#set document(title: "11.2 Particle Conservation Laws", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 11.2#h(0.6em)Particle Conservation Laws Conservation laws are critical to an understanding of particle physics. Strong evidence exists that energy, momentum, and angular momentum are all conserved in all particle interactions. The annihilation of an electron and positron at rest, for example, cannot produce just one photon because this violates the conservation of linear momentum. As discussed in Relativity, the special theory of relativity modifies definitions of momentum, energy, and other familiar quantities. In particular, the relativistic momentum of a particle differs from its classical momentum by a factor #math.equation(block: false, alt: "γ equals 1 / the square root of 1 minus open parenthesis v / c close parenthesis squared")[$γ = 1 "/" sqrt(1 − attach(( v "/" c ), t: 2))$] that varies from 1 to #math.equation(block: false, alt: "∞ ,")[$∞ ,$] depending on the speed of the particle. In previous chapters, we encountered other conservation laws as well. For example, charge is conserved in all electrostatic phenomena. Charge lost in one place is gained in another because charge is carried by particles. No known physical processes violate charge conservation. In the next section, we describe three less-familiar conservation laws: baryon number, lepton number, and strangeness. These are by no means the only conservation laws in particle physics. === Baryon Number Conservation No conservation law considered thus far prevents a neutron from decaying via a reaction such as #math.equation(block: true, alt: "n → e to the power plus plus e to the power − .")[$"n" → "e"^(+) + "e"^("−") .$] This process conserves charge, energy, and momentum. However, it does not occur because it violates the law of baryon number conservation. This law requires that the total baryon number of a reaction is the same before and after the reaction occurs. To determine the total baryon number, every elementary particle is assigned a #strong[baryon number] #emph[B]. The baryon number has the value #math.equation(block: false, alt: "B equals plus 1")[$B = + 1$] for baryons, –1 for antibaryons, and 0 for all other particles. Returning to the above case (the decay of the neutron into an electron-positron pair), the neutron has a value #math.equation(block: false, alt: "B equals plus 1 ,")[$B = + 1 ,$] whereas the electron and the positron each has a value of 0. Thus, the decay does not occur because the total baryon number changes from 1 to 0. However, the proton-antiproton collision process #math.equation(block: true, alt: "p plus p − → p plus p plus p − plus p − ,")[$"p" + limits("p")^("−") → "p" + "p" + limits("p")^("−") + limits("p")^("−") ,$] #emph[does] satisfy the law of conservation of baryon number because the baryon number is zero before and after the interaction. The baryon number for several common particles is given in . #figure(table( columns: 7, align: left, inset: 6pt, table.header([Particle name], [Symbol], [Lepton number #math.equation(block: false, alt: "open parenthesis L sub e close parenthesis")[$( L_(e) )$]], [Lepton number #math.equation(block: false, alt: "open parenthesis L sub μ close parenthesis")[$( L_(μ) )$]], [Lepton number #math.equation(block: false, alt: "open parenthesis L sub τ close parenthesis")[$( L_(τ) )$]], [Baryon number (#emph[B])], [Strange-ness number]), [Electron], [#math.equation(block: false, alt: "e to the power −")[$"e"^("−")$]], [1], [0], [0], [0], [0], [Electron neutrino], [#math.equation(block: false, alt: "ν sub e")[$ν_(e)$]], [1], [0], [0], [0], [0], [Muon], [#math.equation(block: false, alt: "μ to the power −")[$μ^("−")$]], [0], [1], [0], [0], [0], [Muon neutrino], [#math.equation(block: false, alt: "ν sub μ")[$ν_(μ)$]], [0], [1], [0], [0], [0], [Tau], [#math.equation(block: