SOLUTION: 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

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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.