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Chapter 30: Problem 46

Which of the following reactions obey the conservation of baryon number? (a) \(\mathrm{p}+\mathrm{p} \rightarrow \mathrm{p}+\mathrm{e}^{+},\) (b) \(\mathrm{p}+\mathrm{n} \rightarrow 2 \mathrm{e}^{+}+\mathrm{e}^{-}\) (c) \(\mathrm{p} \rightarrow \mathrm{n}+\mathrm{e}^{-}+\bar{\nu}_{\mathrm{e}},(\mathrm{d}) \mathrm{p}+\overline{\mathrm{p}} \rightarrow 2 \gamma\)

Short Answer

Expert verified
Reactions (c) and (d) obey baryon number conservation.

Step by step solution

01

Identify Initial and Final Particles in Each Reaction

First, we will list the particles involved in both the initial and final states of each reaction to evaluate the baryon numbers. The baryon number is 1 for protons (\(\text{p}\)) and neutrons (\(\text{n}\)), -1 for antiparticles of baryons (like \(\overline{\text{p}}\)), and 0 for all other particles (electrons, positrons, photons, etc.).
02

Analyze Reaction (a)

In reaction (a), we have \(\text{p} + \text{p} \rightarrow \text{p} + \text{e}^+\). Initially, the baryon number is \(1 + 1 = 2\). In the final state, the baryon number is \(1 + 0 = 1\). Since the initial baryon number (2) does not equal the final baryon number (1), conservation of baryon number is not obeyed.
03

Analyze Reaction (b)

In reaction (b), \(\text{p} + \text{n} \rightarrow 2 \text{e}^+ + \text{e}^-\). The initial baryon number is \(1 + 1 = 2\). The final state consists of electrons and positrons, each with a baryon number of 0, so the final baryon number is \(0 + 0 + 0 = 0\). As the baryon number changes from 2 to 0, baryon number conservation is not obeyed.
04

Analyze Reaction (c)

In reaction (c), \(\text{p} \rightarrow \text{n} + \text{e}^- + \bar{u}_{\text{e}}\). The initial baryon number is 1 (from the proton). The final state has baryon number \(1 + 0 + 0 = 1\). Since the initial and final baryon numbers are both 1, the conservation of baryon number is obeyed.
05

Analyze Reaction (d)

In reaction (d), \(\text{p} + \overline{\text{p}} \rightarrow 2 \gamma\). The initial baryon number is \(1 + (-1) = 0\). The photon \(\gamma\) has a baryon number of 0, so the final baryon number is \(0\). Since both initial and final baryon numbers are 0, the conservation of baryon number is obeyed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Baryon Number
The baryon number is a conserved quantity in particle physics, which means it remains unchanged in isolated systems. Baryons are a type of particle, including protons and neutrons, which always have a baryon number of 1. Antibaryons, like antiprotons, have a baryon number of -1. Other particles such as electrons, positrons, and photons have a baryon number of 0.

The conservation of baryon number is an important principle. It indicates that in any reaction, the total baryon number before the reaction must equal the total baryon number after the reaction. This principle helps us understand why certain reactions occur in nature and why some do not. If a reaction fails to conserve baryon number, it is not possible under standard conditions without external input.
Proton
Protons are positively charged particles found in the nucleus of an atom. They are stable baryons with a baryon number of 1. This positive charge is equal and opposite to that of an electron's negative charge.

Protons are fundamental to the structure of atoms, playing a crucial role in determining the element's identity and properties. They are involved in various atomic and subatomic processes, including nuclear reactions. Due to their positive charge, they are repelled from each other in the nucleus but are kept together by the strong nuclear force.
  • Positive Charge: Equal to +1 in elementary charge.
  • Baryon Number: Always 1, indicating its presence as a baryon.
  • Stability: Generally stable unless influenced by external forces.
Neutron
Neutrons, like protons, are baryons with a baryon number of 1. However, unlike protons, neutrons do not carry any electric charge, making them neutrally charged particles. They reside in the nucleus of an atom alongside protons, contributing to the atomic mass.

