so
14 = 35e^-0.003 t
0.4 = e^-0.003 t
ln 0.4 = -0.916 = -0.003 t
t = 305
m = M e^-kt
m/M = 1/2 when t = 230
0.5 = e^-230 k
ln 0.5 = -230 k = -0.693
k = 0.003
so
14 = 35e^-0.003 t
0.4 = e^-0.003 t
ln 0.4 = -0.916 = -0.003 t
t = 305
Answered by
oobleck
well duh. I tried to cut too many corners, and even then didn't catch my error!
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