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?

Question

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[latex]_{Pu-236}[/latex] (initial) = 1.35 mg;
t = 5.00 yr; t[latex]_{1/2}[/latex] = 2.86 yr
FIND m[latex]_{Pu-236}[/latex] (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[latex]_{1/2} = \frac{0.693}{k}[/latex] [19.1]

ln [latex]\frac{N_{t}}{N_{0}}[/latex]= -kt [19.3]
Use the value of the rate constant, the initial mass of Pu-236, and the time along with integrated rate law to find the final mass of Pu-236. Since the mass of the Pu-236 (m[latex]_{Pu-236}[/latex]) is directly propor-tional to the number of atoms (N), and since the integrated rate law contains the ratio (N[latex]_{t}/N_{0}[/latex]), you can substitute the initial and final masses for the initial and final number of atoms.

CONCEPTUAL PLAN
t[latex]_{1/2}[/latex] → k.

. t[latex]_{1/2} = \frac{0.693}{k}[/latex]

k, m[latex]_{Pu-236}[/latex](initial), t → m[latex]_{Pu-236}[/latex](final)

. ln [latex]\frac{N_{t}}{N_{0}}[/latex]= -kt

SOLVE Follow your plan. Begin by finding the rate constant from the half-life. Solve the integrated rate law for N[latex]_{t}[/latex] 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.

Accepted Answer

t[latex]_{1/2} = \frac{0.693}{k}[/latex]
k = [latex]\frac{0.693}{t_{1/2}}= \frac{0.693}{2.86 yr}[/latex]
= 0.24[latex]\underline{2}[/latex]3/yr
ln [latex]\frac{N_{t}}{N_{0}}[/latex]= -kt

[latex]\frac{N_{t}}{N_{0}}= e^{-kt}[/latex]
N[latex]_{t} = N_{0}e^{-kt}[/latex]
N[latex]_{t} = 1.35 mg \left[e^{-(0.24\underline{2}3/\cancel{yr} )(5.00 \cancel{yr})}\right] [/latex]
N[latex]_{t }= 1.35 mg \left[e^{-(0.24\underline{2}3/\cancel{yr} )(5.00 \cancel{yr})}\right][/latex]

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.)

Question 19.4: Radioactive Decay Kinetics Plutonium-236 is an alpha emitte...

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