Chapter 2 Elementary particles
2.1 Classification of Particles 2.2 Leptons 2.3 Quarks 2.4 Hadrons 2.5 Interactive Exercise
2.2 Leptons
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Leptons Till the limit of resolution, 10-18m, at present, the leptons are structureless, with none having an internally excited state. Thus they are regarded as fundamental particles. Leptons are spin-1/2 fermions, not subject to strong interaction
Electron e-, muon μ- and tauon τ- have corresponding neutrinos νe, νμ and ντ. Electron, muon and tauon have electric charge of -e. Neutrinos are neutral. Neutrinos possibly have zero masses. For neutrinos, only weak interactions have been observed so far Antileptons are positron e+, positive muon and tauon (pronounced mu-plus and tau-plus), and antineutrinos:
Neutrinos and antineutrinos differ by the lepton number. Leptons posses lepton numbers Lα=1 (α stands for e, μ or τ), and antileptons have Lα=-1.
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There are three Lepton numbers Lℓ ≡ N(ℓ -) - N(ℓ+) + N(υℓ ) - N(υℓ ) where ℓ = e, μ or τ Lepton numbers are conserved in any interaction Each of these, Le ,Lμ and Lτ , are almost conserved τ-
→
e-
+
υe
+
υτ
Le =
0
→
1
+
-1
+
0
Lμ=
0
→
0
+
0
+
0
Lτ =
1
→
0
+
0
+
1
τ-
Table 1
→
e-
+
υe
+
υμ
0
→
1
+
-1
+
0
Lμ=
0
→
0
+
0
+
1
Lτ =
1
→
0
+
0
+
0
Le =
Forbidden
Table 2
Anything that is not a lepton has Lℓ = 0 υμ + p → μ+ + n Lμ= -1 + 0 → -1 + 0 Neutrinos can not be registered by any detector, there are only indirect indications of their quantities. First indication of neutrino existence came from βdecays of a nucleus N: N(Z,A) → N(Z+1,A) + e- + νe β-decay is nothing but a neutron decay: n → p + e- + νe Dayalbagh Educational Institute
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Necessity of a neutrino existence comes from the apparent energy and angular momentum nonconservation in observed reactions Note that for the sake of the lepton number conservation, electron must be accompanied by an antineutrino and not neutrino! Mass limit for νe can be estimated from the precise measurements of the β-decay: me ≤ Ee ≤ ΔMN – mve The best results are obtained from the tritium decay: 3H
→ 3He + e- + νe
It gives mνe ≤ 15 eV/c2, which usually is considered as a zero mass. An inverse β-decay also takes place: νe + n → e- + p
(1)
or ν e + p → e+ + n
(2)
However, the probability of these processes is very low, therefore to register it one needs a very intense flux of neutrinos.
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Neutrino Beta decay crisis: violation of energy conservation? In the early years of studying nuclear beta decay it was wrongly assumed that the beta decay reaction had the form of (A, Z) → (A, Z + 1) + eThen the electron should have a momentum given by
With M = M (A,Z), m1=M (A,Z + 1) and m2 = me The particular value of some specific beta decay is not important. Important is that the momentum, and hence the energy of the electron should have a fixed value defined by the masses of the three particles involved in the decay. This is not observed as can be seen on the following examples. Experimental Beta Decay Spectra:
Fig 1
Fig 2
From G.J. Neary, Proc. Phys. Soc. (London), A175, 71 (1940) Dayalbagh Educational Institute
From J. R. Reitz, Phys. Rev. 77, 50(1950). 7
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From these and many more examples it is seen that the electrons from beta decay have a continuum of energies ranging from zero to some maximum value. This discrepancy between theory and experiment was the beta decay crisis. One way out was suggested by Niels Bohr: a possible violation of the conservation of energy in nuclear processes. Another way out was proposed by Wolfgang Pauli: the existence of a neutral particle that was not “seen” in beta decay. Pauli called it “neutron”. When soon after that the neutron was discovered by Chadwick. Enrico Fermi gave Pauli's neutral particle the name neutrino: “little neutron” in Italian.