false, alt: "τ to the power −")[$τ^("−")$]], [0], [0], [1], [0], [0], [Tau neutrino], [#math.equation(block: false, alt: "ν sub τ")[$ν_(τ)$]], [0], [0], [1], [0], [0], [Pion], [#math.equation(block: false, alt: "π to the power plus")[$π^(+)$]], [0], [0], [0], [0], [0], [Positive kaon], [#math.equation(block: false, alt: "K to the power plus")[$"K"^(+)$]], [0], [0], [0], [0], [1], [Negative kaon], [#math.equation(block: false, alt: "K to the power −")[$"K"^("−")$]], [0], [0], [0], [0], [–1], [Proton], [p], [0], [0], [0], [1], [0], [Neutron], [n], [0], [0], [0], [1], [0], [Lambda zero], [#math.equation(block: false, alt: "Λ to the power 0")[$"Λ"^(0)$]], [0], [0], [0], [1], [–1], [Positive sigma], [#math.equation(block: false, alt: "Σ to the power plus")[$"Σ"^(+)$]], [0], [0], [0], [1], [–1], [Negative sigma], [#math.equation(block: false, alt: "Σ to the power −")[$"Σ"^("−")$]], [0], [0], [0], [1], [–1], [Xi zero], [#math.equation(block: false, alt: "Ξ to the power 0")[$"Ξ"^(0)$]], [0], [0], [0], [1], [–2], [Negative xi], [#math.equation(block: false, alt: "Ξ to the power −")[$"Ξ"^("−")$]], [0], [0], [0], [1], [–2], [Omega], [#math.equation(block: false, alt: "Ω to the power −")[$"Ω"^("−")$]], [0], [0], [0], [1], [–3], )) #examplebox("Example 1")[Baryon Number Conservation][ Based on the law of conservation of baryon number, which of the following reactions can occur? #math.equation(block: true, alt: "open parenthesis a close parenthesis π to the power − plus p → π to the power 0 plus n plus π to the power − plus π to the power plus; open parenthesis b close parenthesis p plus p − → p plus p plus p −")[$( "a" ) #h(0.2em) π^("−") + "p" → π^(0) + "n" + π^("−") + π^(+) \ ( "b" ) #h(0.2em) "p" + limits("p")^("−") → "p" + "p" + limits("p")^("−")$] Strategy Determine the total baryon number for the reactants and products, and require that this value does not change in the reaction. #strong[Solution] For reaction (a), the net baryon number of the two reactants is #math.equation(block: false, alt: "0 plus 1 equals 1")[$0 + 1 = 1$] and the net baryon number of the four products is #math.equation(block: false, alt: "0 plus 1 plus 0 plus 0 equals 1 .")[$0 + 1 + 0 + 0 = 1 .$] Since the net baryon numbers of the reactants and products are equal, this reaction is allowed on the basis of the baryon number conservation law. For reaction (b), the net baryon number of the reactants is #math.equation(block: false, alt: "1 plus open parenthesis −1 close parenthesis equals 0")[$1 + ( −1 ) = 0$] and the net baryon number of the proposed products is #math.equation(block: false, alt: "1 plus 1 plus open parenthesis −1 close parenthesis equals 1 .")[$1 + 1 + ( −1 ) = 1 .$] Since the net baryon numbers of the reactants and proposed products are not equal, this reaction cannot occur. Significance Baryon number is conserved in the first reaction, but not in the second. Baryon number conservation constrains what reactions can and cannot occur in nature. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ What is the baryon number of a hydrogen nucleus? #solutionbox[ 1 ] ] === Lepton Number Conservation Lepton number conservation states that the sum of lepton numbers before and after the interaction must be the same. There are three different #strong[lepton number]#strong[s]: the electron-lepton number #math.equation(block: false, alt: "L sub e ,")[$L_(e) ,$] the muon-lepton number #math.equation(block: false, alt: "L sub μ ,")[$L_(μ) ,$] and the tau-lepton number #math.equation(block: false, alt: "L sub τ .")[$L_(τ) .$] In any interaction, each of these quantities must be conserved #emph[separately]. For electrons and electron neutrinos, #math.equation(block: false, alt: "L sub e equals 1 ;")[$L_(e) = 1 ;$] for their antiparticles, #math.equation(block: false, alt: "L sub e equals −1 ;")[$L_(e) = −1 ;$] all other particles have #math.equation(block: false, alt: "L sub e equals 0 .")