Problem 31 - Chemistry The Molecular Science

Question

Phosphorus- 32 is used in the form of \(\mathrm{Na}_{2} \mathrm{H}^{32} \mathrm{PO}_{4}\) in the treatment of chronic myeloid leukemia. (a) The isotope decays by emitting a \(\beta\) particle. Write a balanced equation to show this process. (b) The half-life of \({ }^{32} \mathrm{P}\) is 14.3 days. If you begin with \(9.6 \mathrm{mg}\) radioactive \(\mathrm{Na}_{2} \mathrm{H}^{32} \mathrm{PO}_{4},\) calculate what mass (mg) remains after 28.6 days.

Step-by-Step Solution

Answer

(a) \\ \(^{32}_{15}P \rightarrow ^{32}_{16}S + \beta^-\\) (b) 2.4 mg remain after 28.6 days.

Step 1: Write Decay Process Equation

In beta decay, a neutron in the nucleus of an atom transforms into a proton, and a beta particle, which is an electron, is emitted. For phosphorus-32 ((^32 P)), the balanced nuclear equation is:\[^{32}_{15}P \rightarrow ^{32}_{16}S + \beta^-\]This shows that a beta decay results in phosphorus turning into sulfur.

Step 2: Understand Half-life Concept

The half-life of a substance is the time it takes for half of the radioactive nuclei to decay. For (^32 P) , the half-life is given as 14.3 days.

Step 3: Calculate Number of Half-lives

To find the number of half-lives that have passed in 28.6 days, divide the total time by the half-life:\[\frac{28.6}{14.3} = 2\]This means that two half-lives have elapsed.

Step 4: Apply Half-life Decay Formula

The formula for calculating the remaining mass after a certain number of half-lives is (m_t = m_0 \times (0.5)^n), where (m_0) is the initial mass, (m_t) is the remaining mass, and (n) is the number of half-lives. Substituting the values:\[m_t = 9.6 \times (0.5)^2 = 9.6 \times 0.25 = 2.4 \text{ mg}\]

Step 5: Conclusion

After two half-lives, 2.4 mg of the original 9.6 mg of radioactive substance remains. We used the properties of radioactive decay to find this value.

Key Concepts

Beta Decay

Beta decay is a process that changes a neutron into a proton within an atom's nucleus. In this process, the atom emits a beta particle, which is essentially a high-energy electron. This change increases the atomic number by 1, turning the original element into a new element. For phosphorus-32, the beta decay leads to the transformation into sulfur-32. We can write this nuclear reaction as: \[^{32}_{15}P \rightarrow ^{32}_{16}S + \beta^-\] This equation tells us that one phosphorus atom emits a beta particle and becomes a sulfur atom. It’s important to note that the mass number (32) remains the same, while the atomic number increases from 15 (phosphorus) to 16 (sulfur). This small change is crucial in understanding how elements can transform into one another through radioactive decay.

Half-life Calculation

The half-life of a radioactive element is the time it takes for half of the atoms in a sample to decay. For phosphorus-32, the half-life is 14.3 days. This concept is critical as it helps us estimate how long an element remains radioactive. Knowing the half-life allows one to calculate how much of a substance remains after a certain period. To calculate this, we first determine the number of half-lives elapsed. In the given exercise, a period of 28.6 days is used. Since the half-life of phosphorus-32 is 14.3 days, the number of half-lives that have passed is: \[\frac{28.6}{14.3} = 2\]This means two half-lives have passed during this period. Using this number, we can apply the half-life formula to find out how much phosphorus-32 remains.

Radioactive Decay Equation

The radioactive decay equation helps us calculate the remaining mass of a radioactive substance after it has decayed over a given period. The formula is given by: \[m_t = m_0 \times (0.5)^n\] where:

• \(m_0\) is the initial mass of the substance.
• \(m_t\) is the mass remaining after decay.
• \(n\) is the number of half-lives elapsed.

To apply this to phosphorus-32, starting with 9.6 mg, and calculating after 28.6 days, we substitute the known values: \[m_t = 9.6 \times (0.5)^2 = 9.6 \times 0.25 = 2.4 \, \text{mg}\]This is how we find out that out of the initial 9.6 mg of phosphorus-32, 2.4 mg remain after two half-lives. This equation significantly aids in predicting the behavior of radioactive materials over time.