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Chapter 19: Problem 12

Describe the relative penetrating powers of alpha, beta, and gamma radiation.

Short Answer

Expert verified
In summary, the relative penetrating powers of alpha, beta, and gamma radiation are: Alpha radiation is the least penetrating, being stopped by a sheet of paper or clothing and traveling only a few centimeters in air. Beta radiation is more penetrating, requiring several millimeters of plastic, glass, or aluminum foil to block, and can travel up to a meter in air. Gamma radiation is the most penetrating, necessitating dense materials like lead or concrete to block effectively, and can travel long distances through air and penetrate deep into materials.

Step by step solution

01

1. Introduction to alpha, beta, and garma radiation

Alpha, beta, and gamma radiation are the three types of nuclear radiation that are emitted by certain radioactive isotopes during nuclear decay processes. They have different energies, mass, and charges, each of which impacts their relative penetrating powers.
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2. Alpha radiation

Alpha radiation consists of alpha particles, which are made up of two protons and two neutrons (a helium nucleus). They are the least penetrating of the three types of radiation because they are relatively large, heavy, and have a positive charge. This makes them more easily deflected by the electrons surrounding atoms in the materials they are trying to penetrate. As a result, alpha particles only travel a few centimeters through air and can be stopped by something as thin as a sheet of paper or even a layer of clothing.
03

3. Beta radiation

Beta radiation consists of beta particles, which are high-energy, high-speed electrons or positrons. They are much smaller and lighter than alpha particles and carry a negative (electron) or positive (positron) charge. This makes them less easily deflected by atomic electrons and gives them greater penetrating powers than alpha particles. Beta particles can penetrate several millimeters into solid materials, such as plastic or glass, and can travel up to about a meter through air. To completely stop beta radiation, a sheet of plastic, glass, or aluminum foil several millimeters thick is often used.
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4. Gamma radiation

Gamma radiation comprises high-energy photons, which are electromagnetic waves without any mass or charge. This makes them the most penetrating type of radiation since they cannot be deflected by atomic electrons and can pass through the spaces between the atoms in almost any material. Gamma radiation can travel long distances through air and can penetrate several centimeters into concrete, lead, or steel. It takes a significant thickness of dense material such as lead or several meters of concrete to effectively block gamma radiation.
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5. Summary of penetrating powers

In summary, the relative penetrating powers of the three types of nuclear radiation are as follows: - Alpha radiation: Least penetrating; can be stopped by a sheet of paper or a layer of clothing; travels only a few centimeters in air. - Beta radiation: More penetrating than alpha particles; can be stopped by a few millimeters of plastic, glass, or aluminum foil; travels up to a meter in air. - Gamma radiation: Most penetrating; requires dense materials such as lead or several meters of concrete to block effectively; can travel long distances through air and penetrate deep into materials.

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Most popular questions from this chapter

Why are the observed energy changes for nuclear processes so much larger than the energy changes for chemical and physical processes?

The first atomic explosion was detonated in the desert north of Alamogordo, New Mexico, on July \(16,1945 .\) What percentage of the strontium- 90\(\left(t_{1 / 2}=28.9 \text { years) originally produced }\right.\) by that explosion still remains as of July \(16,2017 ?\)

A rock contains 0.688 \(\mathrm{mg}^{206} \mathrm{Pb}\) for every 1.000 \(\mathrm{mg}^{238} \mathrm{U}\) present. Assuming that no lead was originally present, that all the \(^{206}\mathrm{P}\) formed over the years has remained in the rock, and that the number of nuclides in intermediate stages of decay between \(^{238} \mathrm{U}\) and \(^{206 \mathrm{P}} \mathrm{b}\) is negligible, calculate the age of the rock. (For $38 \mathrm{U}, t_{1 / 2}=4.5 \times 10^{9}$ years.)

The mass ratios of 40 \(\mathrm{Ar}\) to 40 \(\mathrm{K}\) also can be used to date geologic materials. Potassium-40 decays by two processes: $$_{19}^{40} \mathrm{K}+_{-1}^{0} \mathrm{e} \longrightarrow_{\mathrm{i} 8}^{40} \mathrm{Ar}(10.7 \%)$$ $$_{19}^{40} \mathrm{K} \longrightarrow_{20}^{40} \mathrm{Ca}+_{-1}^{0} \mathrm{e}(89.3 \%)$$ $$t_{1 / 2}=1.27 \times 10^{9}$$ a. Why are \(^{40}\mathrm{Ar} /^{40} \mathrm{K}\) ratios used to date materials rather than \(^{40}\mathrm{Ca} / 40 \mathrm{K}\) ratios? b. What assumptions must be made using this technique? c. A sedimentary rock has an Ar \(^{40} \mathrm{K}\) ratio of \(0.95 .\) Calculate the age of the rock. d. How will the measured age of a rock compare to the actual age if some \(^{40}\) Ar escaped from the sample?

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