[$L_(e) = 0 .$] Similarly, #math.equation(block: false, alt: "L sub μ equals 1")[$L_(μ) = 1$] for muons and muon neutrinos, #math.equation(block: false, alt: "L sub μ equals −1")[$L_(μ) = −1$] for their antiparticles, and #math.equation(block: false, alt: "L sub μ equals 0")[$L_(μ) = 0$] for all other particles. Finally, #math.equation(block: false, alt: "L sub τ equals 1 , −1")[$L_(τ) = 1 , −1$], or 0, depending on whether we have a tau or tau neutrino, their antiparticles, or any other particle, respectively. Lepton number conservation guarantees that the number of electrons and positrons in the universe stays relatively constant. (#emph[Note:] The total lepton number is, as far as we know, conserved in nature. However, observations have shown variations of family lepton number (for example, #math.equation(block: false, alt: "L sub e close parenthesis")[$L_(e) \)$] in a phenomenon called #emph[neutrino oscillations.]) To illustrate the lepton number conservation law, consider the following known two-step decay process: #math.equation(block: true, alt: "π to the power plus → μ to the power plus plus ν sub μ; μ to the power plus → e to the power plus plus ν sub e plus ν — sub μ .")[$π^(+) → μ^(+) + ν_(μ) \ μ^(+) → "e"^(+) + ν_(e) + limits(ν)^("—")_(μ) .$] In the first decay, all of the lepton numbers for #math.equation(block: false, alt: "π to the power plus")[$π^(+)$] are 0. For the products of this decay, #math.equation(block: false, alt: "L sub μ equals −1")[$L_(μ) = −1$] for #math.equation(block: false, alt: "μ to the power plus")[$μ^(+)$] and #math.equation(block: false, alt: "L sub μ equals 1")[$L_(μ) = 1$] for #math.equation(block: false, alt: "ν sub μ .")[$ν_(μ) .$] Therefore, muon-lepton number is conserved. Neither electrons nor tau are involved in this decay, so #math.equation(block: false, alt: "L sub e equals 0")[$L_("e") = 0$] and #math.equation(block: false, alt: "L sub τ equals 0")[$L_(τ) = 0$] for the initial particle and all decay products. Thus, electron-lepton and tau-lepton numbers are also conserved. In the second decay, #math.equation(block: false, alt: "μ to the power plus")[$μ^(+)$] has a muon-lepton number #math.equation(block: false, alt: "L sub μ equals −1 ,")[$L_(μ) = −1 ,$] whereas the net muon-lepton number of the decay products is #math.equation(block: false, alt: "0 plus 0 plus open parenthesis −1 close parenthesis equals minus 1")[$0 + 0 + ( −1 ) = − 1$]. Thus, the muon-lepton number is conserved. Electron-lepton number is also conserved, as #math.equation(block: false, alt: "L sub e equals 0")[$L_("e") = 0$] for #math.equation(block: false, alt: "μ to the power plus")[$μ^(+)$], whereas the net electron-lepton number of the decay products is #math.equation(block: false, alt: "open parenthesis −1 close parenthesis plus 1 plus 0 equals 0")[$( −1 ) + 1 + 0 = 0$]. Finally, since no taus or tau-neutrinos are involved in this decay, the tau-lepton number is also conserved. #examplebox("Example 2")[Lepton Number Conservation][ Based on the law of conservation of lepton number, which of the following decays can occur? #math.equation(block: true, alt: "open parenthesis a close parenthesis n → p plus e to the power − plus ν − sub e; open parenthesis b close parenthesis π to the power − → μ to the power − plus ν sub μ plus ν − sub μ")[$( "a" ) #h(0.2em) "n" → "p" + "e"^("−") + limits(ν)^("−")_(e) \ ( "b" ) #h(0.2em) π^("−") → μ^("−") + ν_(μ) + limits(ν)^("−")_(μ)$] Strategy Determine the total lepton number for the reactants and products, and require that this value does not change in the reaction. Solution For decay (a), the electron-lepton number of the neutron is 0, and the net electron-lepton number of the decay products is #math.equation(block: false, alt: "0 plus 1 plus open parenthesis −1 close parenthesis equals 