/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 The half-life \(\left(t_{1 / 2}\... [FREE SOLUTION] | 91Ó°ÊÓ

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The half-life \(\left(t_{1 / 2}\right)\) of a radioactive species is the time it takes for half of the species to emit radiation and decay (turn into a different species). If a quantity \(N_{0}\) of the species is present at time \(t=0,\) the amount present at a later time \(t\) is given by the expression \(N=N_{0}\left(\frac{1}{2}\right)^{t / t_{1 / 2}}\) Each decay event involves the emission of radiation. A unit of the intensity of radioactivity is a curie (Ci), defined as \(3.7 \times 10^{10}\) decay events per second. A 300,000 -gallon tank has been storing aqueous radioactive waste since \(1945 .\) The waste contains the radioactive isotope cesium-137 \(\left(^{137} \mathrm{Cs}\right)\), which has a half-life of 30.1 years and a specific radioactivity of 86.58 Cilg. The isotope undergoes beta decay to radioactive barium- 137 , which in turn emits gamma rays and decays to stable (nonradioactive) barium with a half-life of 2.5 minutes. The concentration of \(^{137} \mathrm{Cs}\) in 2013 was \(2.50 \times 10^{-3} \mathrm{g} / \mathrm{L}\) (a) What fraction of the \(^{137} \mathrm{Cs}\) would have to decay for the level of cesium-related radioactivity of the contents to be \(1.00 \times 10^{-3} \mathrm{Ci} / \mathrm{L} ?\) What total mass of cesium ( \(\mathrm{kg}\) ) would that loss represent? In what year would that level be reached? (b) What was the concentration of \(^{137} \mathrm{Cs}\) in the tank (g/L) when the waste was first stored? (c) Explain why the radioactive cesium in the tank poses a significant environmental threat while the radioactive barium does not.

Short Answer

Expert verified
a) Using the given specific radioactivity of 86.58 Ci/g, the concentration of Cs-137 needed for a rate of decay of \(1.00 \times 10^{-3} \) Ci/L is found. The difference between this value and the initial concentration represents the concentration which needs to decay. Multiply this by the volume of the tank for the total mass that must decay. Using the decay equation helps predict the future year when the level of radioactivity will be as desired. b) Re-arranging the decay equation will provide a formula to seek the initial concentration of Cs-137 in 1945. c) The environmental threat of Cs-137 is explained by its much longer half-life, and hence, longer persistence in the environment, as compared to the short-lived Ba-137, which rapidly decays to stable Barium.

Step by step solution

01

Determine fraction of Cs-137 that must decay

We are given the specific radioactivity of Cs-137 as 86.58 Ci/g. This implies that 86.58 Ci of radiation is emitted per gram of Cs-137. We need to find the amount of Cs-137 that need to decay in order to reduce the radioactivity to \(1.00 \times 10^{-3} \) Ci/L. So, we can write an equation for specific activity, i.e, \[1.00 \times 10^{-3} \, \text{Ci/L} = x \, \text{g/L} \times 86.58 \, \text{Ci/g}\] Solving for x (representing mass per volume or concentration), we find the concentration of Cs-137 that needs to remain.
02

Determine total mass of Cs-137 corresponds to the loss

After computing the concentration of Cs-137 that needs to remain in the tank, we subtract this amount from the initial concentration (which is 2.50 \( \times 10^{-3} \) g/L) to find out how much Cs-137 needs to decay. Then, we multiply this concentration by the volume of the tank to find out the total mass in Kg.
03

Determine the year level will be reached

Now we use the decay equation \(N = N_{0}(\frac{1}{2})^{t / t_{1 / 2}}\) where \(N_0\) represents the initial quantity, \(N\) is the quantity after time \(t\), \(t_{1 / 2}\) is the half-life. We know the values \(N\), \(N_0\) and \(t_{1 / 2}\), and we need to solve this equation for \(t\). This will give us the time elapsed since 1945, and we can calculate the current year by adding this time to 1945.
04

