Asked by Freud Mark
Suppose f(t) = P0e^-0.3t grams of radioactive substance are present after t seconds. If 200 grams of substance are present initially, how much is present after 4 seconds? What is the half-life of the substance?
Answers
Answered by
Reiny
f(4) = 200 e^(-.3(4))
= ...
for 2nd part:
half of 200 is 100
100 = 200 e^(-.3t)
.5 = e^(-.3t)
take ln of both sides
ln .5 = -.3t lne , remember lne = 1
etc
= ...
for 2nd part:
half of 200 is 100
100 = 200 e^(-.3t)
.5 = e^(-.3t)
take ln of both sides
ln .5 = -.3t lne , remember lne = 1
etc
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