0")[$0 + 1 + ( −1 ) = 0$]. Since the net electron-lepton numbers before and after the decay are the same, the decay is possible on the basis of the law of conservation of electron-lepton number. Also, since there are no muons or taus involved in this decay, the muon-lepton and tauon-lepton numbers are conserved. For decay (b), the muon-lepton number of the #math.equation(block: false, alt: "π to the power −")[$π^("−")$] is 0, and the net muon-lepton number of the proposed decay products is #math.equation(block: false, alt: "1 plus 1 plus open parenthesis −1 close parenthesis equals 1")[$1 + 1 + ( −1 ) = 1$]. Thus, on the basis of the law of conservation of muon-lepton number, this decay cannot occur. Significance Lepton number is conserved in the first reaction, but not in the second. Lepton number conservation constrains what reactions can and cannot occur in nature. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ What is the lepton number of an electron-positron pair? #solutionbox[ 0 ] ] === Strangeness Conservation In the late 1940s and early 1950s, cosmic-ray experiments revealed the existence of particles that had never been observed on Earth. These particles were produced in collisions of pions with protons or neutrons in the atmosphere. Their production and decay were unusual. They were produced in the strong nuclear interactions of pions and nucleons, and were therefore inferred to be hadrons; however, their decay was mediated by the much more slowly acting weak nuclear interaction. Their lifetimes were on the order of #math.equation(block: false, alt: "10 to the power −10")[$10^(−10)$] to #math.equation(block: false, alt: "10 to the power −8 s ,")[$10^(−8) #h(0.2em) "s" ,$] whereas a typical lifetime for a particle that decays via the strong nuclear reaction is #math.equation(block: false, alt: "10 to the power −23 s .")[$10^(−23) #h(0.2em) "s" .$] These particles were also unusual because they were always produced in pairs in the pion-nucleon collisions. For these reasons, these newly discovered particles were described as #emph[strange]. The production and subsequent decay of a pair of strange particles is illustrated in and follows the reaction #math.equation(block: true, alt: "π to the power − plus p → Λ to the power 0 plus K to the power 0 .")[$π^("−") + "p" → "Λ"^(0) + "K"^(0) .$] The lambda particle then decays through the weak nuclear interaction according to #math.equation(block: true, alt: "Λ to the power 0 → π to the power − plus p ,")[$"Λ"^(0) → π^("−") + "p" ,$] and the kaon decays via the weak interaction #math.equation(block: true, alt: "K to the power 0 → π to the power plus plus π to the power − .")[$"K"^(0) → π^(+) + π^("−") .$] #figure(figph[Figure a shows a photograph with a black background and a white pattern of swirls and lines on it. There is a bright white spot on the top left. Figure b shows the same pattern as a line drawing. It is labeled in various places with names of particles.], alt: "Figure a shows a photograph with a black background and a white pattern of swirls and lines on it. There is a bright white spot on the top left. Figure b shows the same pattern as a line drawing. It is labeled in various places with names of particles.", caption: [The interactions of hadrons. (a) Bubble chamber photograph; (b) sketch that represents the photograph.]) To rationalize the behavior of these strange particles, particle physicists invented a particle property conserved in strong interactions but not in weak interactions. This property is called #strong[strangeness] and, as the name suggests, is associated with the presence of a strange quark. The strangeness of a particle is equal to the number of strange quarks of the particle. Strangeness conservation requires the total strangeness of a reaction or decay (summing the strangeness of all the particles) is the same before and after the interaction. Strangeness conservation is not absolute: It is conserved in strong interactions and electromagnetic interactions but not in weak interactions. The strangeness number for several common particles is given in . #examplebox("Example 3")[Strangeness Conservation][ (a) Based on the conservation of strangeness, can the following reaction occur? #math.equation(block: true, alt: "π to the power − plus p → K to the power plus plus K to the power − plus n .")