Problem 32 - 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 \(^{32} P\) is 14.3 days. If you begin with 4.8 \(\mu g\) 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) \(^ {32}_{15}P \rightarrow ^{32}_{16}S + \beta^- + \overline{\nu}_e \) . (b) After 28.6 days, 1.2 \(\mu g\) remains.

Step 1: Understand Beta Decay

In beta decay, a neutron is converted into a proton, resulting in the emission of a beta particle (an electron) and an antineutrino. This changes the original element into a new element with the same mass number but an increased atomic number by 1.

Step 2: Write Beta Decay Equation for Phosphorus-32

Phosphorus-32 (\(^{32}_{15}P\)) decays by beta particle emission. According to the process, one neutron in the phosphorus is converted to a proton, increasing the atomic number by 1. Thus, \(^{32}_{15}P\) becomes \(^{32}_{16}S\): \[ ^{32}_{15}P \rightarrow ^{32}_{16}S + \beta^- + \overline{u}_e \]

Step 3: Calculate Number of Half-Lives

The half-life of \(^ {32} P\) is given as 14.3 days. The total time for decay is 28.6 days. Divide the total time by the half-life to determine the number of half-lives: \( \text{Number of half-lives} = \frac{28.6}{14.3} = 2 \).

Step 4: Use Half-Life Formula to Determine Remaining Mass

Each half-life reduces the remaining mass of a substance by half. Start with 4.8 \(\mu g\). After 1 half-life (14.3 days), the mass is \(\frac{4.8}{2} = 2.4 \mu g\). After 2 half-lives (28.6 days), the mass is \(\frac{2.4}{2} = 1.2 \mu g\).

Step 5: Conclusion

After 28.6 days, which is 2 half-lives of phosphorus-32, the remaining mass of radioactive \(^{32} P\) in the form of \(\mathrm{Na}_{2}\mathrm{HPO}_{4} \) is 1.2 \(\mu g\).

Key Concepts

Phosphorus-32

Phosphorus-32 is a radioactive isotope commonly used in medical treatments like chronic myeloid leukemia. This isotope has 32 nucleons: 15 protons and 17 neutrons. During its decay process, phosphorus-32 undergoes beta decay, transforming into a different element.
Phosphorus-32 is advantageous in treatments because of its radioactive properties, allowing it to target and destroy specific cells such as cancer cells. This medical utility makes it a valuable tool in nuclear medicine. Its decay involves straightforward concepts like the emission of beta particles, which students typically encounter in chemistry and physics courses.
In the case of phosphorus-32's decay, understanding its initial composition—15 protons and an atomic mass of 32—is crucial. This information helps in comprehending the changes it undergoes during decay, like the increase in atomic number when it becomes sulfur-32 during beta decay.

Half-life Calculation

The concept of half-life is essential in nuclear chemistry and physics.
It refers to the time required for half of a radioactive substance to decay. For phosphorus-32, the half-life is 14.3 days.
This means every 14.3 days, the amount of phosphorus-32 will reduce by half, continuing until very little remains.

• Example: If you start with 4.8 micrograms of phosphorus-32, you will have 2.4 micrograms left after the first 14.3 days.
• After another 14.3 days, totaling 28.6 days, you will be left with 1.2 micrograms.

Understanding half-life helps in planning medical treatments or experiments using radioactive materials. It's a predictable and consistent measure, making it a reliable tool for determining safe handling times and dosages in medical applications.

Radioactive Decay Equation

Writing a balanced radioactive decay equation involves representing the transformation of an unstable isotope into another element. Each decay process must conserve both mass number and atomic number, although they may redistributes among the new formed particles.
For phosphorus-32 undergoing beta decay, the equation illustrates its transformation into sulfur-32. The decay can be represented as follows:
\[ ^{32}_{15}\mathrm{P} \rightarrow ^{32}_{16}\mathrm{S} + \beta^- + \overline{u}_e \]
In this reaction:

• The phosphorus-32 atom loses a neutron, which turns into a proton. This conversion increases the atomic number from 15 to 16, forming sulfur-32.
• The beta particle (\( \beta^- \)) emitted is essentially an electron, representing the change from neutron to proton.
• The antineutrino (\( \overline{u}_e \)) is a nearly massless particle emitted during this process, balancing the equation.

Understanding and writing these equations helps explain how elements transform and how decay processes conserve certain properties, despite changes in atomic composition. This is fundamental in mastering nuclear transformations and their applications.