Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Why is it important that radioisotopes used as diagnostic tools in nuclear medicine produce gamma radiation when they decay? Why are alpha emitters not used as diagnostic tools?

Short Answer

Expert verified
Gamma radiation-emitting radioisotopes are used as diagnostic tools in nuclear medicine due to their high penetration ability and low ionizing potential, providing detailed imaging with minimal harm to the patient. Alpha emitters are not suitable for this purpose because of their limited penetration abilities and highly ionizing nature, which can cause significant damage to cells and tissues.

Step by step solution

01

Radioisotopes in Nuclear Medicine

Radioisotopes are used as diagnostic tools in nuclear medicine because of their ability to provide valuable information about the structure and function of various organs and tissues in the body. They can be incorporated within a molecule that can interact and target a specific organ or location in the body, which in turn can be visualized and examined using imaging devices like gamma cameras or PET scans.
02

Properties of Gamma Radiation

Gamma radiation is a form of electromagnetic radiation, with a very high energy and frequency levels, placing it at the uppermost part of the electromagnetic spectrum. It is highly penetrating, and its rays can pass through most materials, including human tissue, without causing much damage. It is also characterized by its low ionizing potential, meaning that it is less likely to cause significant damage or disruptions to cells and tissue as it passes through them. This makes gamma radiation ideal for diagnostic purposes, as it allows for detailed imaging without causing significant harm to the patient.
03

Why Gamma Radiation is Used

Gamma radiation, when produced by radioisotopes used in nuclear medicine, is valuable as a diagnostic tool mainly because of its penetration ability and low ionizing potential. Since gamma rays can easily pass through human tissue and other materials, they can be detected by the appropriate imaging devices, providing high-resolution images of the target organ or tissue. This allows for accurate and non-invasive diagnosis of various medical conditions, leading to appropriate treatments that can improve patient outcomes.
04

Properties of Alpha Radiation

Alpha radiation, on the other hand, consists of positively charged particles known as alpha particles. These particles contain two protons and two neutrons, making them much larger and heavier than gamma rays. This larger size and mass result in alpha radiation having a much lower penetration ability as compared to gamma radiation. Consequently, alpha particles can be stopped by relatively thin layers of materials, even including our skin or a sheet of paper.
05

Why Alpha Emitters are Not Used

Due to the limited penetration potential of alpha radiation, alpha emitters are not suitable for use as diagnostic tools in nuclear medicine. The low penetration ability of alpha particles would hinder their ability to pass through human tissue and organs, making it very difficult to detect and visualize the target area using imaging devices, especially when comparing it to gamma radiation's functionality. Additionally, alpha radiation is highly ionizing and can cause significant damage and disruptions to cells and tissues that it directly interacts with – which is another reason why it is not suitable for diagnostic purposes in nuclear medicine. In conclusion, gamma radiation-emitting radioisotopes are used as diagnostic tools in nuclear medicine because of their high penetration ability and low ionizing potential, which provides detailed imaging while minimizing harm to the patient. Conversely, alpha emitters are not suitable for this purpose due to their limited penetration abilities and highly ionizing nature.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A sample of an alpha emitter having an activity of 0.18 Ci is stored in a 25.0 -mL sealed container at \(22^{\circ} \mathrm{C}\) for 245 days. (a) How many alpha particles are formed during this time? (b) Assuming that each alpha particle is converted to a helium atom, what is the partial pressure of helium gas in the container after this 245 -day period?

Complete and balance the following nuclear equations by supplying the missing particle: \begin{equation}\begin{array}{l}{\text { (a) }_{98}^{252} \mathrm{Cf}+_{5}^{10} \mathrm{B} \longrightarrow 3_{0}^{1} \mathrm{n}+?} \\\ {\text { (b) }_{1}^{2} \mathrm{H}+_{2}^{3} \mathrm{He} \longrightarrow_{2}^{4} \mathrm{He}+?}\\\ {\text { (c) }_{1}^{1} \mathrm{H}+_{5}^{11} \mathrm{B} \longrightarrow 3?} \\ {\text { (d) }_{53}^{122} \mathrm{I}\longrightarrow_{54}^{122} \mathrm{Xe}+?}\\\ {\text { (e) }_{26}^{59} \mathrm{Fe}\longrightarrow_{-1}^{0} \mathrm{e}+?}\end{array}\end{equation}

The energy from solar radiation falling on Earth is \(1.07 \times 10^{16} \mathrm{kJ} / \mathrm{min.}\) (a) How much loss of mass from the Sun occurs in one day from just the energy falling on Earth? (\mathbf{b} )If the energy released in the reaction \begin{equation}^{235} \mathrm{U}+_{0}^{1} \mathrm{n} \longrightarrow_{56}^{141} \mathrm{Ba}+_{36}^{92} \mathrm{Kr}+3_{0}^{1} \mathrm{n} \end{equation}\(\left(^{235} \mathrm{U}\right.\)nuclear mass,234.9935 amu; \(^{235} \mathrm{Ba}.\) nuclear mass, 140.8833 amu; \(^{92} \mathrm{Kr}\) nuclear mass, 91.9021 amu) is taken as typical of that occurring in a nuclear reactor, what mass of uranium-235 is required to equal 0.10\(\%\) of the solar energy that falls on Earth in 1.0 day?

Write balanced nuclear equations for the following processes: \((\mathbf{a})\) rubidium-90 undergoes beta emission; \((\mathbf{b})\) selenium- 72 undergoes electron capture; \((\mathbf{c})\) krypton-76 undergoes positron emission; \((\mathbf{d})\) radium-226 emits alpha radiation.

In 1930 the American physicist Ernest Lawrence designed the first cyclotron in Berkeley, California. In 1937 Lawrence bombarded a molybdenum target with deuterium ions, producing for the first time an element not found in nature. What was this element? Starting with molybdenum-96 as your reactant, write a nuclear equation to represent this process.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free