/*! 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 95 The radionuclide \({ }^{32} \mat... [FREE SOLUTION] | 91Ó°ÊÓ

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The radionuclide \({ }^{32} \mathrm{P}\left(T_{1 / 2}=14.28 \mathrm{~d}\right)\) is often used as a tracer to follow the course of biochemical reactions involving phosphorus. (a) If the counting rate in a particular experimental setup is initially 3050 counts/s, how much time will the rate take to fall to 170 counts/s? (b) A solution containing \({ }^{32} \mathrm{P}\) is fed to the root system of an experimental tomato plant, and the \({ }^{32} \mathrm{P}\) activity in a leaf is measured 3.48 days later. By what factor must this reading be multiplied to correct for the decay that has occurred since the experiment began?

Short Answer

Expert verified
(a) 55.60 days, (b) Correction factor is 1.188.

Step by step solution

01

Understand the decay formula

The decay of a radioactive substance can be described by the formula: \[N(t) = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}\]where \(N(t)\) is the remaining quantity at time \(t\), \(N_0\) is the initial quantity, and \(T_{1/2}\) is the half-life.
02

Calculate decay time (Part a)

Using the formula, we need to solve for \(t\) such that the initial rate of 3050 counts/s reduces to 170 counts/s.\[3050 \left( \frac{1}{2} \right)^{t/14.28} = 170\]Divide both sides by 3050:\[ \left( \frac{1}{2} \right)^{t/14.28} = \frac{170}{3050}\]This simplifies to:\[ \left( \frac{1}{2} \right)^{t/14.28} = 0.0557\]Take the logarithm of both sides to solve for \(t\):\[ \frac{t}{14.28} \log \left( \frac{1}{2} \right) = \log(0.0557)\]Solving for \(t\):\[ t = 14.28 \times \frac{\log(0.0557)}{\log(0.5)}\]\[ t \approx 55.60 \text{ days}\]
03

Determine decay factor (Part b)

The correction factor for decay can be calculated by the formula given the time elapsed and the half-life.\[ \text{Correction factor} = \left( \frac{1}{2} \right)^{-t/T_{1/2}} \]Here, \(t = 3.48\) days. Substituting the values:\[ \text{Correction factor} = \left( \frac{1}{2} \right)^{-3.48/14.28}\]Calculating gives:\[ \text{Correction factor} \approx 1.188 \]
04

Final Answer

The time for the rate to decrease from 3050 counts/s to 170 counts/s is approximately 55.60 days. For part b, the correction factor for 3.48 days decay is approximately 1.188.

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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 and is crucial for understanding how radionuclides like \(^{32} \text{P}\) change over time. Half-life, represented as \(T_{1/2}\), refers to the time it takes for half of the radioactive atoms in a sample to decay. This term is different for each radioactive isotope and is a key characteristic in identifying how quickly they lose their radioactivity.
When we say that the half-life of \(^{32} \text{P}\) is 14.28 days, it implies that if you start with a certain amount of \(^{32} \text{P}\), only half of that amount will remain after 14.28 days.
This decay continues exponentially, meaning that half of the remaining material decays in the next half-life period, further reducing the quantity.
Understanding the half-life allows scientists to predict how long a substance will remain active, which is particularly useful when using radioactive tracers in experiments.
radiation counting rate
The radiation counting rate is a measure of how much radiation is being emitted from a radioactive source at any given time. It's often measured in counts per second (cps), which tells us how frequently radiation events (like disintegrations of atoms) are being detected by a sensor.
In our given problem, the counting rate began at 3050 cps and decreased to 170 cps. This change occurs because radioactive materials decay over time, thus decreasing their emission rate. By understanding the initial and desired counting rates, and using the concept of half-life, we can determine how long it will take to reach a particular level of radioactivity.
This is crucial in experimental settings where researchers need to track the decay and ensure measurements are made under predictable conditions. The goal is to track precisely how the radioactivity decreases, providing insights into both the rate of decay and the effectiveness of the isotope as a tracer.
tracer in biochemical reactions
Tracers are powerful tools in biochemical research, allowing scientists to study processes that would otherwise be invisible. These tracers, like \(^{32} \text{P}\), are often radioactive elements that can be incorporated into molecules within an organism.
When introduced into a system, they "trace" the physical and chemical changes, enabling researchers to follow a compound's journey through different biochemical pathways. Radioactive \(^{32} \text{P}\) is particularly useful in studies involving phosphorus cycles in plants and animals, as it mimics the natural element's behavior but allows detection through its radioactive emissions.
For instance, in the exercise example, \(^{32} \text{P}\) is introduced to a tomato plant's root system. By measuring the radiation from leaves after a set period, researchers can understand how phosphorus moves through and is utilized by the plant.
This observed behavior helps in research fields like nutrition, pharmacology, and environmental science, offering insights into everything from nutrient absorption to pollutant behavior in organisms.

