Question 21.SE.7: If we start with 1.000 g of strontium-90, 0.953 g will remai...
If we start with 1.000 g of strontium-90, 0.953 g will remain after 2.00 yr. (a) What is the half-life of strontium-90? (b) How much strontium-90 will remain after 5.00 yr?
Analyze (a) We are asked to calculate a half-life, t_{1/2}, based on data that tell us how much of a radioactive nucleus has decayed in a time interval t = 2.00 yr and the information N_0 = 1.000 g, N_t = 0.953 g. (b) We are asked to calculate the amount of a radionuclide remaining after a given period of time.
Plan (a) We first calculate the rate constant for the decay, k, and then we use that to compute t_{1/2}. (b) We need to calculate N_t, the amount of strontium present at time t, using the initial quantity, N_0, and the rate constant for decay, k, calculated in part (a).
Solve
(a) Equation 21.20 is solved for the decay constant, k, and then Equation 21.19 is used to calculate half-life,t_{1/2}:
k=\frac{0.693}{t_{1 / 2}} (21.19)
\ln \frac{N_t}{N_0}=-k t (21.20)
\begin{aligned}k &=-\frac{1}{t}\ln \frac{N_t}{N_0}=-\frac{1}{2.00\, yr}\ln \frac{0.953 \,g}{1.000 \,g}\\&=-\frac{1}{2.00\, yr}(-0.0481)=0.0241 \,yr ^{-1}\\t_{1 / 2}&=\frac{0.693}{k}=\frac{0.693}{0.0241 \,yr ^{-1}}=28.8 \,yr\end{aligned}
(b) Again using Equation 21.20, with k = 0.0241 yr^{-1}, we have:
\ln \frac{N_t}{N_0}=-k t=-\left(0.0241 \,\cancel{yr ^{-1}}\right)(5.00 \,\cancel{yr} )=-0.120
N_t/N_0 is calculated from ln(N_t/N_0) = -0.120 using the e^x or INV LN function of a calculator:
\frac{N_t}{N_0}=e^{-0.120}=0.887
Because N_0 = 1.000 g,we have:
N_t=(0.887) N_0=(0.887)(1.000 \,g )=0.887 \,g