Asked by Ryan kustin
The half-life of Kryponite-123 is 40 years. Suppose we have a 700-mg sample. Find the decay k, and the expected amount of Kryptonite left after 30 years.
I need help setting this problem up.
I need help setting this problem up.
Answers
Answered by
Reiny
since you are dealing with "half-life" , we can use the formula
amount = c (1/2)^(t/40)
amount = 700 (1/2)^(t/40)
when t = 40, amount = 350
so when t = 30
amount = 700 (1/2)^(30/40)
= 700 (1/2)^.75 = 416.22.. mg
reading your question again, I gather you want to use
amount = 700 e^(kt)
350 = 700 e^(40k)
.5 = e^(40k)
take ln of both sides
ln .5 = 40k
k = ln.5/40 = -.017328...
so when t = 30
amount = 700 e^(30(.017328)) = 416.22.. (same as above)
amount = c (1/2)^(t/40)
amount = 700 (1/2)^(t/40)
when t = 40, amount = 350
so when t = 30
amount = 700 (1/2)^(30/40)
= 700 (1/2)^.75 = 416.22.. mg
reading your question again, I gather you want to use
amount = 700 e^(kt)
350 = 700 e^(40k)
.5 = e^(40k)
take ln of both sides
ln .5 = 40k
k = ln.5/40 = -.017328...
so when t = 30
amount = 700 e^(30(.017328)) = 416.22.. (same as above)
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