Problem 33 - 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 one month)?

Step-by-Step Solution

Answer

(a) \(^{32}_{15} \text{P} \rightarrow ^{32}_{16} \text{S} + \beta^- + \overline{\nu}_e\). (b) 1.2 \(\mu\text{g}\) remains after 28.6 days.

Step 1: Understanding Beta Decay

Phosphorus-32 decays by beta (\( \beta \)) emission. In a beta decay, a neutron in the nucleus is converted into a proton, an electron (beta particle), and an antineutrino. The atomic number increases by 1, but the mass number remains the same. Therefore, \(^{32}_{15} \text{P} \rightarrow ^{32}_{16} \text{S} + \beta^- \).

Step 2: Writing the Balanced Nuclear Equation

The balanced equation for the beta decay of \(^{32}_{15} \text{P}\) is:\[^{32}_{15} \text{P} \rightarrow ^{32}_{16} \text{S} + \beta^- + \overline{u}_e\] where \(\beta^-\) represents the emitted beta particle (electron) and \(\overline{u}_e\) is the antineutrino.

Step 3: Understanding Half-life and Calculating Time Periods

The half-life of \(^{32} \text{P}\) is 14.3 days. In this problem, you are asked to find the remaining mass after 28.6 days, which is exactly two half-lives (28.6 / 14.3 = 2).

Step 4: Applying Half-life Formula

One full half-life decreases the amount of radioactive substance by half. After one half-life (14.3 days), 50% of \(^{32} \text{P}\) remains. After two half-lives (28.6 days), 25% of the original \(^{32} \text{P}\) remains. Use the formula \(\text{Remaining Mass} = \text{Initial Mass} \times \left(\frac{1}{2}\right)^n\), where \(n\) is the number of half-lives.So, \(\text{Remaining Mass} = 4.8 \times \left(\frac{1}{2}\right)^2 = 1.2 \, \mu \text{g}\).

Step 5: Verifying the Calculation

To ensure accuracy, re-calculate the decay: Initially, you have 4.8 \(\mu\text{g}\) and after one half-life (14.3 days), it reduces to 2.4 \(\mu\text{g}\). After another 14.3 days, it further reduces to 1.2 \(\mu\text{g}\).

Key Concepts

Phosphorus-32

Phosphorus-32 (\(^32\text{P}\)) is a radioactive isotope of phosphorus. It is commonly used in medical and scientific applications, particularly due to its ability to emit beta particles. In therapies such as the treatment of chronic myeloid leukemia, Phosphorus-32 plays a crucial role due to its radioactive nature, which helps in targeting and destroying cancerous cells.

As an isotope, Phosphorus-32 has a specific number of protons and neutrons. It possesses 15 protons and 17 neutrons in its nucleus, making it distinct from the non-radioactive Phosphorus-31. Its radioactive properties come from this imbalance between protons and neutrons, allowing it to undergo beta decay to reach a more stable configuration. Understanding this decay process is critical in utilizing Phosphorus-32 safely and effectively in various applications.

Nuclear Equation

The nuclear equation represents the transformation Phosphorus-32 undergoes through beta decay. In this process, a neutron is converted into a proton, an electron (which is the beta particle), and an antineutrino. This conversion results in an increase in the atomic number by one while maintaining the same mass number, as the total nucleon count remains unchanged.

The balanced nuclear equation for this beta decay is as follows:\[^{32}_{15} \text{P} \rightarrow ^{32}_{16} \text{S} + \beta^- + \overline{u}_e\] Here:

• \(^{32}_{15} \text{P}\) is the original phosphorus-32 isotope.
• \(^{32}_{16} \text{S}\) is sulfur-32, the stable product of the decay.
• \(\beta^-\) is the beta particle, an electron emitted during the decay.
• \(\overline{u}_e\) refers to the antineutrino that accompanies the beta particle.

This concise nuclear equation helps in predicting the products of the decay and understanding the fundamental changes in atomic structure during this process.

Half-life Calculation

The concept of half-life is pivotal in understanding radioactive decay. The half-life of an isotope is defined as the time required for half of the radioactive atoms in a sample to decay. For Phosphorus-32, this period is 14.3 days.

Here's how half-life calculations work for Phosphorus-32:

• After one half-life (14.3 days), 50% of the initial quantity remains.
• After two half-lives (28.6 days), only 25% of the initial quantity remains.

The formula to determine the remaining mass of a radioactive sample is:\[\text{Remaining Mass} = \text{Initial Mass} \times \left(\frac{1}{2}\right)^n\]where \(n\) is the number of half-lives. In this exercise, starting with 4.8 \(\mu\text{g}\) of Phosphorus-32, after 28.6 days, which is equivalent to two half-lives, the remaining mass is calculated as:\[\text{Remaining Mass} = 4.8 \times \left(\frac{1}{2}\right)^2 = 1.2 \, \mu\text{g}\]This calculation demonstrates how quickly radioactive isotopes can diminish and the importance of understanding half-lives for practical applications.

Radioactive Isotope

Radioactive isotopes, also known as radioisotopes, are atoms whose nuclei are unstable due to an imbalance in the number of protons and neutrons. This instability leads to the emission of radiation as the isotope attempts to reach a more stable form. Phosphorus-32 is an example of a radioactive isotope, widely used in medicine and industry.

Key characteristics of radioactive isotopes include:

• They emit radiation in the form of alpha, beta, or gamma rays.
• These emissions occur at a rate determined by the isotope's half-life.
• They are utilized in various applications including medical treatments, nuclear power generation, and radiocarbon dating.

The use of radioactive isotopes, like Phosphorus-32, must be carefully managed to ensure safety. Understanding their properties helps in harnessing their benefits effectively while minimizing any potential risks.