Find the initial concentration of Cs-137

To find initial concentration, we must consider the time that has passed since the initial moment when the tank was filled up. We use the decay equation inversely, setting \(N_0\) as the variable we seek, while \(N\), \(t\), and \(t_{1 / 2}\) are known values. We know the time that passed since the tank was filled (\(t = 2013 - 1945\)) and we apply it to find \(N_0\).
05

Explain environmental implications of Cs-137 and Ba-137 decay

This doesn't involve a numerical solution but requires understanding of radioactive decay. While \(^{137}Cs\) and \(^{137}Ba\) are both radioactive, only Cs poses an environmental threat, because the half-life of Ba is very short (2.5 minutes), therefore it decays rapidly to stable, non-radioactive Barium shortly after its formation.

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Key Concepts

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

half-life
The concept of half-life is a fundamental aspect of radioactive decay. It is the time required for half of the radioactive nuclei in a sample to decay. In simpler terms, it's how long it takes for half of the radioactive material to lose its radioactivity.
Imagine you have a substance with 100 radioactive atoms. After one half-life, only 50 of those atoms would remain radioactive. After two half-lives, only 25 would remain, and so on.
Half-life varies greatly among different substances. For example, cesium-137 has a relatively long half-life of 30.1 years. This means it remains radioactive for many years, increasing its potential environmental impact.
Mathematically, the decay of a radioactive substance over time can be described by the equation:
  • \( N = N_{0} \left(\frac{1}{2}\right)^{t/t_{1/2}} \)
Where:
  • \( N \) is the remaining quantity of the substance.
  • \( N_{0} \) is the initial quantity.
  • \( t \) represents the time elapsed, and
  • \( t_{1/2} \) is the half-life.
cesium-137
Cesium-137, often denoted as \( ^{137} \text{Cs} \), is a radioactive isotope of cesium. It is a byproduct of nuclear fission processes involving uranium and plutonium, such as those occurring in nuclear reactors and during nuclear weapon explosions.
This isotope is of significant concern due to its properties and environmental persistence.
  • **Half-Life**: One of the key characteristics of cesium-137 is its half-life of 30.1 years. This means that it takes over three decades for half of any given amount to decay.
  • **Decay Products**: It primarily undergoes beta decay, transforming into barium-137m. This decay process involves the emission of beta particles and gamma radiation, making cesium-137 a potent source of radioactivity.
  • **Environmental Impact**: Due to its long half-life and the penetrating nature of its gamma radiation, cesium-137 is a significant environmental hazard. It can spread readily in the environment, contaminating soil and water.
  • **Usage and Risks**: It is often found in medical radioisotope therapy and is used in industrial gauges. However, improper handling or accidental release can lead to widespread contamination and pose health risks.
specific radioactivity
Specific radioactivity refers to the activity per unit mass of a radioactive material. It provides a measure of how active a particular amount of a substance is, which is crucial for assessing potential exposure and risks.
For cesium-137, the specific activity is a measure of how many decay events occur per second per gram of this substance, often expressed in curies per gram (Ci/g). In this context:
  • Cesium-137 has a specific radioactivity of 86.58 Ci/g. This means that each gram of cesium-137 emits a significant amount of radiation, amounting to 86.58 curies worth of decay events per second.
  • Specific radioactivity helps determine how long a particular substance will remain hazardous. High specific radioactivity implies a high decay rate, leading to intense emissions of radiation over a short period. For cesium-137, even small quantities can be highly radioactive.
  • Understanding specific radioactivity is important for safe handling and storage of radioactive materials, and for the calculation of how much material can present a danger to living organisms or the environment.
The concept is crucial when calculating allowable concentrations of radioactive substances in environmental and safety standards, ensuring that exposure remains within manageable limits.

