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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
Chapter 42, Problem 66a

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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Understand the problem: A free neutron decays into a proton and an electron (n → p⁺ + e⁻). We are tasked with finding the total kinetic energy of the proton and electron after the decay, using relativistic energy and momentum conservation principles.
Step 1: Write the energy conservation equation. The total energy before the decay is the rest energy of the neutron, which is given by Eₙ = mₙc², where mₙ is the mass of the neutron and c is the speed of light. After the decay, the total energy is the sum of the rest energies and kinetic energies of the proton and electron: Eₙ = Eₚ + Eₑ, where Eₚ = mₚc² + Kₚ and Eₑ = mₑc² + Kₑ.
Step 2: Write the momentum conservation equation. Since the neutron is initially at rest, its momentum is zero. After the decay, the momentum of the proton and electron must cancel each other out: pₚ = -pₑ. Use the relativistic momentum-energy relation: E² = (pc)² + (mc²)² for both the proton and electron.
Step 3: Solve for the kinetic energies. Subtract the rest energy of the neutron from the total energy to find the combined kinetic energy of the proton and electron: K_total = Eₙ - (mₚc² + mₑc²). Use the known masses of the neutron, proton, and electron to calculate this difference.
Step 4: Distribute the total kinetic energy between the proton and electron. Since the proton is much more massive than the electron, it will have a smaller velocity and thus a smaller kinetic energy. Use the momentum conservation equation (pₚ = -pₑ) and the relativistic energy-momentum relation to determine the individual kinetic energies of the proton and electron.

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Key Concepts

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

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 in unstable nuclei to achieve a more stable configuration. The emitted electron carries away energy, and the proton remains in the nucleus, contributing to the overall charge and stability of the atom.
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Conservation of Energy and Momentum

In any physical process, the total energy and momentum before and after the event must remain constant. In the context of beta decay, the initial energy of the neutron at rest is converted into the kinetic energy of the emitted electron and the proton. This principle allows us to calculate the kinetic energies of the decay products by applying the conservation laws.
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Relativistic Energy

Relativistic energy takes into account the effects of special relativity, particularly at high velocities close to the speed of light. In beta decay, the kinetic energy of the emitted particles must be calculated using the relativistic energy-momentum relation, which combines rest mass energy and kinetic energy. This ensures accurate predictions of the particles' behavior and energy distribution after the decay.
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Related Practice
Textbook Question

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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Textbook Question

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.

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Textbook Question

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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Textbook Question

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.

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Textbook Question

A sample contains radioactive atoms of two types, A and B. Initially there are five times as many A atoms as there are B atoms. Two hours later, the numbers of the two atoms are equal. The half-life of A is 0.50 hour. What is the half-life of B?

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