Problem 42 - Chemistry and Chemical Reactivity

Question

Phosphorus-32 is used in the form of \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) in the treatment of chronic myeloid leukemia, among other things. (a) The isotope decays by \(\beta\) -particle emission. Write a balanced equation for this process. (b) The half-life of \(^{32} \mathrm{P}\) is 14.3 days. If you begin with \(4.8 \mu \mathrm{g}\) of radioactive \(^{32} \mathrm{P}\) in the form of \(\mathrm{Na}_{2} \mathrm{HPO}_{4},\) what mass remains after 28.6 days (about 1 month)?

Step-by-Step Solution

Answer

(a) \( \,_{15}^{32}\mathrm{P} \rightarrow \,_{16}^{32}\mathrm{S} + \beta^- + \overline{\nu}_e \). (b) 1.2 μg remains.

Step 1: Understand Beta Decay Process

Phosphorus-32 undergoes beta decay to become an isotope of sulfur. In beta decay, a neutron in the nucleus is transformed into a proton and an electron (beta particle) is emitted. The atomic number increases by 1, but the mass number remains the same.

Step 2: Write the Balanced Equation for Beta Decay

The balanced equation for the beta decay of phosphorus-32 is:\[ \underline{\phantom{xxx}}_{15}^{32}\mathrm{P} \rightarrow \,_{16}^{32}\mathrm{S} + \beta^- + \overline{u}_e \] Here, phosphorus-32 (P) decays to sulfur-32 (S) with the emission of a beta particle (\beta^-) and an antineutrino (\overline{u}_e).

Step 3: Identify Initial and Time Parameters

We are given that the initial mass of phosphorus-32 is 4.8 μg, and we need to find the remaining mass after 28.6 days. The half-life of phosphorus-32 is given as 14.3 days.

Step 4: Calculate Number of Half-life Periods

Calculate the number of half-life periods as follows:\[ \text{Number of half-lives} = \frac{\text{Total time elapsed}}{\text{Half-life}} = \frac{28.6\, \text{days}}{14.3\, \text{days/half-life}} = 2 \] This tells us that 2 half-life periods have elapsed.

Step 5: Apply Exponential Decay Formula

Using the concept of half-lives, the formula for the remaining mass after a given number of half-lives is:\[ N = N_0 \times \left(\frac{1}{2}\right)^n \]where \( N_0 \) is the initial mass, \( n \) is the number of half-life periods, and \( N \) is the remaining mass.

Step 6: Calculate Remaining Mass

Plug in the values to the formula:\[ N = 4.8 \, \mu\mathrm{g} \times \left(\frac{1}{2}\right)^2 \]\[ N = 4.8 \, \mu\mathrm{g} \times 0.25 = 1.2 \, \mu \mathrm{g} \]The remaining mass of phosphorus-32 after 28.6 days is 1.2 μg.

Key Concepts

Phosphorus-32

Phosphorus-32 is a radioactive isotope of phosphorus, commonly represented as \(_{15}^{32}\mathrm{P}\). It is used in a variety of applications, including research and medical treatments. One of its notable uses is in the form of \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) for the treatment of certain types of cancers, like chronic myeloid leukemia. This isotope is valuable for such treatments because of its radioactive properties, which allow it to target cancer cells effectively.
Understanding phosphorus-32 is crucial for comprehending radioactive decay processes, especially given its role in medical applications. It is also an essential isotope used in tracers in molecular biology and biochemistry to follow biochemical processes.

Beta Decay

Beta decay is a type of radioactive decay where a beta particle is emitted from an atomic nucleus. Phosphorus-32 undergoes beta decay, which is a process that transforms a neutron into a proton within the nucleus, emitting an electron (known as a beta particle) and an antineutrino. This changes the phosphorus atom into a sulfur atom, moving it from \(_{15}^{32}\mathrm{P}\) to \(_{16}^{32}\mathrm{S}\).

• The atomic number increases by one, as a neutron becomes a proton.
• The mass number remains unchanged despite the emission.
• The emitted beta particle and antineutrino carry away the excess energy.

Beta decay is key for understanding nuclear stability and transformation. The balanced equation for beta decay of phosphorus-32 is: \[ \underline{\phantom{xxx}}_{15}^{32}\mathrm{P} \rightarrow \,_{16}^{32}\mathrm{S} + \beta^- + \overline{u}_e \]This equation succinctly communicates the transformation phosphorus-32 undergoes during decay.

Half-life

The half-life of a radioactive isotope is the time it takes for half of the material to decay. For phosphorus-32, the half-life is 14.3 days. This means after 14.3 days, only half of the original amount of phosphorus-32 would remain. Understanding half-life is crucial for calculating how much of a radioactive substance remains over a specific period.
For example, if you start with \(4.8 \mu \mathrm{g}\) of phosphorus-32, after 28.6 days (two half-life periods), the mass would be reduced to \(1.2 \mu \mathrm{g}\). The calculation involves dividing the time elapsed by the half-life to determine the number of half-lives that have passed, and applying the formula: \[ N = N_0 \times \left(\frac{1}{2}\right)^n \]This exponential decay formula helps predict the remaining mass after a given time. It's an essential concept for studying radioactive materials and their decay over time.

Radioactive Isotopes

Radioactive isotopes, also known as radioisotopes, are variants of chemical elements that have unstable nuclei. They emit radiation as they decay into more stable forms. Phosphorus-32 is one such isotope, known for its application in medicine and research. These isotopes are key to understanding many natural and artificial processes as they undergo changes that are predictable by their half-life and decay type.

• They have applications in medical diagnostics and treatments, often used in the form of tracers or treatment agents.
• Their decay processes help scientists study chemical and biological pathways.
• Radioactive isotopes can vary in their stability, half-life, and the type of radiation they emit.

Understanding radioactive isotopes is essential for fields like nuclear medicine, environmental science, and even archaeology. Their ability to shed light on processes that are otherwise invisible makes them invaluable tools in both scientific research and practical applications.