Question 19.4: Radioactive Decay Kinetics Plutonium-236 is an alpha emitte...
Radioactive Decay Kinetics
Plutonium-236 is an alpha emitter with a half-life of 2.86 years. If a sample initially contains 1.35 mg of Pu-236, what mass of Pu-236 is present after 5.00 years?
| SORT You are given the initial mass of Pu-236 in a sample and asked to find the mass after 5.00 years. | GIVEN m_{Pu-236} (initial) = 1.35 mg; t = 5.00 yr; t_{1/2} = 2.86 yr FIND m_{Pu-236} (final) |
| STRATEGIZE You can use the integrated rate law (Equation 19.3) to solve this problem. However, you must determine the value of the rate constant (k) from the half-life expression (Equation 19.1).
t_{1/2} = \frac{0.693}{k} [19.1] ln \frac{N_{t}}{N_{0}}= -kt [19.3] |
CONCEPTUAL PLAN t_{1/2} → k. . t_{1/2} = \frac{0.693}{k} k, m_{Pu-236}(initial), t → m_{Pu-236}(final) . ln \frac{N_{t}}{N_{0}}= -kt |
| SOLVE Follow your plan. Begin by finding the rate constant from the half-life. Solve the integrated rate law for N_{t} and substitute the values of the rate constant, the initial mass of Pu-236, and the time into the solved equation. Calculate the final mass of Pu-236. | |
t_{1/2} = \frac{0.693}{k}
k = \frac{0.693}{t_{1/2}}= \frac{0.693}{2.86 yr}
= 0.24\underline{2}3/yr
ln \frac{N_{t}}{N_{0}}= -kt
\frac{N_{t}}{N_{0}}= e^{-kt}
N_{t} = N_{0}e^{-kt}
N_{t} = 1.35 mg \left[e^{-(0.24\underline{2}3/\cancel{yr} )(5.00 \cancel{yr})}\right]
N_{t }= 1.35 mg \left[e^{-(0.24\underline{2}3/\cancel{yr} )(5.00 \cancel{yr})}\right]
CHECK The units of the answer (mg) are correct. The magnitude of the answer (0.402 mg) is about one-third of the originalmass (1.35 mg), which seems reasonable given that the amount of time is between one and two half-lives. (One half-life would result in one-half of the original mass, and two half-lives would result in one-fourth of the original mass.)