Neutrons are essential for the stability of most atomic nuclei. A balanced number of neutrons and protons contribute to a stable nucleus. Outside the nucleus, free neutrons are unstable and decay into a proton, an electron, and an antineutrino through beta decay. This process highlights conservation laws, including baryon number conservation.
  • Charge: Neutral, with no net charge.
  • Baryon Number: 1, indicating its identity as a baryon.
  • Role: Stability in atomic nuclei, crucial for isotopes.
Antimatter
Antimatter is composed of antiparticles, which mirror regular particles but with opposite properties. This includes antiprotons, which have a negative baryon number of -1, unlike their positive counterparts, protons.

Antimatter is fascinating because when it meets regular matter, they annihilate each other, releasing energy in the form of photons (gamma rays). This kind of matter-antimatter interaction obeys conservation laws, such as the conservation of baryon number, energy, and momentum.
  • Antiproton: Carries a negative baryon number (-1), opposite to a proton.
  • Annihilation: Interaction with matter leads to complete energy conversion.
  • Conservation Laws: Must be satisfied in antimatter interactions, ensuring total baryon number remains constant.

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Most popular questions from this chapter

Calcium- 47 is a \(\beta^{-}\) emitter with a half-life of 4.5 days. If a bone sample contains \(2.24 \mathrm{~g}\) of this isotope, at what rate will it decay?

A photon with a wavelength of \(3.50 \times 10^{-13} \mathrm{~m}\) strikes a deuteron, splitting it into a proton and a neutron. (a) Calculate the kinetic energy released in this interaction. (b) Assuming the two particles share the energy equally, and taking their masses to be \(1.00 \mathrm{u},\) calculate their speeds after the photodisintegration.

(a) Using the empirical formula for the radius of a nucleus, show that the volume of a nucleus is directly proportional to its nucleon number \(A\). (b) Give a reasonable argument concluding that the mass \(m\) of a nucleus of nucleon number \(A\) is approximately \(m=m_{\mathrm{p}} A,\) where \(m_{\mathrm{p}}\) is the mass of a proton. (c) Use the results of parts (a) and (b) to show that all nuclei should have about the same density. Then calculate this density in \(\mathrm{kg} / \mathrm{m}^{3}\), and compare it with the density of lead (which is \(\left.11.4 \mathrm{~g} / \mathrm{cm}^{3}\right)\) and a neutron star (about \(10^{17} \mathrm{~kg} / \mathrm{m}^{3}\) ).

\(\mathrm{A}^{60} \mathrm{Co}\) source with activity \(15.0 \mathrm{Ci}\) is embedded in a tumor that has a mass of \(0.500 \mathrm{~kg}\). The Co source emits gamma-ray photons with average energy of \(1.25 \mathrm{MeV}\). Half the photons are absorbed in the tumor, and half escape. (a) What energy is delivered to the tumor per second? (b) What absorbed dose (in rad) is delivered per second? (c) What equivalent dose (in rem) is delivered per second if the \(\mathrm{RBE}\) for these gamma rays is \(0.70 ?\) (d) What exposure time is required for an equivalent dose of 200 rem?

(a) If a chest X-ray delivers \(0.25 \mathrm{mSv}\) to \(5.0 \mathrm{~kg}\) of tissue, how many total joules of energy does this tissue receive? (b) Natural radiation and cosmic rays deliver about \(0.10 \mathrm{mSv}\) per year at sea level. Assuming an \(\mathrm{RBE}\) of \(1,\) how many rem and rads is this dose, and how many joules of energy does a 75-kg person receive in a year? (c) How many chest X-rays like the one in part (a) would it take to deliver the same total amount of energy to a \(75-\mathrm{kg}\) person as she receives from natural radiation in a year at sea level, as described in part (b)?
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