Thus according to Pauli the beta decay reaction must be written as (A, Z) → (A, Z + 1) + e- + v
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Enrico Fermi formulated the theory of nuclear beta decay assuming that the neutrino was a neutral particle with spin ½ and a very small mass, much smaller than the mass of the electron. He made this theory as similar to quantum electrodynamics as possible. There were two important differences: There was nothing like the photon as a carrier of the weak force; The coupling strength that determined the weak interaction was given by a new constant G; this constant is nowadays called the Fermi constant (A,Z+1) (A,Z)
eG ν
Fig 3
Comparison of the theoretical with experimental spectra soon showed that the neutrino mass was unmeasurably small. Eventually it was thought that the neutrino mass was exactly zero. Another problem had to do with the possible existence of an antineutrino. In other words, the question arose: does an antineutrino exist that is distinct from the neutrino? The answer can come only from experiment. If the neutrino is identical with the antineutrino, then a neutrinoless double-beta decay should be possible. Such a neutrino is called Majorana neutrino (the other kind is the Dirac neutrino). Dayalbagh Educational Institute
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A Majorana neutrino must also have a nonzero mass. The neutrinoless double-beta decay is lepton number violating. p
n Wv
ee-
Wn
p
Fig 4
To date there is no evidence for neutrinoless DBD The negative result of neutrinoless DBD means that the neutrino is distinct from the antineutrino. All experimental evidence is that the neutrino can be assigned a lepton number identical with that of the electron, and the antineutrino a lepton number identical with that of the positron. With this assignment the lepton number is empirically known to be conserved in all reactions. Searches for lepton number violation are an important and active area of research. Accepting that the neutrino is a Dirac particle, we must finally write the β decay equation as (A, Z) → (A, Z + 1) + e- + v or for β+ decay (A, Z) → (A, Z - 1) + e+ + v Dayalbagh Educational Institute
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Detection of the neutrino: the ReinesCowan experiment However successful the Fermi theory of beta decay was, the conclusive experimental test that the neutrino was not just a hypothetical particle had to be an experiment in which the neutrino could be demonstrated to cause a reaction. Such an experiment of direct neutrino observation was first carried out by Frederick Reines and Clyde L. Cowan. The source of neutrinos had to be of a sufficient intensity which became available only in the late 1940s – early 1950s as a result of the construction of highflux nuclear reactors.
Schematic diagram of the neutrino detection experiment of Reines and Cowan Antineutrinos from a nuclear reactor cause the inverse beta decay reaction v + p → n + e+ The positron slows down and annihilates with an atomic electron into a pair of photons. The photons travel back-to-back; each photon has an energy equal to the rest energy of the electron, about 0.5 MeV. Passing through liquid scintillator, the photons give light pulses which are seen be arrays of PM tubes. Dayalbagh Educational Institute
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The detector vessel is filled with water in which cadmium chloride is dissolved. The neutrons from the inverse beta decay are captured by cadmium nuclei: n + 108Cd → 109Cd* → 109Cd + γ The photon from this reaction is detected 5μs after the photon pair: delayed coincidence. Photomultiplier Port from Nuclear reactor Neutrino Flux 1013/cm2s
+
e -e
-
+ n 109
Cd
Delayed coincident detection of γ from 109 Cd with pair of γ's from e+-e- annihilation.
