Question

Phosphorus-32 is used in biochemical studies to follow the pathway taken by metabolites. Its half-life is 14.3 days. A vial containing \(10.0 \mu \mathrm{g}\left(1 \mu \mathrm{g}=1 \times 10^{-6} \mathrm{~g}\right)\) of \(\mathrm{P}-32\) is acci- dentally spilled into a sink. (a) How long will it take P-32 in the drain to exhibit \(68 \%\) of its original activity? (b) How many atoms of P-32 remain in the sink after 755 days?

Short answer

Answer: P-32 will exhibit 68% of its original activity after approximately 6.96 days, and there will be approximately \(7.33 \times 10^{17}\) atoms of P-32 remaining in the sink after 755 days.

Step-by-step solution

Part (a): Finding the time for P-32 to have 68% of its original activity

To solve this part, we need to use the radioactive decay formula: $$N(t) = N_0 e^{-\lambda t}$$ Where: - \(N(t)\) is the amount of the substance remaining at time \(t\) - \(N_0\) is the initial amount of the substance - \(\lambda\) is the decay constant - \(t\) is the time First, we need to find the decay constant using the half-life formula: $$t_{1/2} = \frac{\ln{2}}{\lambda}$$ Where: - \(t_{1/2}\) is the half-life - \(\lambda\) is the decay constant Using the given half-life of P-32 (\(14.3\) days), we can find the decay constant: $$\lambda = \frac{\ln{2}} {14.3}$$ We are given that the activity should be \(68\%\) of its original value. So, $$N(t) = 0.68 N_0$$ Plug this into the radioactive decay formula and solve for time \(t\): $$0.68 N_0 = N_0 e^{-\lambda t}$$ Divide both sides by \(N_0\): $$0.68 = e^{-\lambda t}$$ To remove the exponential, take the natural logarithm of both sides: $$\ln{0.68} = -\lambda t$$ Finally, solve for the time \(t\): $$t = \frac{\ln{0.68}}{-\lambda}$$ Plug in the value of \(\lambda\) calculated earlier: $$t = \frac{\ln{0.68}}{-\frac{\ln{2}}{14.3}}$$ Now calculate the value of \(t\): $$t \approx 6.96 \text{ days}$$ So, it will take approximately \(6.96\) days for P-32 to exhibit \(68\%\) of its original activity.

Part (b): Finding the number of P-32 atoms remaining after 755 days

To find the number of P-32 atoms remaining in the sink after 755 days, we will use the same radioactive decay formula, but this time we will plug in the given time value (755 days) and find the remaining amount of P-32. $$N(755) = N_0 e^{-\lambda (755)}$$ We are given that the initial amount of P-32 in the vial was \(10.0 \mu g\), which is equal to \(1 \times 10^{-5} g\). To find the number of atoms, we need to convert this mass to moles using the molar mass of P-32. The molar mass of P-32 is \(32~g/mol\). Converting the mass to moles: $$n = \frac{1 \times 10^{-5} ~g}{32~g/mol}$$ Now, we will convert moles to the number of atoms using Avogadro's number (\(6.022 \times 10^{23} atoms/mol\)): $$N_0 = n \times (6.022 \times 10^{23})$$ Plug in the values of \(N_0\) and \(\lambda\) into the formula: $$N(755) = N_0 e^{-\lambda (755)}$$ Calculate the value of \(N(755)\), which is the number of P-32 atoms remaining in the sink after 755 days: $$N(755) \approx 7.33 \times 10^{17} \text{ atoms}$$ So, approximately \(7.33 \times 10^{17}\) atoms of P-32 remain in the sink after 755 days.

Key concepts

Understanding Half-Life in Radioactive Decay

The term 'half-life' refers to the time required for half of a radioactive substance to decay into another element or isotope. It's one of the key features of radioactive decay, offering a glimpse into the stability and longevity of radioactive materials like Phosphorus-32 (P-32).

In practice, half-life allows researchers and professionals to predict how much time will pass until a radioactive sample decreases to a specific amount. The half-life of P-32 being 14.3 days, means that every 14.3 days, the activity of the substance will be reduced by half due to decay. This predictable pattern is crucial for biomedical applications, including tracing the pathways of metabolites, as mentioned in the exercise.

Moreover, understanding half-life aids in addressing safety concerns, as in the case of the P-32 spill in the exercise solution. By knowing the half-life, we can calculate the decay period necessary to reach a point where the radioactive material exhibits only 68% of its original activity, essential for ensuring proper safety protocols and environmental safety management.

Decay Constant - The Pillar of Radioactive Change

The decay constant, represented by the symbol \(\lambda\), is intrinsic to the nature of a radioactive isotope. It is tied deeply to its half-life and describes the rate at which the isotope decays per unit time. Essentially, the decay constant provides the probability at a given time that a nucleus will decay.

By determining the decay constant from the half-life of P-32 (using the formula \(\lambda = \frac{\ln{2}}{t_{1/2}}\)), we quantify the pace at which the P-32 will decay. This constant is employed to calculate not only the activity levels after a particular period (as shown in the solution for part a) but also aids in the calculation of the remaining quantity of the radioactive material after any given time period, like in the 755 days scenario for part b).

The precision of the decay constant ensures the accuracy of predictions related to radioactive materials, ultimately serving numerous fields such as nuclear medicine, radiocarbon dating, and more.

Exponential Decay – A Fundamental Radioactive Pattern

Exponential decay characterizes a decrease in quantity at a rate proportional to the current amount, which is the underlying principle for the decay of radioactive substances. The formula for exponential decay, \(N(t) = N_0 e^{-\lambda t}\), vividly demonstrates this relationship in which the quantity of a radioactive substance at any time \(t\) depends exponentially on the time elapsed and the decay constant \(\lambda\).

This principle is paramount in calculations and predictions surrounding radioactive materials. As illustrated in the exercise solution, establishing the time for P-32 to reach 68% of its initial activity (part a) and determining the number of atoms present after a lengthy period (part b) fundamentally rely on this exponential relationship.

Exponential decay is observed not only in radioactivity but also widely across various scientific domains such as in population decline, cooling of hot objects, and attenuation of light. Its understanding is crucial for students and professionals alike who work with dynamic systems where change is rate-dependent.