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Ch 42: Nuclear Physics
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 42: Nuclear PhysicsProblem 33
Chapter 42, Problem 33

What is the total energy (in MeV) released in the beta-minus decay of ³H?

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Identify the beta-minus decay process: In beta-minus decay, a neutron in the nucleus converts into a proton, emitting an electron (beta particle) and an antineutrino. For ³H (tritium), the reaction is: ³H → ³He + e⁻ + ν̅ₑ.
Determine the mass difference between the parent nucleus (³H) and the daughter nucleus (³He). The energy released in the decay is related to this mass difference via Einstein's equation, E = Δm * c², where Δm is the mass difference and c is the speed of light.
Look up the atomic masses of ³H and ³He. Note that the mass of the emitted electron is already included in the atomic mass of ³H, so you do not need to account for it separately. Use the relation Δm = m(³H) - m(³He).
Convert the mass difference Δm from atomic mass units (u) to energy using the conversion factor 1 u = 931.5 MeV/c². The energy released is then E = Δm * 931.5 MeV.
The total energy released in the beta-minus decay of ³H is the value of E calculated in the previous step. This energy is shared between the kinetic energy of the emitted beta particle, the antineutrino, and the recoil of the daughter nucleus.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Beta-minus Decay

Beta-minus decay is a type of radioactive decay in which a neutron in an atomic nucleus is transformed into a proton, emitting an electron (beta particle) and an antineutrino. This process increases the atomic number of the element by one while keeping the mass number the same, resulting in the formation of a new element.
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Energy Release in Nuclear Reactions

In nuclear reactions, energy is released due to the conversion of mass into energy, as described by Einstein's equation E=mc². The total energy released during beta decay can be calculated by considering the mass difference between the initial and final states of the nucleus, which is converted into energy during the decay process.
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Units of Energy (MeV)

The mega-electronvolt (MeV) is a unit of energy commonly used in nuclear and particle physics. It is defined as one million electronvolts, where one electronvolt is the energy gained by an electron when accelerated through a potential difference of one volt. MeV is particularly useful for expressing the energy changes in nuclear reactions, including decay processes.
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Related Practice
Textbook Question

Particle accelerators fire protons at target nuclei so that investigators can study the nuclear reactions that occur. In one experiment, the proton needs to have 20 MeV of kinetic energy as it impacts a 207Pb nucleus. With what initial kinetic energy (in MeV) must the proton be fired toward the lead target? Assume the nucleus stays at rest. Hint: The proton is not a point particle.

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Textbook Question

A 50 kg laboratory worker is exposed to 20 mJ of beta radiation. What is the dose equivalent in mrem?

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Textbook Question

The doctors planning a radiation therapy treatment have determined that a 100 g tumor needs to receive 0.20 J of gamma radiation. What is the dose in grays?

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Textbook Question

An unstable nucleus undergoes alpha decay with the release of 5.52 MeV of energy. The combined mass of the parent and daughter nuclei is 452 u. What was the mass number of the parent nucleus?

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Textbook Question

For those that are not stable, identify both the decay mode and the daughter nucleus.

Textbook Question

Identify the unknown isotope XX in the following decays. X→40K+e++ν\(\text{X}\]\rightarrow\)^{40}\(\text{K}\)+\(\text{e}\)^{+}+\(\nu\)

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