Water target with scintillator plus CdCl2
Fig 5
Muons were first observed in 1936, in cosmic rays. Cosmic rays have two components:
1) primaries, which are high-energy particles coming from the outer space, mostly hydrogen nuclei 2) secondaries, the particles which are produced in collisions of primaries with nuclei in the Earth atmosphere; muons belong to this component Muons are 200 times heavier than electrons and are very penetrating particles. Electromagnetic properties of muon are identical to those of electron (upon the proper account of the mass difference)
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Muon decay crisis: one or two types of neutrinos? In the late 50s, after the discovery of parity violation in the weak interactions, the hypothesis of an intermediate vector boson was gaining ground. Such a heavy boson – heavier than the proton –was thought to be a candidate for the carrier of the weak interaction. First calculations were carried out and they soon led to difficulties. One of the problems was a serious discrepancy between the calculated and observed rates of muon decays into electron and photon and into an electron and neutrino and antineutrino. vμ μ eWve
Fig 6
Muon decay: μ- → e-vμve Feynman diagrams of hypothetical decay μ- → e-γ assuming there is only one kind of Neutrino. W-
a) μ-
v γ
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eFig 7 (a) 13
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b) -
μ
W
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γ
-
v
c) e
-
μ-
We-
v γ
Fig 7 (b) and (c)
This can be calculated without detailed knowledge of the W IVB and compared with the decay μ- → e-vv In the ratio of the decay rates of these processes all W parameters cancel and one gets a very simple answer: Γ (μ- → e- γ ) 1 e2 ~ ~ 10-4 Γ (μ → e vv ) 24π ħc The experimental upper limit is less than 10-11. This discrepancy gave rise to the two-neutrino hypothesis: υe ≠ υμ
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Experimental Test of the Two-Neutrino hypothesis 1) Make a neutrino beam from pion decays into muons. 2) pass the neutrino beam through a detector which must a)have enough dense mass for some neutrinos to interact and produce a muon and/or an electron; b)show the difference between electrons and muons. Then, if you see about equal numbers of electrons and muons produced, you conclude that there is only one type of neutrino. But if you see only muons produced, then you know that the muon neutrino is different from the electron neutrino. Such an experiment was first proposed by Bruno Pontecorvo in 1959 (ЖЭТФ т. 37, вып. 6, с. 1751-1757 (1959)) and was carried out at the Brookhaven National Laboratory in 1962. The BNL experiment: a beam of 15 GeV protons struck a beryllium target and produced mostly pions. After 21 meters about 10% of the pions had decayed according to π- → μ- ν; π+ → μ+ ν Then the pions, muons and neutrinos passed into a 13.5 m steel absorber that absorbed all pions and muons, leaving only the neutrinos from the pion decay.
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The detector consisted of ten 1-ton modules, each module consisting of nine 1 inch thick aluminium plates separated by ¾ inch gaps filled with a gas mixture. There was a high voltage between the aluminium plates, so the ionization from a charged particle passing through the gas gives rise to sparks between the plates (such a detector is called spark chamber). Muons passing through the spark chambers leave practically straight tracks. Electrons produce showers which can be cleanly distinguished from the muon tracks. The result of the experiment was that there were only muons produced, and hence muon neutrinos are not identical with electron neutrinos.
