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Question 250489: Use the formula N = Ie^(kt), where N is the number of items in terms of the initial population I, at time t, and k is the growth constant equal to the percent of growth per unit of time. A certain radioactive isotope decays at a rate of 0.275% annually. Determine the half-life of this isotope, to the nearest year.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! N = Ie^(kt)
Let I = 1
If in 1 year it decays by .275%, then at the end of the first year it will be reduced from 1to 1 - .00275 = .99725
your formula becomes:
.99725 = 1 * e^(k*1) which becomes:
.99725 = 1 * e^k
which is the same as:
.99725 = e^k
take the log of both sides to get:
log(.99725) = log(e^k)
this becomes:
log(.99725) = k*log(e)
divide both sides by log(e) to get:
log(.99725) / log(e) = k
solve for k to get:
k = -.002753788
now that you have found k, you can find the half life.
to find the half life, let I = 1 and let N = .5 and substitute in general equation of N = I * e^(kt) to get:
.5 = 1 * e^(-.002753788*t)
this is the same as:
.5 = e^(-.002753788*t)
take the log of both sides to get:
log(.5) = log(e^(-.002753788*t)
this becomes:
log(.5) = -.002753788*t*log(e)
divide both sides of this equation by -.002753788*log(e) to get:
t = log(.5)/(-.002753788*log(e))
solve for t to get:
t = 251.7067876 years
The half life of this radioactive isotope is 251.70657867 years.
plug that value in the original equation to get:
N = 1 * e^(kt) becomes
N = 1 * e^(-.002753788*251.7067876)
Solve for N to get:
N = .5
.5 is one half of 1 so the half life of the isotope is equal to 251.7067876 years.
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