Problema Solution
A certain radioactive isotope decays at a rate of .15% annually. Determine the half-life of this isotope to the nearest year. Use the formula A = Ie^k^t where A is the amount at time t, I is the initial amount, and k is the decay constant unique to the substance.
Answer provided by our tutors
Let's plug the information given into the formula A = Ie^k^t
After t = 1 year, you're left with 1 - 0.0015 = 0.9985% of your original amount of isotope, or
A(1) = 0.9985I
A(1) = I e^(k *1)
0.9985 I = I e^(k * 1 yr) divide both sides by I
0.9985 = e^k
e^k = 0.9985
........
........
k = -0.00150113
The half life is the time, T, that will give you a value of A(T) = 1/2 * I or
A(T) = I * e^( - 0.00150113 * T)
1/2 * I = I * e^( - 0.00150113 * T)
e^( - 0.00150113 * T) = 1/2
........
........
T = 461.75 years
The half-life of this isotope is 461.75 years.