Question

Calculations Involving Radioactive Decay and Time If we start with \(1.000 \mathrm{~g}\) of strontium- \(90,0.953 \mathrm{~g}\) will remain after \(2.00 \mathrm{yr}\). (a) What is the half-life of strontium- 90 ? (b) How much strontium- 90 will remain after \(5.00 \mathrm{yr}\) ? (c) What is the initial activity of the sample in becquerels and curies? SOLUTION (a) Analyze 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 \(t=2.00 \mathrm{yr}\) and the information \(N_{0}=1.000 \mathrm{~g}, N_{t}=0.953 \mathrm{~g}\). Plan We first calculate the rate constant for the decay, \(k\), and then use that to compute \(t_{1 / 2}\). Solve Equation \(21.19\) is solved for the decay constant, \(k\), and then Equation \(21.20\) is used to calculate half-life, \(t_{1 / 2}\) : $$ \begin{aligned} k &=-\frac{1}{t} \ln \frac{N_{\mathrm{f}}}{N_{0}}=-\frac{1}{2.00 \mathrm{yr}} \ln \frac{0.953 \mathrm{~g}}{1.000 \mathrm{~g}} \\ &=-\frac{1}{2.00 \mathrm{yr}}(-0.0481)=0.0241 \mathrm{yr}^{-1} \\ t_{1 / 2} &=\frac{0.693}{k}=\frac{0.693}{0.0241 \mathrm{yr}^{-1}}=28.8 \mathrm{yr} \end{aligned} $$ (b) Analyze We are asked to calculate the amount of a radionuclide remaining after a given period of time. Plan We need to calculate \(N_{b}\) 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 Again using Equation \(21.19\), with \(k=0.0241 \mathrm{yr}^{-1}\), we have $$ \ln \frac{N_{t}}{N_{0}}=-k t=-\left(0.0241 \mathrm{yr}^{-1}\right)(5.00 \mathrm{yr})=-0.120 $$ \(N_{t} / N_{0}\) is calculated from \(\ln \left(N_{t} / N_{0}\right)=-0.120\) using the \(e^{x}\) or INV LN function of a calculator: $$ \frac{N_{t}}{N_{0}}=e^{-1.120}=0.887 $$ Because \(N_{0}=1.000 \mathrm{~g}\), we have $$ N_{t}=(0.887) N_{0}=(0.887)(1.000 \mathrm{~g})=0.887 \mathrm{~g} $$ (c) Analyze We are asked to calculate the activity of the sample in becquerels and curies. Plan We must calculate the number of disintegrations per atom per second and then multiply by the number of atoms in the sample. Solve The number of disintegrations per atom per second is given by the decay constant, \(k\) : $$ k=\left(\frac{0.0241}{\mathrm{yr}}\right)\left(\frac{1 \mathrm{yr}}{365 \text { days }}\right)\left(\frac{1 \text { day }}{24 \mathrm{~h}}\right)\left(\frac{\mathrm{lh}}{3600 \mathrm{~s}}\right)=7.64 \times 10^{-10} \mathrm{~s}^{-1} $$ To obtain the total number of disintegrations per second, we calculate the number of atoms in the sample. We multiply this quantity by \(k\), where we express \(k\) as the number of disintegrations per atom per second, to obtain the number of disintegrations per second: $$ \begin{aligned} \left(1.000 \mathrm{~g}^{90} \mathrm{Sr}\right)\left(\frac{1 \mathrm{~mol}^{90} \mathrm{Sr}}{90 \mathrm{~g}^{90} \mathrm{Sr}}\right)\left(\frac{6.022 \times 10^{23} \mathrm{atoms} \mathrm{Sr}}{1 \mathrm{~mol}^{90} \mathrm{Sr}}\right)=6.7 \times 10^{21} \text { atoms }^{90} \mathrm{Sr} \\ \text { Total disintegrations/s } &=\left(\frac{7.64 \times 10^{-10} \text { disintegrations }}{\text { atom }{ }^{2} \mathrm{~s}}\right)\left(6.7 \times 10^{21} \text { atoms }\right) \\ &=5.1 \times 10^{12} \text { disintegrations/s } \end{aligned} $$ Because \(1 \mathrm{~Bq}\) is one disintegration per second, the activity is \(5.1 \times 10^{12} \mathrm{~Bq}\). The activity in curies is given by $$ \left(5.1 \times 10^{12} \text { disintegrations/s }\right)\left(\frac{1 \mathrm{Ci}}{3.7 \times 10^{10} \text { disintegrations/s }}\right)=1.4 \times 10^{2} \mathrm{Ci} $$ We have used only two significant figures in products of these calculations because we do not know the atomic weight of \({ }^{90} \mathrm{Sr}\) to more than two significant figures without looking it up in a special source.

Short answer

The half-life of strontium-90 is \(28.8 \mathrm{yr}\). After \(5.00 \mathrm{yr}\), \(0.887 \mathrm{~g}\) of strontium-90 will remain. The initial activity of the sample is \(5.1 \times 10^{12} \mathrm{~Bq}\) or \(1.4 \times 10^{2} \mathrm{Ci}\).

