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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?

Short Answer

Expert verified
a. The mass of Na鈧傗們鈦窼O鈧 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 by step solution

01

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 脳 \(3.7\times10^{10}\) decay events/s = \(3.7\times10^{8}\) decay events/s
02

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 脳 decay events per second per mol) = \(3.7\times10^{8} \text{ decay events/s}\) / (\(6.022\times10^{23} \text{ atoms/mol}\) 脳 \(3.7\times10^{10}\) decay events/s per mol) = \(10^{-9}\, \) mol
03

Calculating mass of Na鈧傗們鈦窼O鈧

Now, let's find the mass of Na鈧傗們鈦窼O鈧 with an activity of 10 mCi. First, calculate the molar mass of Na鈧傗們鈦窼O鈧: Na: 22.99 u (2 atoms) O: 16.00 u (4 atoms) S: 32.07 u (1 atom) Molar Mass = 2 脳 22.99 u + 1 脳 32.07 u + 4 脳 16.00 u = 142.04 g/mol Mass = Moles 脳 Molar Mass = \(10^{-9}\, \text{mol}\) 脳 142.04 g/mol = \(1.42\times10^{-7}\, \) g The mass of Na鈧傗們鈦窼O鈧 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:
04

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 N鈧: 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 N鈧 = 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 = \(2.87\,h \cdot \log_{\frac{1}{2}}{(\frac{0.01}{100})}\) 鈮 19.372 h It takes 19.372 hours for 99.99% of a sample of Sulfur-38 to decay.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 鈥 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 鈥 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 鈥 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.

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Most popular questions from this chapter

Which do you think would be the greater health hazard: the release of a radioactive nuclide of Sr or a radioactive nuclide of Xe into the environment? Assume the amount of radioactivity is the same in each case. Explain your answer on the basis of the chemical properties of \(\mathrm{Sr}\) and Xe. Why are the chemical properties of a radioactive substance important in assessing its potential health hazards?

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 1.7 -mL sample of water containing tritium was injected into a 150 -lb person. The total activity of radiation injected was \(86.5 \mathrm{mCi}\). After some time to allow the tritium activity to equally distribute throughout the body, a sample of blood plasma containing \(2.0 \mathrm{mL}\) water at an activity of \(3.6 \mu \mathrm{Ci}\) was removed. From these data, calculate the mass percent of water in this 150 -lb person.

The most significant source of natural radiation is radon-222. \(^{222} \mathrm{Rn},\) a decay product of \(^{238} \mathrm{U},\) is continuously generated in the earth's crust, allowing gaseous Rn to seep into the basements of buildings. Because \(^{222} \mathrm{Rn}\) is an \(\alpha\) -particle producer with a relatively short half-life of 3.82 days, it can cause biological damage when inhaled. a. How many \(\alpha\) particles and \(\beta\) particles are produced when \(^{238} \mathrm{U}\) decays to \(^{222} \mathrm{Rn} ?\) What nuclei are produced when \(^{222} \mathrm{Rn}\) decays? b. Radon is a noble gas so one would expect it to pass through the body quickly. Why is there a concern over inhaling \(^{222} \mathrm{Rn} ?\) c. Another problem associated with \(^{222} \mathrm{Rn}\) is that the decay of \(^{222} \mathrm{Rn}\) produces a more potent \(\alpha\) -particle producer \(\left(t_{1 / 2}=\right.\) 3.11 min) that is a solid. What is the identity of the solid? Give the balanced equation of this species decaying by \(\alpha\) particle production. Why is the solid a more potent \(\alpha\) -particle producer? d. The U.S. Environmental Protection Agency (EPA) recommends that \(^{222}\) Rn levels not exceed 4 pCi per liter of air (1 \(\mathrm{Ci}=1\) curie \(=3.7 \times 10^{10}\) decay events per second; \(1 \mathrm{pCi}=1 \times 10^{-12} \mathrm{Ci}\). Convert \(4.0 \mathrm{pCi}\) per liter of air into concentrations units of \(^{222} \mathrm{Rn}\) atoms per liter of air and moles of \(^{222}\) Rn per liter of air.

What are transuranium elements and how are they synthesized?

A chemist wishing to do an experiment requiring \(^{47} \mathrm{Ca}^{2+}\) (half-life \(=4.5\) days) needs \(5.0 \mu \mathrm{g}\) of the nuclide. What mass of \(^{47} \mathrm{CaCO}_{3}\) must be ordered if it takes \(48 \mathrm{h}\) for delivery from the supplier? Assume that the atomic mass of \(^{47} \mathrm{Ca}\) is \(47.0 \mathrm{u}\)

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