A \(0.20-\mathrm{mL}\) sample of a solution containing \({ }_{1}^{3} \mathrm{H}\) that produces \(3.7 \times\) \(10^{3}\) cps is injected into the bloodstream of an animal. After allowing circulatory equilibrium to be established, a \(0.20-\mathrm{mL}\) sample of blood is found to have an activity of 20 . cps. Calculate the blood volume of the animal.

Short Answer

Expert verified
The blood volume of the animal is 0.01086 mL.

Step by step solution

01

Understand the given information

We are given the following information: - Initial volume of radioactive solution (\(V_1\)) = 0.20 mL - Initial activity of radioactive solution (\(A_1\)) = \(3.7 \times 10^3\) cps (counts per second) - Final activity of the same volume of blood after reaching circulatory equilibrium (\(A_2\)) = 20 cps It is important to note that the volume of the blood sample, \(V_2\) = 0.20 mL, is the same as the initial volume of the radioactive solution. This will be helpful as we proceed with the calculations.
02

Apply the principle of radioactive dilution

According to the principle of radioactive dilution, the total activity of the system remains constant, meaning that the product of the concentration of radioactive isotopes and the volume it occupies remains constant. This can be written mathematically as: \[A_1 \cdot V_1 = A_2 \cdot V_2\] We will now use this equation to find the blood volume of the animal.
03

Solve for the unknown blood volume, \(V_2\)

Given that the volume of the blood sample is equal to the initial volume of the radioactive solution (as mentioned above), we can substitute \(V_2 = V_1\) into our equation: \[A_1 \cdot V_1 = A_2 \cdot V_1\] Since our goal is to find the total blood volume, let's solve for \(V_1\): \[V_1 = \frac{A_2}{A_1 - A_2} \cdot V_1\] Now, substitute the given values: \[V_1 = \frac{20 \,\text{cps}}{3.7 \times 10^3\,\text{cps} - 20\,\text{cps}} \cdot 0.20\,\text{mL}\]
04

Calculate the blood volume

Perform the calculation: \[V_1 = \frac{20}{3.7 \times 10^3 - 20} \cdot 0.20 = \frac{20}{3.68 \times 10^3}\cdot 0.20 = 0.01086 \, \text{mL}\] So the blood volume of the animal is 0.01086 mL.

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Key Concepts

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

Radioactive Isotope Concentration
Understanding the concentration of radioactive isotopes in a solution is crucial in various scientific calculations and experiments, especially in the field of nuclear chemistry. Radioactive isotope concentration refers to the amount of a radioactive substance within a given volume. This is commonly measured in counts per second (cps), indicating how many atoms are decaying and thereby emitting radiation each second.

In our exercise scenario, the initial concentration is provided, and the concentration after dilution through the animal's bloodstream is measured. To calculate the change in concentration, one must understand the principle of radioactive dilution. This principle states that when a solution containing radioactive isotopes is diluted, the total activity (the product of concentration and volume) remains unchanged if there is no decay over the course of the experiment.

Therefore, by using the equation: \[A_1 \times V_1 = A_2 \times V_2\] we can relate the initial concentration (activity) of the radioactive substance to its concentration after dilution. This equation is pivotal in calculating parameters such as the blood volume of an organism.
Blood Volume Calculation
The calculation of an organism's blood volume is a practical application of the aforementioned concepts of radioactive isotope concentration. The concept revolves around injecting a known volume of a radioactive tracer and then, after dilution within the bloodstream, measuring the activity in a blood sample.

The blood volume is crucial for diagnostic and research purposes in both human medicine and veterinary fields. To calculate blood volume accurately, scientists use the formula derived from the conservation of activity during the dilution process: \[V_{blood} = \frac{A_1}{A_2} \times V_{sample}\] where:
  • \(V_{blood}\) is the total blood volume,
  • \(A_1\) is the initial activity of the injected solution,
  • \(A_2\) is the activity measured in the blood sample after equilibrium, and
  • \(V_{sample}\) is the volume of the blood sample.
By applying this formula, we can determine the circulatory blood volume using a small sample post-injection, which is a method known for its accuracy and minimal invasiveness.
Nuclear Chemistry Applications
The use of radioactive materials and the principles of nuclear chemistry have widespread applications in modern science and industry. These applications range from medical diagnostics and treatments, such as the use of radioactive tracers in PET scans and the treatment of cancer with radiotherapy, to environmental tracking where isotopes can trace the path of pollutants.

Moreover, in research, radioactive isotopes are invaluable tools for understanding complex biochemical pathways and reactions by labeling specific molecules and tracking their progress through these systems. Nuclear chemistry also plays a significant role in energy production through nuclear reactors and, increasingly, in the study of sustainability and long-term disposal of nuclear waste.

With every application, the fundamental principles that govern the behavior of radioactive substances, such as decay rates, isotope concentration, and the conservation of activity, are utilized to ensure precise and reliable results in scientific studies.

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Most popular questions from this chapter

During the research that led to production of the two atomic bombs used against Japan in World War II, different mechanisms for obtaining a supercritical mass of fissionable material were investigated. In one type of bomb, a "gun" shot one piece of fissionable material into a cavity containing another piece of fissionable material. In the second type of bomb, the fissionable material was surrounded with a high explosive that, when detonated, compressed the fissionable material into a smaller volume. Discuss what is meant by critical mass, and explain why the ability to achieve a critical mass is essential to sustaining a nuclear reaction.

Write balanced equations for each of the processes described below. a. Chromium- 51 , which targets the spleen and is used as a tracer in studies of red blood cells, decays by electron capture. b. Iodine-131, used to treat hyperactive thyroid glands, decays by producing a \(\beta\) particle. c. Phosphorus- 32, which accumulates in the liver, decays by \(\beta\) particle production.

A recent study concluded that any amount of radiation exposure can cause biological damage. Explain the differences between the two models of radiation damage, the linear model and the threshold model.

Photosynthesis in plants can be represented by the following overall reaction: $$6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \stackrel{\text { Light }}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g)$$ Algae grown in water containing some \({ }^{18} \mathrm{O}\) (in \(\mathrm{H}_{2}{ }^{18} \mathrm{O}\) ) evolve oxygen gas with the same isotopic composition as the oxygen in the water. When algae growing in water containing only \({ }^{16} \mathrm{O}\) were furnished carbon dioxide containing \({ }^{18} \mathrm{O}\), no \({ }^{18} \mathrm{O}\) was found to be evolved from the oxygen gas produced. What conclusions about photosynthesis can be drawn from these experiments?

During World War II, tritium \(\left({ }^{3} \mathrm{H}\right)\) was a component of fluorescent watch dials and hands. Assume you have such a watch that was made in January 1944 . If \(17 \%\) or more of the original tritium was needed to read the dial in dark places, until what year could you read the time at night? (For \({ }^{3} \mathrm{H}, t_{1 / 2}=12.3 \mathrm{yr}\).)

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