[$π^("−") + "p" → "K"^(+) + "K"^("−") + "n" .$] (b) The following decay is mediated by the weak nuclear force: #math.equation(block: true, alt: "K to the power plus → π to the power plus plus π to the power 0 .")[$"K"^(+) → π^(+) + π^(0) .$] Does the decay conserve strangeness? If not, can the decay occur? Strategy Determine the strangeness of the reactants and products and require that this value does not change in the reaction. Solution + The net strangeness of the reactants is #math.equation(block: false, alt: "0 plus 0 equals 0 ,")[$0 + 0 = 0 ,$] and the net strangeness of the products is #math.equation(block: false, alt: "1 plus open parenthesis −1 close parenthesis plus 0 equals 0 .")[$1 + ( −1 ) + 0 = 0 .$] Thus, the strong nuclear interaction between a pion and a proton is not forbidden by the law of conservation of strangeness. Notice that baryon number is also conserved in the reaction. + The net strangeness before and after this decay is 1 and 0, so the decay does not conserve strangeness. However, the decay may still be possible, because the law of conservation of strangeness does not apply to weak decays. Significance Strangeness is conserved in the first reaction, but not in the second. Strangeness conservation constrains what reactions can and cannot occur in nature. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ What is the strangeness number of a muon? #solutionbox[ 0 ] ] === Summary - Elementary particle interactions are governed by particle conservation laws, which can be used to determine what particle reactions and decays are possible (or forbidden). - The baryon number conservation law and the three lepton number conversation law are valid for all physical processes. However, conservation of strangeness is valid only for strong nuclear interactions and electromagnetic interactions. === Conceptual Questions What are six particle conservation laws? Briefly describe them. #solutionbox[ Conservation energy, momentum, and charge (familiar to classical and relativistic mechanics). Also, conservation of baryon number, lepton number, and strangeness—numbers that do not change before and after a collision or decay. ] In general, how do we determine if a particle reaction or decay occurs? Why might the detection of particle interaction that violates an established particle conservation law be considered a #emph[good] thing for a scientist? #solutionbox[ It means that the theory that requires the conservation law is not understood. The failure of a long-established theory often leads to a deeper understanding of nature. ] === Problems Which of the following decays cannot occur because the law of conservation of lepton number is violated? #math.equation(block: true, alt: "open parenthesis a close parenthesis n → p plus e to the power −, open parenthesis e close parenthesis π to the power − → e to the power − plus υ − sub e; open parenthesis b close parenthesis μ to the power plus → e to the power plus plus υ sub e, open parenthesis f close parenthesis μ to the power − → e to the power − plus υ − sub e plus υ sub μ; open parenthesis c close parenthesis π to the power plus → e to the power plus plus υ sub e plus υ − sub μ, open parenthesis g close parenthesis Λ to the power 0 → π to the power − plus p; open parenthesis d close parenthesis p → n plus e to the power plus plus υ sub e, open parenthesis h close parenthesis