Question

The curie (Ci) is a commonly used unit for measuring nuclear radioactivity: 1 curie of radiation is equal to \(3.7 \times 10^{10}\) decay events per second (the number of decay events from 1 g radium in 1 s). a. What mass of \(\mathrm{Na}_{2}^{38} \mathrm{SO}_{4}\) has an activity of 10.0 \(\mathrm{mCi}^{2}\) Sulfur- 38 has an atomic mass of 38.0 \(\mathrm{u}\) and a half-life of 2.87 \(\mathrm{h} .\) b. How long does it take for 99.99\(\%\) of a sample of sulfur- 38 to decay?

Short answer

(a) The mass of \(\mathrm{Na}_{2}^{38}\mathrm{SO}_{4}\) with 10.0 mCi activity is calculated as follows: 1. Convert the activity to Ci: \(10.0 \times 10^{-3}\,\mathrm{Ci}\) 2. Calculate the total decay events: \(3.7 \times 10^7\,\mathrm{decay\, events/s}\) 3. Find the number of decay events per second for 1 mole of Sulfur-38: \(N_{1\,\mathrm{mole}}\) 4. Calculate the number of moles of \(\mathrm{S}^{38}\) in the compound: \(n\) 5. Find the molar mass of \(\mathrm{Na}_{2}^{38}\mathrm{SO}_{4}\): Molar mass 6. Determine the mass of the compound: Mass = n * Molar mass (b) To find the time for 99.99% decay of Sulfur-38, use the decay formula: \(0.0001 = \left(\frac{1}{2}\right)^{\frac{t}{2.87}}\) Solve for t to get the time it takes for 99.99% of a sample of sulfur-38 to decay.

Step-by-step solution

(a) Finding the mass of \(\mathrm{Na}_{2}^{38}\mathrm{SO}_{4}\)

First, let's convert 10.0 mCi to Ci by dividing by 1000: \(10.0\,\mathrm{mCi} = 10.0 \times 10^{-3}\,\mathrm{Ci}\) Now, we will calculate the number of decay events per second for this activity: \(N = (10.0 \times 10^{-3}\,\mathrm{Ci}) \times (3.7 \times 10^{10}\,\mathrm{decay\, events/s}\) Total decay events = \(10.0 \times 10^{-3}\) * \(3.7 \times 10^{10}\) = \(3.7 \times 10^7\) Now, we will use the fact that 1 mole of Sulfur-38 will have an atomic mass of 38.0 g, and with Avogadro's number (\(6.02 \times 10^{23}\, \mathrm{atoms}\)), we can find the number of decay events per second for 1 mole of Sulfur-38, \(N_{1\,\mathrm{mole}} = \frac{3.7 \times 10^7\,\mathrm{decay\,events/s}}{(6.02 \times 10^{23}\,\mathrm{atoms})}\) Now, we need to find the number of moles of \(\mathrm{S}^{38}\) in the compound with 10.0 mCi activity. \(n = \frac{3.7 \times 10^7}{N_{1\,\mathrm{mole}}}\) Consider the molar mass of \(\mathrm{Na}_{2}^{38}\mathrm{SO}_{4}\) as, Molar mass = 2 x (mass of Na) + mass of Sulfur-38 + 4 x (mass of Oxygen) g/mol Finally, the mass of \(\mathrm{Na}_{2}^{38}\mathrm{SO}_{4}\) with 10.0 mCi activity is given by: Mass = n * Molar mass

(b) Time for 99.99% decay of Sulfur-38

We are given the half-life of Sulfur-38, so we will use that information to find the time it takes for 99.99% of a sample to decay. The decay formula is given by: \(N = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{t_{\frac{1}{2}}}}\) \(N\) = final amount \(N_0\) = initial amount \(t_{\frac{1}{2}}\) = half-life \(t\) = time Here, we want 99.99% decay, meaning only 0.01% of the initial amount remains after time t. So, \(\frac{N}{N_0} = 0.0001\), as we are given \(t_{\frac{1}{2}} = 2.87\,\mathrm{h}\) We can find the time t. \(0.0001 = \left(\frac{1}{2}\right)^{\frac{t}{2.87}}\) Now solve for t.

Key concepts

Curie (Ci) unit

In the world of nuclear radioactivity, the Curie (Ci) unit is an essential measurement of activity. Named after famed scientist Marie Curie, it is used to quantify radioactivity by counting the number of nuclear decay events that occur each second. One Curie is equal to a staggering \(3.7 \times 10^{10}\) decay events per second. This is the number of decays emitted from one gram of radium. To better understand, think of the Curie as a way to express how 'active' or decay-prone a radioactive sample might be.
For smaller quantities, like milligrams or micrograms, we often use milliCuries (mCi) and microCuries (µCi), which represent one-thousandth and one-millionth of a Curie respectively. These smaller units help to gauge smaller levels of radioactivity common in laboratory settings. Remember, when handling or measuring radioactivity, it's crucial to convert these smaller units into Curie for accuracy in calculations.

Sulfur-38

Sulfur-38 is a radioactive isotope of sulfur, not commonly found in nature. Isotopes are variants of a particular chemical element that differ in neutron number, although all isotopes of the same element have the same number of protons.

• Sulfur-38 has an atomic mass of 38 u.
• It specifically contains 16 protons (the same as any sulfur atom) and 22 neutrons (unlike the more stable isotope, Sulfur-32, which contains 16 neutrons).
• This gives Sulfur-38 its radioactive nature.

Producing Sulfur-38 usually involves neutron activation, where stable isotopes capture neutrons and transform into radioactive ones. Because of its radioactive properties, Sulfur-38 is utilized in tracing and labeling applications in research settings—notably in nuclear medicine.
By understanding isotopes like Sulfur-38, scientists can explore nuclear reactions and their applications in a safe manner, leading to advancements in fields like medical diagnostics.

Half-life

The concept of half-life is integral to understanding radioactive decay. It serves as a measure of the time required for half of a radioactive substance to decay. The half-life of a radioactive isotope indicates how stable it is.
Sulfur-38, for example, has a half-life of 2.87 hours. This means that if you start with a certain amount of Sulfur-38, after 2.87 hours, you'd have only half of the original amount due to decay. This continual halving is what makes half-life a useful measure in predicting the behavior of radioactive substances over time.
In practical terms, the half-life helps determine how quickly a sample will become less radioactive. Whether assessing nuclear waste, medical isotopes, or other radioactive materials, understanding half-lives guides safe handling and usage. Calculating the time it takes for a significant portion, like 99.99%, of material to decay involves using exponential decay formulas, which leverage the half-life to provide insights into decay progressions.