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College Physics (Urone)
Found in: Page 1181
College Physics (Urone)

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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Short Answer

Suppose one food irradiation plant uses a \({}^{{\rm{137}}}{\rm{Cs}}\) source while another uses an equal activity of \({}^{{\rm{60}}}{\rm{Co}}\). Assuming equal fractions of the \({\rm{\gamma }}\) rays from the sources are absorbed, why is more time needed to get the same dose using the \({}^{{\rm{137}}}{\rm{Cs}}\) source?

Cobalt contains \({\rm{\gamma }}\) rays that are more energetic (frequency).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Define radiation

The emission or transmission of energy in the form of waves or particles across space or a material medium is known as radiation.

Step 2: Explanation

Look at food irradiation using \({}^{{\rm{137}}}{\rm{Cs}}\) and \({}^{{\rm{60}}}{\rm{Co}}\) as sources. Y rays are created from these sources when an equivalent amount of activity is utilised. Even yet, irradiating the same dosage from a \({}^{{\rm{137}}}{\rm{Cs}}\) source takes longer than from a \({}^{{\rm{60}}}{\rm{Co}}\) source.

Cobalt and caesium are two of the most well-known sources of \({\rm{\gamma }}\) rays, according to the literature. As \({\rm{\gamma }}\) rays are electromagnetic radiation, their radiation dosage at any given period is solely determined by their energy (frequency). So, because issue statement states that caesium takes longer than cobalt, we may safely assume that gamma rays from cobalt contain more energy than gamma rays from caesium.

According to the literature, the energy of \({\rm{\gamma }}\) rays from cobalt can range between \({\rm{1}}{\rm{.173}}\) and \({\rm{1}}{\rm{.332 MeV}}\), while the energy of \({\rm{\gamma }}\) rays from caesium is \({\rm{0}}{\rm{.662 MeV}}\).

In any instance, cobalt will always provide a higher radiation dose for the same level of source activity when exposed for the same period of time.

Therefore, cobalt has \({\rm{\gamma }}\) rays with a greater energy level (frequency).

Most popular questions for Physics Textbooks

The naturally occurring radioactive isotope \(^{{\rm{232}}}{\rm{Th}}\) does not make good fission fuel, because it has an even number of neutrons; however, it can be bred into a suitable fuel (much as \(^{{\rm{238}}}{\rm{U}}\) is bred into\(^{239}P\)).

(a) What are Z and N for\(^{{\rm{232}}}{\rm{Th}}\)?

(b) Write the reaction equation for neutron captured by \(^{{\rm{232}}}{\rm{Th}}\) and identify the nuclide \(^AX\)produced in\(n{ + ^{232}}Th{ \to ^A}X + \gamma \).

(c) The product nucleus \({\beta ^ - }\)decays, as does its daughter. Write the decay equations for each, and identify the final nucleus.

(d) Confirm that the final nucleus has an odd number of neutrons, making it a better fission fuel.

(e) Look up the half-life of the final nucleus to see if it lives long enough to be a useful fuel.

(a) Estimate the years that the deuterium fuel in the oceans could supply the energy needs of the world. Assume world energy consumption to be ten times that of the United States which is \(8 \times {10^9}J/y\) and that the deuterium in the oceans could be converted to energy with an efficiency of \(32\% \). You must estimate or look up the amount of water in the oceans and take the deuterium content to be \(0.015\% \) of natural hydrogen to find the mass of deuterium available. Note that approximate energy yield of deuterium is \(3.37 \times {10^{14}}J/kg\).

(b) Comment on how much time this is by any human measure. (It is not an unreasonable result, only an impressive one.)

The energy produced by the fusion of a \(1.00 - kg\) mixture of deuterium and tritium was found in Example Calculating Energy and Power from Fusion. Approximately how many kilograms would be required to supply the annual energy use in the United States?

Suppose a person swallows some radioactive material by accident. What information is needed to be able to assess possible damage?

(a) Calculate the energy released in the neutron-induced fission (similar to the spontaneous fission in Example\(32.3\)) \(n{ + ^{238}}U{ \to ^{96}}Sr{ + ^{140}}Xe + 3n\), given \(m{(^{96}}Sr) = 95.921750{\rm{ }}u\) and \(m{(^{140}}Xe) = 139.92164{\rm{ }}u\).

(b) This result is about \(6{\rm{ }}MeV\) greater than the result for spontaneous fission. Why?

(c) Confirm that the total number of nucleons and total charge are conserved in this reaction.
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