K to the power plus → μ to the power plus plus υ sub μ")[$( "a" ) #h(0.2em) "n" → "p" + "e"^("−") & & & ( "e" ) #h(0.2em) π^("−") → "e"^("−") + limits(υ)^("−")_("e") \ ( "b" ) #h(0.2em) μ^(+) → "e"^(+) + υ_("e") & & & ( "f" ) #h(0.2em) μ^("−") → "e"^("−") + limits(υ)^("−")_("e") + υ_(μ) \ ( "c" ) #h(0.2em) π^(+) → "e"^(+) + υ_("e") + limits(υ)^("−")_(μ) & & & ( "g" ) #h(0.2em) "Λ"^(0) → π^("−") + "p" \ ( "d" ) #h(0.2em) "p" → "n" + "e"^(+) + υ_("e") & & & ( "h" ) #h(0.2em) "K"^(+) → μ^(+) + υ_(μ)$] #solutionbox[ a, b, and c ] Which of the following reactions cannot because the law of conservation of strangeness is violated? #math.equation(block: true, alt: "open parenthesis a close parenthesis p plus n → p plus p plus π to the power −, open parenthesis e close parenthesis K to the power − plus p → Ξ to the power 0 plus K to the power plus plus π to the power −; open parenthesis b close parenthesis p plus n → p plus p plus K to the power −, open parenthesis f close parenthesis K to the power − plus p → Ξ to the power 0 plus π to the power − plus π to the power −; open parenthesis c close parenthesis K to the power − plus p → K to the power − plus ∑ to the power plus, open parenthesis g close parenthesis π to the power plus plus p → Σ to the power plus plus K to the power plus; open parenthesis d close parenthesis π to the power − plus p → K to the power plus plus ∑ to the power −, open parenthesis h close parenthesis π to the power − plus n →")[$( "a" ) #h(0.2em) "p" + "n" → "p" + "p" + π^("−") & & ( "e" ) #h(0.2em) "K"^("−") + "p" → "Ξ"^(0) + "K"^(+) + π^("−") \ ( "b" ) #h(0.2em) "p" + "n" → "p" + "p" + "K"^("−") & & ( "f" ) #h(0.2em) "K"^("−") + "p" → "Ξ"^(0) + π^("−") + π^("−") \ ( "c" ) #h(0.2em) "K"^("−") + "p" → "K"^("−") + ∑^(+) & & ( "g" ) #h(0.2em) π^(+) + "p" → Σ^(+) + "K"^(+) \ ( "d" ) #h(0.2em) π^("−") + "p" → "K"^(+) + ∑^("−") & & ( "h" ) #h(0.2em) π^("−") + "n" → "K"^("−") + "Λ"^(0)$] Identify one possible decay for each of the following antiparticles: (a) #math.equation(block: false, alt: "n −")[$limits(n)^("−")$], (b) #math.equation(block: false, alt: "Λ to the power 0 —")[$limits("Λ"^(0))^("—")$], (c) #math.equation(block: false, alt: "Ω to the power plus")[$"Ω"^(+)$], (d) #math.equation(block: false, alt: "K to the power −")[$"K"^("−")$], and (e) #math.equation(block: false, alt: "Σ −")[$limits(Σ)^("−")$]. #solutionbox[ a. #math.equation(block: false, alt: "p − sub e to the power plus v e")[$limits(p)^("−")_(e) + v e$]; b. #math.equation(block: false, alt: "p − π to the power plus")[$limits(p)^("−") π^(+)$] or #math.equation(block: false, alt: "p − π to the power 0")[$limits(p)^("−") π^(0)$]; c. #math.equation(block: false, alt: "Ξ to the power 0 — π to the power 0")[$limits("Ξ"^("0"))^("—") π^(0)$] or #math.equation(block: false, alt: "Λ to the power 0 — K to the power plus")[$limits(Λ^("0"))^("—") "K"^(+)$]; d. #math.equation(block: false, alt: "μ to the power minus v − sub μ")[$μ^(−) limits(v)^("−")_(μ)$] or #math.equation(block: false, alt: "π to the power − π to the power 0")[$π^("−") π^(0)$]; e. #math.equation(block: false, alt: "p − π to the power 0")[$limits(p)^("−") π^(0)$] or #math.equation(block: false, alt: "n − π to the power −")[$limits(n)^("−") π^("−")$] ] Each of the following strong nuclear reactions is forbidden. Identify a conservation law that is violated for each one. #math.equation(block: true, alt: "(a) p plus p − → p plus n plus p −; (b) p plus n → p plus p − plus n plus π to the power plus; (c) π to the power − plus p → Σ to the power plus plus K to the power −; (d) K to the power − plus p → Λ to the power 0 plus n")[$"(a)" #h(0.2em) "p" + limits("p")^("−") → "p" + "n" + limits("p")^("−") \ "(b)" #h(0.2em) "p" + "n" → "p" + limits("p")^("−") + "n" + π^(+) \ "(c)" #h(0.2em) π^("−") + "p" → Σ^(+) + "K"^("−") \ "(d)" #h(0.2em) "K"^("−") + "p" → "Λ"^(0) + "n"$]