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Most popular questions from this chapter

Things were going smoothly at the Breaux Bridge Drug Co. pilot plant during the midnight to 8 a.m. shift until Therèse Lagniappe, the reactor operator, let the run instruction sheet get too close to the Coleman stove that was being used to heat water to prepare Lagniappe's bihourly cup of Community Coffee. What followed ended in a total loss of the run sheet, the coffee, and a substantial portion of the novel Lagniappe was writing. Remembering the less than enthusiastic reaction she got the last time she telephoned her supervisor in the middle of the night, Lagniappe decided to rely on her memory of the required flow-rate settings. The two liquids being fed to a stirred-tank reactor were circulostoic acid (CSA: \(M W=75, S G=0.90\) ) and flubitol (FB: \(M W=90, S G=0.75\) ). The product from the system was a popular over-the-counter drug that simultaneously cures high blood pressure and clumsiness. The molar ratio of the two feed streams had to be between 1.05 and 1.10 mol CSA/mol FB to keep the contents of the reactor from forming a solid plug. At the time of the accident, the flow rate of CSA was 45.8 L min. Lagniappe set the flow of flubitol to the value she thought had been in the run sheet: 55.2 L/min. Was she right? If not, how would she have been likely to learn of her mistake? (Note: The reactor was stainless steel, so she could not see the contents.)

The chemical reactor shown below has a cover that is held in place by a series of bolts. The cover is made of stainless steel ( \(\mathrm{SG}=8.0\) ), is 3 inches thick, has a diameter of 24 inches, and covers and seals an opening 20 inches in diameter. During turnaround, when the reactor is taken out of service for cleaning and repair, the cover was removed by an operator who thought the reactor had been depressurized using a standard venting procedure. However, the pressure gauge had been damaged in an earlier process upset (the reactor pressure had exceeded the upper limit of the gauge), and instead of being depressurized completely, the vessel was under a gauge pressure of 30 psi. (a) What force ( \(\left(\mathrm{b}_{\mathrm{f}}\right)\) were the bolts exerting on the cover before they were removed? (Hint: Don't forget that a pressure is exerted on the top of the cover by the atmosphere.) What happened when the last bolt was removed by the operator? Justify your prediction by estimating the initial acceleration of the cover upon removal of the last bolt. (b) Propose an alteration in the turnaround procedure to prevent recurrence of an incident of this kind.

A housing development is served by a water tower with the water level maintained between 20 and 30 meters above the ground, depending on demand and water availability. Responding to a resident's complaint about the low flow rate of water at his kitchen sink, a representative of the developer measured the water pressure at the tap above the kitchen sink and at the junction between the water main (a pipe connected to the bottom of the water tower) and the feed pipe to the house. The junction is \(5 \mathrm{m}\) below the level of the kitchen tap. All water valves in the house were turned off. (a) If the water level in the tower was 25 m above tap level, what should be the gauge pressures (kPa) at the tap and junction? (b) Suppose the pressure measurement at the tap was lower than your estimate in Part (a), but the measurement at the junction was as predicted. State a possible explanation. (c) If pressure measurements corresponded to the predictions in Part (a), what else could be responsible for the low water flow to the sink?

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A thermostat control with dial markings from 0 to 100 is used to regulate the temperature of an oil bath. A calibration plot on logarithmic coordinates of the temperature, \(T(\text { ( } \mathrm{F} \text { ), versus the dial setting, } R\), is a straight line that passes through the points \(\left(R_{1}=20.0, T_{1}=110.0^{\circ} \mathrm{F}\right)\) and \(\left(R_{2}=40.0, T_{2}=250.0^{\circ} \mathrm{F}.\right)\) (a) Derive an equation for \(T\left(^{\circ} \mathrm{F}\right)\) in terms of \(R\). (b) Estimate the thermostat setting needed to obtain a temperature of \(320^{\circ} \mathrm{F}\). (c) Suppose you set the thermostat to the value of \(R\) calculated in Part (b), and the reading of athermocouple mounted in the bath equilibrates at \(295^{\circ} \mathrm{F}\) instead of \(320^{\circ} \mathrm{F}\). Suggest several possible explanations.

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