Question 250489
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.