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Ch 42: Nuclear Physics
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796당신이 사용하는 게 아니라요?교과서 변경
42장, 문제 66b

It might seem strange that in beta decay the positive proton, which is repelled by the positive nucleus, remains in the nucleus while the negative electron, which is attracted to the nucleus, is ejected. To understand beta decay, let's analyze the decay of a free neutron that is at rest in the laboratory. We'll ignore the antineutrino and consider the decay n → p⁺ + e⁻. The analysis requires the use of relativistic energy and momentum, from Chapter 36. Write the equation that expresses the conservation of relativistic energy for this decay. Your equation will be in terms of the three masses mn, mp and me and the relativistic factors yp and ye.

검증된 단계별 안내
1
Step 1: Begin by recalling the principle of conservation of energy. In beta decay, the total energy before the decay (energy of the neutron) must equal the total energy after the decay (energy of the proton and electron). This includes both rest energy and kinetic energy.
Step 2: Write the total energy of the neutron before the decay. Since the neutron is at rest, its total energy is simply its rest energy, which is given by Eₙ = mₙc², where mₙ is the mass of the neutron and c is the speed of light.
Step 3: Write the total energy of the proton and electron after the decay. For the proton, the total energy is Eₚ = γₚmₚc², where γₚ is the relativistic factor for the proton and mₚ is its rest mass. For the electron, the total energy is Eₑ = γₑmₑc², where γₑ is the relativistic factor for the electron and mₑ is its rest mass.
Step 4: Apply the conservation of energy principle. Equate the total energy before and after the decay: mₙc² = γₚmₚc² + γₑmₑc². This equation expresses the conservation of relativistic energy for the decay process.
Step 5: Simplify the equation if needed. Note that γₚ and γₑ are the relativistic factors for the proton and electron, respectively, and are defined as γ = 1 / √(1 - v²/c²), where v is the velocity of the particle. This equation can now be used to analyze the energy distribution in the decay process.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
4m
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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Beta Decay

Beta decay is a type of radioactive decay in which a neutron is transformed into a proton, emitting an electron (beta particle) and an antineutrino. This process occurs when a neutron in the nucleus undergoes a transformation, resulting in a change in the atomic number of the element. Understanding beta decay is crucial for analyzing nuclear reactions and the stability of atomic nuclei.
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Conservation of Energy and Momentum

In physics, the conservation of energy states that the total energy in a closed system remains constant over time. Similarly, the conservation of momentum states that the total momentum of a system remains unchanged if no external forces act on it. These principles are fundamental in analyzing particle decays, as they allow us to relate the initial and final states of the particles involved in the decay process.
추천 영상:
가이드 코스
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Conservation Of Momentum

Relativistic Factors

Relativistic factors, often denoted by the symbol gamma (γ), account for the effects of relativity at high velocities. They are defined as γ = 1 / √(1 - v²/c²), where v is the velocity of the particle and c is the speed of light. In the context of beta decay, these factors are essential for accurately calculating the energy and momentum of the emitted particles, as their speeds can approach the speed of light.
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교과서 질문

All the very heavy atoms found in the earth were created long ago by nuclear fusion reactions in a supernova, an exploding star. The debris spewed out by the supernova later coalesced into the gases from which the sun and the planets of our solar system were formed. Nuclear physics suggests that the uranium isotopes ²³⁵U and ²³⁸U should have been created in roughly equal numbers. Today, 99.28% of uranium is ²³⁸U and only 0.72% is ²³⁵U. How long ago did the supernova occur?

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There is evidence that low-energy x rays have an RBE slightly greater than 1. Suppose that 10 keV photons with an RBE of 1.2 are used to make a chest x ray. A 60 kg person receives a 0.30 mSv dose from a chest x ray that exposes 25% of the patient's body. How many x ray photons are absorbed in the patient's body?

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교과서 질문

It might seem strange that in beta decay the positive proton, which is repelled by the positive nucleus, remains in the nucleus while the negative electron, which is attracted to the nucleus, is ejected. To understand beta decay, let's analyze the decay of a free neutron that is at rest in the laboratory. We'll ignore the antineutrino and consider the decay n → p⁺ + e⁻. The analysis requires the use of relativistic energy and momentum, from Chapter 36. What is the total kinetic energy, in MeV, of the proton and electron?

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교과서 질문

It might seem strange that in beta decay the positive proton, which is repelled by the positive nucleus, remains in the nucleus while the negative electron, which is attracted to the nucleus, is ejected. To understand beta decay, let's analyze the decay of a free neutron that is at rest in the laboratory. We'll ignore the antineutrino and consider the decay n → p⁺ + e⁻. The analysis requires the use of relativistic energy and momentum, from Chapter 36. Write the equation that expresses the conservation of relativistic momentum for this decay. Let v represent speed, rather than velocity, then write any minus signs explicitly.

119
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