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Most popular questions from this chapter

Under certain rare circumstances, a nucleus can decay by emitting a particle more massive than an alpha particle. Consider the decays $${ }^{223} \mathrm{Ra} \rightarrow{ }^{209} \mathrm{~Pb}+{ }^{14} \mathrm{C} \quad \text { and } \quad{ }^{223} \mathrm{Ra} \rightarrow{ }^{219} \mathrm{Rn}+{ }^{4} \mathrm{He}$$ Calculate the \(Q\) value for the (a) first and (b) second decay and determine that both are energetically possible. (c) The Coulomb barrier height for alpha-particle emission is \(30.0 \mathrm{MeV}\). What is the barrier height for \({ }^{14} \mathrm{C}\) emission? (Be careful about the nuclear radii.) The needed atomic masses are $$ \begin{aligned} &\begin{array}{llll} { }^{223} \mathrm{Ra} & 223.01850 \mathrm{u} & { }^{14} \mathrm{C} & 14.00324 \mathrm{u} \end{array}\\\ &{ }^{209} \mathrm{~Pb} \quad 208.98107 \mathrm{u} \quad{ }^{4} \mathrm{He} \quad 4.00260 \mathrm{u}\\\ &{ }^{219} \mathrm{Rn} \quad 219.00948 \mathrm{u} \end{aligned} $$

Some radionuclides decay by capturing one of their own atomic electrons, a \(K\) -shell electron, say. An example is $${ }^{49} \mathrm{~V}+\mathrm{e}^{-} \rightarrow{ }^{49} \mathrm{Ti}+\nu, \quad T_{1 / 2}=331 \mathrm{~d}$$ Show that the disintegration energy \(Q\) for this process is given by $$Q=\left(m_{\mathrm{V}}-m_{\mathrm{Ti}}\right) c^{2}-E_{K}$$ where \(m_{\mathrm{v}}\) and \(m_{\mathrm{Ti}}\) are the atomic masses of \({ }^{49} \mathrm{~V}\) and \({ }^{49} \mathrm{Ti},\) respectively, and \(E_{K}\) is the binding energy of the vanadium \(K\) -shell electron. (Hint: Put \(\mathbf{m}_{\mathrm{V}}\) and \(\mathbf{m}_{\mathrm{Ti}}\) as the corresponding nuclear masses and then add in enough electrons to use the atomic masses.)

A radium source contains \(1.00 \mathrm{mg}\) of \({ }^{226} \mathrm{Ra},\) which decays with a half-life of \(1600 \mathrm{y}\) to produce \({ }^{222} \mathrm{Rn},\) a noble gas. This radon isotope in turn decays by alpha emission with a half-life of \(3.82 \mathrm{~d}\). If this process continues for a time much longer than the half-life of \({ }^{222} \mathrm{Rn}\), the \({ }^{222}\) Rn decay rate reaches a limiting value that matches the rate at which \({ }^{222} \mathrm{Rn}\) is being produced, which is approximately constant because of the relatively long half-life of \({ }^{226} \mathrm{Ra}\). For the source under this limiting condition, what are (a) the activity of \({ }^{226} \mathrm{Ra},\) (b) the activity of \({ }^{222} \mathrm{Rn},\) and \((\mathrm{c})\) the total mass of \({ }^{222} \mathrm{Rn} ?\)

Cancer cells are more vulnerable to \(\mathrm{x}\) and gamma radiation than are healthy cells. In the past, the standard source for radiation therapy was radioactive \({ }^{60} \mathrm{Co},\) which decays, with a half-life of \(5.27 \mathrm{y},\) into an excited nuclear state of \({ }^{60} \mathrm{Ni}\). That nickel isotope then immediately emits two gamma-ray photons, each with an approximate energy of \(1.2 \mathrm{MeV}\). How many radioactive \({ }^{60} \mathrm{Co}\) nuclei are present in a \(6000 \mathrm{Ci}\) source of the type used in hospitals? (Energetic particles from linear accelerators are now used in radiation therapy.)

When aboveground nuclear tests were conducted, the explosions shot radioactive dust into the upper atmosphere. Global air circulations then spread the dust worldwide before it settled out on ground and water. One such test was conducted in October 1976 . What fraction of the \({ }^{90} \mathrm{Sr}\) produced by that explosion still existed in October 2006 ? The half-life of \({ }^{90} \mathrm{Sr}\) is \(29 \mathrm{y}\).

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