Problem 30 - Chemistry An Atoms First Approach

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 \(1 \mathrm{s}\) ). a. What mass of \(\mathrm{Na}_{2}^{38} \mathrm{SO}_{4}\) has an activity of \(10.0 \mathrm{mCi} ?\) Sulfur-38 has an atomic mass of 38.0 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?

Step-by-Step Solution

Answer

a. The mass of Na₂₃⁸SO₄ with an activity of 10 mCi is \(1.42\times10^{-7}\, \) g. b. It takes 19.372 hours for 99.99% of a sample of Sulfur-38 to decay.

Step 1: Finding decay events per second of 10 mCi

Convert 10 mCi into decay events per second: 1 mCi = \(1\times10^{-3}\) Ci So, 10 mCi = \(10\times10^{-3}\) Ci Now, 1 Ci = \(3.7\times10^{10}\) decay events per second So, 10 mCi = \(10\times10^{-3}\) Ci × \(3.7\times10^{10}\) decay events/s = \(3.7\times10^{8}\) decay events/s

Step 2: Calculating moles of Sulfur-38

Now, let's convert decay events to moles. 1 mole of Sulfur-38 contains Avogadro's number of atoms, which is \(6.022\times10^{23}\) atoms/mol. Moles of Sulfur-38 = (decay events per second) / (Avogadro's Number × decay events per second per mol) = \(3.7\times10^{8} \text{ decay events/s}\) / (\(6.022\times10^{23} \text{ atoms/mol}\) × \(3.7\times10^{10}\) decay events/s per mol) = \(10^{-9}\, \) mol

Step 3: Calculating mass of Na₂₃⁸SO₄

Now, let's find the mass of Na₂₃⁸SO₄ with an activity of 10 mCi. First, calculate the molar mass of Na₂₃⁸SO₄: Na: 22.99 u (2 atoms) O: 16.00 u (4 atoms) S: 32.07 u (1 atom) Molar Mass = 2 × 22.99 u + 1 × 32.07 u + 4 × 16.00 u = 142.04 g/mol Mass = Moles × Molar Mass = \(10^{-9}\, \text{mol}\) × 142.04 g/mol = \(1.42\times10^{-7}\, \) g The mass of Na₂₃⁸SO₄ with an activity of 10 mCi is \(1.42\times10^{-7}\, \) g. b. Time for 99.99% of the sample of Sulfur-38 to decay:

Step 4: Applying the decay formula to find time

The decay formula is given by: \( N(t) = N_0 \cdot \frac{1}{2}^{\frac{t}{t_{1/2}}}\) Where, N(t): Number of atoms remaining at time t N₀: initial number of atoms t: time \(t_{1/2}\): half-life of the radioactive isotope Now, we need to find the time when 99.99% of a sample of sulfur-38 has decayed. In other words, we want to find the time when only 0.01% remain. Let N₀ = 100%. We want N(t) = 0.01%. \( 0.01\% = 100\% \cdot \frac{1}{2}^{\frac{t}{2.87\,h}}\) Divide both sides by 100 to find the ratio: \( \frac{0.01\%}{100\%} = \frac{1}{2}^{\frac{t}{2.87\,h}}\) Now solve for t: \( \frac{t}{2.87\,h} = \log_{\frac{1}{2}}{(\frac{0.01}{100})}\) t = \(2.87\,h \cdot \log_{\frac{1}{2}}{(\frac{0.01}{100})}\) ≈ 19.372 h It takes 19.372 hours for 99.99% of a sample of Sulfur-38 to decay.

Key Concepts

Curie Unit

When we talk about measuring radioactivity, one of the classic units we use is the curie unit. Named after the pioneering scientists Marie and Pierre Curie, the curie (Ci) is fundamental in understanding the intensity of radioactivity. One curie is defined as the amount of any radioactive substance that undergoes \(3.7 \times 10^{10}\) decay events per second.

This figure wasn't chosen at random — it's approximately the activity of one gram of radium, an element studied extensively by the Curies. To make this more tangible, imagine watching a fireworks display: each spark that flies off and extinguishes corresponds to a decay event. In one second, for something to be one curie, you'd have to visualise \(3.7 \times 10^{10}\) such sparks! That's a lot of activity happening very rapidly.

In practical situations like the original exercise, we often deal with smaller units like the millicurie (mCi), which is one-thousandth of a curie. The ability to convert between these units is crucial in fields such as nuclear medicine, radiography, and environmental radioactivity monitoring.

Radioactive Decay

Radioactive decay is a spontaneous process by which an unstable atomic nucleus loses energy by emitting radiation. This can be alpha particles, beta particles, gamma rays, or other particles. Each decay event results in the nucleus getting closer to a stable state.

Imagine you have a box full of fresh apples; over time, some apples start to deteriorate. Similarly, in a sample of a radioactive element, atoms are continually decaying into another element or isotope. We measure this activity in decay events per second, and as described earlier, this is where the curie comes into play.

It's important to recognize that each isotope has its particular decay path, which occurs at a rate defined by its half-life — the time it takes for half of a sample to decay. This concept of half-life leads to our next major topic but before we get there, remember that understanding radioactive decay is not just about numbers but also about the inherent unpredictability of each atom's decay and the statistical nature of large numbers of these events.

Half-Life Calculation

The half-life of a radioactive isotope is a measure of how quickly it decays. Specifically, it's the time required for half of the atoms in a sample to transform through radioactive decay. Every isotope has a characteristic half-life that remains constant, regardless of the amount of the substance. This property allows us to perform calculations to determine how long it will take for a specific amount of radioactive material to decrease to a certain level.

For instance, dealing with a substance like Sulfur-38, as in our textbook problem, with a known half-life, calculations can predict not just when half of it will decay, but also when nearly all of it will be gone — such as the 99.99% decay mentioned in the problem. This is done through logarithmic relationships in the decay formula, where the desired percentage of remaining substance guides the calculation of time passed.

Understanding how to calculate half-life is essential for a variety of practical applications including dating archaeological finds, determining the proper dosage for medical treatments, and ensuring the safety around nuclear waste disposal. It enables us to forecast the future behavior of the radioactive material and make informed decisions based on that forecast.