Question 24.5: Finding the Number of Radioactive Nuclei Problem Strontium-9......
Finding the Number of Radioactive Nuclei
Problem Strontium-90 is a radioactive byproduct of nuclear reactors that behaves biologically like calcium, the element above it in Group 2A(2). When ^{90}Sr is ingested by mammals, it is found in their milk and eventually in the bones of those drinking the milk. If a sample of ^{90}Sr has an activity of 1.2×10^{12} d/s, what are the activity and the fraction of nuclei that have decayed after 59 yr (t_{1/2} of ^{90}Sr = 29 yr)?
Plan The fraction of nuclei that have decayed is the change in number of nuclei, expressed as a fraction of the starting number. The activity of the sample (𝒜) is proportional to the number of nuclei (𝒩), so we know that
\text{Fraction decayed }= \frac{𝒩_0 − 𝒩_t}{𝒩_0} = \frac{𝒜_0 − 𝒜_t}{𝒜_0}We are given 𝒜_0 (1.2×10^{12} d/s), so we find 𝒜_t from the integrated form of the first-order rate equation (Equation 24.3), in which t is 59 yr. To solve that equation, we first need k, which we can calculate from the given t_{1/2} (29 yr) using Equation 24.4.
\ln \frac{𝒩_t}{𝒩_0} = −kt\text{ or } 𝒩_t = 𝒩_0e^{−kt} \text{ and }\ln \frac{𝒩_0}{𝒩_t} = kt (24.3)
\ln \frac{𝒩_0}{\frac{1}{2}𝒩_0} = kt_{1/2}\text{ so } t_{1/2} = \frac{\ln 2}{k} (24.4)
Solution Calculating the decay constant k:
t_{1/2} = \frac{\ln 2}{k}\text{ so }k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{29 yr} = 0.024 yr^{−1}
Applying Equation 24.3 to calculate 𝒜_t, the activity remaining at time t:
\ln \frac{𝒩_0}{𝒩_t} = \ln \frac{𝒜_0}{𝒜_t} = kt \text{ or }\ln 𝒜_0 − \ln 𝒜_t = kt
So, \ln 𝒜_t = −kt + \ln 𝒜_0 = −(0.024 yr^{−1} × 59 yr) + \ln (1.2×10^{12} d/s)
\ln 𝒜_t = −1.4 + 27.81 = 26.4
𝒜_t = 2.9×10^{11} d/s
(All the data contain two significant figures, so we retained two in the answer.) Calculating the fraction decayed:
\text{Fraction decayed =} \frac{𝒜_0 − 𝒜_t}{𝒜_0} = \frac{1.2×10^{12} d/s − 2.9×10^{11} d/s}{1.2×10^{12} d/s} = 0.76
Check The answer is reasonable: t is about 2 half-lives, so 𝒜_t should be about \frac{1}{4}𝒜_0, or about 0.3×10^{12}; therefore, the activity should have decreased by about \frac{3}{4}.
Comment 1. A substitution in Equation 24.3 is useful for finding 𝒜_t, the activity at time t: 𝒜_t = 𝒜_0e^{−kt}.
2. Another way to find the fraction of activity (or nuclei) remaining uses the number of half-lives (t/t_{1/2}). By combining Equations 24.3 and 24.4 and substituting (\ln 2)/t_{1/2} for k, we obtain
Inverting the ratio gives
\ln \frac{𝒩_t}{𝒩_0} = \ln \left( \frac{1}{2}\right)^{\frac{t}{t_{1/2}}}
Taking the antilog gives
\text{Fraction remaining = }\frac{𝒩_t}{𝒩_0} = \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} = \left( \frac{1}{2}\right)^{\frac{59}{29}} = 0.24
So, \text{Fraction decayed = 1.00 − 0.24 = 0.76}