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Half-life of radioactive substance

by gmmstr827
Tags: halflife, radioactive, substance
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gmmstr827
#1
Feb8-11, 12:45 PM
P: 86
1. The problem statement, all variables and given/known data

If 20% of a radioactive substance disappears in 70 days, what is its half-life?

2. Relevant equations

y = C*e^(k*t)
where t is time in days
k is the constant of proportionality?
y is the current amount of substance


3. The attempt at a solution

20% disappears, so 100% - 20% = 80% remaining = 0.80 after 70 days
.8*C = C*e^(70*k)
Solve for k
k = -3.188 * 10^-3

Plug k value back into equation to find t at the substance's half-life (.5)
.5*C = C*e^((-3.188 * 10^-3)*t)
Solve for t
t = 217.44

217.44 days is the substance's half-life.

Is that correct?
Thank you.
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timthereaper
#2
Feb8-11, 02:38 PM
P: 343
It matches what I get.


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