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.
Ch 42: Nuclear Physics
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Capitolo 42, Problema 33
What is the total energy (in MeV) released in the beta-minus decay of ³H?
Guida verificata passo dopo passo1
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.

Risposta video verificata per un problema simile:
Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Concetti chiave
Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.
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.
Video consigliato:
Percorso guidato
Amplitude Decay in an LRC Circuit
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.
Video consigliato:
Percorso guidato
Energy Released by Flashbulb
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.
Video consigliato:
Percorso guidato
Unit Conversions
Pratica correlata
Domanda del libro di testo
53
views
Domanda del libro di testo
A 50 kg laboratory worker is exposed to 20 mJ of beta radiation. What is the dose equivalent in mrem?
57
views
Domanda del libro di testo
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?
44
views
Domanda del libro di testo
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?
65
views
Domanda del libro di testo
For those that are not stable, identify both the decay mode and the daughter nucleus.
71
views
Domanda del libro di testo
Identify the unknown isotope in the following decays.
66
views