Step-by-step solution

Calculate the decay constant (k)

We will first calculate the decay constant using the given data in the following equation: $$k = -\frac{1}{t} \ln \frac{N_{\mathrm{f}}}{N_{0}}$$ Substitute the given values: $$k = -\frac{1}{2.00 \mathrm{yr}} \ln \frac{0.953 \mathrm{~g}}{1.000 \mathrm{~g}} = 0.0241 \mathrm{yr}^{-1}$$

Determine the half-life of strontium-90

Now, we will calculate the half-life using the decay constant, k, calculated in step 1: $$t_{1 / 2} = \frac{0.693}{k} = \frac{0.693}{0.0241 \mathrm{yr}^{-1}} = 28.8 \mathrm{yr}$$

Calculate the amount of strontium-90 remaining after 5 years

Using Equation \(N_t = N_0 e^{-kt}\), we'll calculate the remaining amount of strontium-90 after 5 years: $$N_{t} = N_{0} e^{-k t} = (1.000 \mathrm{~g}) e^{-0.120}= 0.887 \mathrm{~g}$$

Calculate the initial activity in becquerels

First, we convert decay constant k into the number of disintegrations per atom per second: $$k = 7.64 \times 10^{-10} \mathrm{~s}^{-1}$$ Now, we will calculate the total number of atoms in the sample and multiply by k to get the number of disintegrations per second: $$\text{Total disintegrations/s} = 5.1 \times 10^{12} \text{ disintegrations/s}$$ Therefore, the initial activity in becquerels is: $$A = Nk = 5.1 \times 10^{12} \mathrm{~Bq}$$

Calculate the initial activity in curies

Finally, we convert the activity from becquerels to curies using the conversion factor (1 Ci = 3.7 x 10^10 disintegrations/s): $$A = \left(5.1 \times 10^{12} \text{ disintegrations/s} \right) \left(\frac{1 \mathrm{Ci}}{3.7 \times 10^{10} \text{ disintegrations/s }}\right) = 1.4 \times 10^{2} \mathrm{Ci}$$

Key concepts

Half-Life of Strontium-90

Understanding the half-life of a radioactive element is crucial in predicting how long it will remain radioactive. The half-life of an isotope like strontium-90 is the time required for half of the radioactive nuclei in a given sample to undergo decay. In our exercise, the half-life of strontium-90 was calculated using the decay constant, with the formula
\[ t_{1/2} = \frac{0.693}{k} \]
where \( k \) is the rate constant. It was found that strontium-90 has a half-life of approximately 28.8 years, which means that after 28.8 years, half of the original amount of strontium-90 would have decayed into other elements or isotopes.

Rate Constant

The rate constant, often symbolized by \( k \), is a crucial parameter in the equations that describe radioactive decay. It helps to determine the speed at which a radioactive substance decays. For strontium-90, the rate constant was derived using the natural logarithm of the ratio of the remaining mass to the initial mass of the isotope over a time period. From the given data,
\[ k = -\frac{1}{t} \ln \frac{N_{f}}{N_{0}} \]
and by plugging in the values, the rate constant \( k \) for strontium-90 was found to be 0.0241 \( \text{yr}^{-1} \). This means that each year, approximately 2.41% of the remaining strontium-90 atoms will decay.

Decay Constant

The decay constant represents the probability per unit time that a given radioactive nucleus will decay. It is a natural extension of the rate constant in the context of nuclear decay. The decay constant for strontium-90, derived from our calculations, is
\[ 7.64 \times 10^{-10} \text{s}^{-1} \]
This value indicates the number of disintegrations per atom per second. When you have the decay constant, you can easily calculate the number of radioactive decays (disintegrations) that will occur in a second by multiplying the number of atoms by this constant.

Initial Activity in Becquerels

Activity is a term used in nuclear physics to describe the rate at which a sample of radioactive material decays. Measured in becquerels (Bq), one becquerel is defined as one decay per second. To calculate the initial activity of our strontium-90 sample, we multiplied the decay constant by the total number of atoms in the sample, as follows:
\[ A = Nk \]
Where \( A \) is the activity in becquerels, \( N \) is the number of atoms, and \( k \) is the decay constant. Given the initial gram amount of strontium-90, the activity was found to be
\[ 5.1 \times 10^{12} \text{ Bq} \]
which signifies a highly radioactive sample.

Initial Activity in Curies

The curie (Ci) is a traditional unit of radioactivity which is based on the activity of one gram of radium-226. The conversion factor used for curies to becquerels is one curie equals 3.7 x 10^10 disintegrations per second. Keeping in mind the conversion from becquerels to curies, we applied the following calculation in our exercise:
\[ A = \left(5.1 \times 10^{12} \text{ disintegrations/s} \right) \left(\frac{1 \text{ Ci}}{3.7 \times 10^{10} \text{ disintegrations/s }}\right) \]
The resulting activity of the strontium-90 sample in curies was
\[ 1.4 \times 10^{2} \text{ Ci} \]
indicating it as a significantly radioactive material.