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Chapter 21: Q42 E (page 1206)

A\({}_{\rm{4}}^{\rm{7}}{\rm{Be}}\) atom (mass\({\rm{ = 7}}{\rm{.0169 amu}}\)) decays into a\({}_{\rm{3}}^{\rm{7}}{\rm{Li}}\) atom (mass\({\rm{ = 7}}{\rm{.0160 amu}}\)) by electron capture. How much energy (in millions of electron volts, \({\rm{MeV}}\)) is produced by this reaction?

Short Answer

Expert verified

The energy produced by the reaction \(_{\rm{4}}^{\rm{7}}{\rm{Be + }}_{{\rm{ - 1}}}^{\rm{0}}{\rm{e}} \to _{\rm{3}}^{\rm{7}}{\rm{Li}}\)is \({\rm{1}}{\rm{.297 MeV}}\).

Step by step solution

01

Concept Introduction

Radioactive decay is known as the process of an unstable atomic nucleus losing energy through radiation. The term "radioactive" refers to a substance that contains unstable nuclei. Alpha decay (\({\rm{\alpha - }}\)decay), beta decay (\({\rm{\beta - }}\)decay), and gamma decay (\({\rm{\gamma - }}\)decay) are three of the most prevalent types of decay, all of which entail the emission of one or more particles.

02

Finding the formula

The reaction is

\(_{\rm{4}}^{\rm{7}}{\rm{Be + }}_{{\rm{ - 1}}}^{\rm{0}}{\rm{e}} \to _{\rm{3}}^{\rm{7}}{\rm{Li}}\)

The energy can be obtained using the formula:

\(\begin{array}{c}{\rm{E = m}} \cdot {{\rm{c}}^{\rm{2}}}\\{\rm{m = m( reactants ) - m( products )}}\\{\rm{m = (7}}{\rm{.0169 amu + 0}}{\rm{.0005 amu) - 7}}{\rm{.0160 amu}}\\{\rm{m = 1}}{\rm{.4}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{amu}}\end{array}\)

Converting\({\rm{amu}}\)in to\({\rm{kg}}\)–

\(\begin{array}{c}{\rm{1 amu = 1}}{\rm{.6605}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 27}}}}{\rm{\;kg}}\\{\rm{m = 1}}{\rm{.4}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{amu \times 1}}{\rm{.6605}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 27}}}}{\rm{\;kg}} \cdot {\rm{am}}{{\rm{u}}^{{\rm{ - 1}}}}\\{\rm{m = 2}}{\rm{.325}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 30}}}}{\rm{\;kg}}\end{array}\)

03

Calculation for Energy

Substituting the values and calculating:

\(\begin{array}{c}{\rm{E = m}} \cdot {{\rm{c}}^{\rm{2}}}\\{\rm{E = 2}}{\rm{.325}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 30}}}}{\rm{\;kg \times 2}}{\rm{.99}} \cdot {\rm{1}}{{\rm{0}}^{\rm{8}}}{\rm{\;m}}{{\rm{s}}^{{\rm{ - }}{{\rm{1}}^{\rm{2}}}}}\\{\rm{E = 2}}{\rm{.078}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 13}}}}{\rm{\;J}}\end{array}\)

Converting energy from\({\rm{J}}\)to\({\rm{MeV}}\):

\(\begin{array}{c}{\rm{1 MeV = 1}}{\rm{.602}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 13}}}}{\rm{\;J}}\\{\rm{E = 2}}{\rm{.078}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 1}}}}{\rm{13\;J \times }}\frac{{{\rm{1 MeV}}}}{{{\rm{1}}{\rm{.602}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 13}}}}{\rm{\;J}}}}\\{\rm{E = 1}}{\rm{.297 MeV}}\end{array}\)

Therefore, the value for energy is obtained as\({\rm{E = 1}}{\rm{.297 MeV}}\).

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Most popular questions from this chapter

Iodine that enters the body is stored in the thyroid gland from which it is released to control growth and metabolism. The thyroid can be imaged if iodine\({\rm{ - 131}}\)is injected into the body. In larger doses, I\({\rm{ - 133}}\)is also used as a means of treating cancer of the thyroid. I\({\rm{ - 131}}\)has a half-life of\({\rm{8}}{\rm{.70}}\)days and decays by\({\rm{\beta - }}\)emission. (a) Write an equation for the decay. (b) How long will it take for\({\rm{95}}{\rm{.0\% }}\)of a dose of I\({\rm{ - 131}}\)to decay?

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Question: Why is electron capture accompanied by the emission of an\({\rm{X - }}\)ray?

Write a balanced equation for each of the following nuclear reactions:

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(d) strontium\({\rm{ - 90}}\)decays into yttrium\({\rm{ - 90}}\).
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