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Chapter 29: Problem 32

The nucleus in a \({ }_{7}^{12} \mathrm{~N}\) atom captures one of the atom's electrons, changing the nucleus to \({ }_{6}^{12} \mathrm{C}\) and emitting a neutrino. What is the total energy of the emitted neutrino? [Hint:You can use the classical expression for the kinetic energy of the ${ }_{6}^{12} \mathrm{C}$ atom and the extremely relativistic expression for the kinetic energy of the neutrino.]

Short Answer

Expert verified
Answer: The total energy of the emitted neutrino is approximately 16.834 MeV.

Step by step solution

01

Calculate the mass difference between the initial and final states.

We will use the atomic mass unit (u) to calculate the mass difference. 1u = 931.5 MeV/c^2. Mass of \({ }_{7}^{12} \mathrm{~N}\): 12.018613u Mass of \({ }_{6}^{12} \mathrm{C}\): 12.000000u Mass of electron captured: 0.000549u Initial mass = \({ }_{7}^{12} \mathrm{~N}\) = 12.018613u Final mass = \({ }_{6}^{12} \mathrm{C}\) + captured electron = 12.000000u + 0.000549u = 12.000549u Mass difference = Initial mass - Final mass = 12.018613u - 12.000549u = 0.018064u Now, convert the mass difference to energy using E=mc^2. Energy released = 0.018064u × 931.5 MeV/c^2 = 16.834 MeV
02

Find the kinetic energy of the \({ }_{6}^{12} \mathrm{C}\) atom.

Using the classical expression for kinetic energy, KE = (1/2)mv^2, where m is the mass of \({ }_{6}^{12} \mathrm{C}\) and v is its velocity. Given that the velocity is not provided, we can simplify by noting that the kinetic energy of the \({ }_{6}^{12} \mathrm{C}\) atom will be relatively small compared to the energy released in the reaction. Thus, the energy conservation equation can be approximated to match the energy released in the reaction.
03

Calculate the kinetic energy of the neutrino.

Using the energy conservation, the total energy of the emitted neutrino can be approximated as the energy released during the reaction. Total energy of neutrino ≈ Energy released in the reaction ≈ 16.834 MeV

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Most popular questions from this chapter

The activity of a sample containing radioactive \(^{108} \mathrm{Ag}\) is $6.4 \times 10^{4}\( Bq. Exactly 12 min later, the activity is \)2.0 \times 10^{3} \mathrm{Bq} .\( Calculate the half-life of \)^{108} \mathrm{Ag} .$
A radioactive sample has equal numbers of \(^{15} \mathrm{O}\) and $^{19} \mathrm{O}$ nuclei. Use the half-lives found in Appendix B to determine how long it will take before there are twice as many \(^{15} \mathrm{O}\) nuclei as \(^{19} \mathrm{O} .\) What percent of the \(^{19} \mathrm{O}\) nuclei have decayed during this time?
A water sample is found to have \(0.016 \%\) deuterium content (that is, $0.016 \%\( of the hydrogen nuclei in the water are \)^{2} \mathrm{H}$ ). If the fusion reaction \(\left(^{2} \mathrm{H}+^{2} \mathrm{H}\right)\) yields 3.65 MeV of energy on average, how much energy could you get from \(1.00 \mathrm{L}\) of the water? (There are two reactions with approximately equal probabilities; one yields \(4.03 \mathrm{MeV}\) and the other \(3.27 \mathrm{MeV} .\) ) Assume that you are able to extract and fuse \(87.0 \%\) of the deuterium in the water. Give your answer in kilowatt hours.
Show that \(c^{2}=931.494\) MeV/u. [Hint: Start with the conversion factors to SI units for MeV and atomic mass units.]
Radioactive iodine, \(^{131} \mathrm{I},\) with a half-life of $8.0252 \mathrm{d},$ is used in some forms of medical diagnostics. (a) If the initial activity of a sample is \(64.5 \mathrm{mCi},\) what is the mass of $^{131} \mathrm{I}$ in the sample? (b) What will the activity be 4.5 d later?
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