The Third Neutrino: ντ After the discovery of the third charged lepton, the tau lepton, it was natural to assume that there was also a third neutral lepton, the tau neutrino, which must be different from the electron and muon neutrinos. Indirect evidence of its existence came from the LEP experiments at CERN where it was shown, by comparison with precision calculations, that the Z boson peak could be quantitatively described only with 3 light neutrinos. But as always, indirect evidence is thought to be unsatisfactory: one wants direct evidence. In the present case this is similar to the direct evidence of the existence of the neutrino (the Reines-Cowan experiment) and to the two-neutrino experiment at BNL: one is looking for a reaction induced by a tau neutrino. Dayalbagh Educational Institute
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Design of the DONUT Experiment The General Idea: The general concept behind the DONUT experiment (Direct Observation of the NU Tau) was to create a beam of tau neutrinos that would interact to form tau leptons, which could then be observed. The set-up is as follows: An 800 GeV beam of protons from the TeVatron collides into the beam dump, a large block of tungsten, to produce the charm particle Ds (a meson comprised of charm and strange quarks). About 7% of Ds mesons decay to an anti-tau neutrino and a tau lepton:
All tau decays go into a tau neutrino + other particles. About 35% of tau decays are to a charged lepton and a tau neutrino: The neutrinos pass through 36m of shielding to a detector consisting of emulsion targets followed by a spectrometer. In the emulsion targets some tau neutrinos interact with a nucleon to produce a tau lepton, which is detectable by the various levels of the spectrometer. Only a tau neutrino can produce a tau lepton; tau leptons are recognized by their decay products. Dayalbagh Educational Institute
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The muon neutrinos produced in pion and kaon decays contribute about 23% to the total number of observed reactions. This is background. The mean energy of the neutrinos is 56 GeV for the taus and 54 GeV for the muon and electron neutrinos. There is 1 tau neutrino per 50,000 primary protons on the tungsten target per m2 and 10 times more muon and electron neutrinos. There are several uncertainties in the calculations of the expected number of neutrinos. For example, there is a 20% uncertainty in the Ds production and a 15% uncertainty due to the Ds branching ratio, i.e. the probability of the particle to decay to a specific final state. To remove all particles other than neutrinos, the beam is passed through a shield consisting of 36 meters of steel. Only neutrinos can pass through the shielding; this neutrino beam contains tau neutrinos.
Steel shielding Beam dump
Ds
τ
ντ
νμ ντ
τ
Emulsion target where some neutrinos interact; only tau neutrinos can produce tau leptons.
Fig 8 Dayalbagh Educational Institute
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Located 36 m from the beam dump sits the hybrid emulsion spectrometer, measuring 16 m in length, comprised of 6 sections. DONUT scientists had two difficulties to overcome in building a target/detector that had the resolution power to observe the tau neutrino: First, the lifetime of the tau, partner to the tau neutrino, is extremely short: Ď„t ≅ 0.3ps its mean decay length under the conditions of DONUT is ≈ 2.3 mm. Second, the tau neutrino is extremely non-interacting. So they had to build a detector with very fine resolution and sufficient target density to allow for events to take place. They chose a hybrid emulsion spectrometer (HES), because the dense emulsion gives plenty of targets for the incoming neutrinos and simultaneously record all events, while the spectrometer provides the necessary data to identify the event products.
Drift Chamber Tracking: The rest of the spectrometer's job is to determine the charge and momentum of the passing particles so that they can be identified as products of interactions and their tracks can be interpolated back to the event vertex. Dayalbagh Educational Institute
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Specifically, drift chambers aid the scintillating fiber trackers in track discrimination. Drift chambers are fixed volumes of gas divided into discrete cells by sheets of closely spaced parallel wires. These wires are held at a high voltage with respect to 'sense' wires which line the middle of the cells at 1mm spacing. When a charged particle enters the drift chamber and ionizes an atom, the freed electron is accelerated through an electric potential to the sense wires, liberating other electrons as it passes through. The sense wires output electrical signals. The time delays between signals as the particle passes through a series of chambers determine the speed of the particle. Drift chambers are designed to have as little impact on the passing particle as possible, thereby not changing its energy.
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The drift chambers are placed in a magnetic field to determine the particle momenta: Knowing that particles traveling perpendicular to a magnetic field B exhibit circular motion, and using the magnetic force law F=q vxB, you come to the expression p = qrB, where p is the momentum and r is the radius of curvature of the particle track. The radius of curvature is measured using the drift chambers, and hence the particle's momentum is calculated.
Electromagnetic Calorimeter: The electromagnetic calorimeter is used to measure the electromagnetic energy produced by electron neutrino events in the emulsion and the energy of electrons produced in tau decays. The calorimeter consists of lead glass (glass that is approximately 25% lead by weight) and scintillating glass attached to photomultiplier tubes. The presence of lead increases the instrument's index of refraction, slowing the speed of light through the glass. When a particle moves faster than the speed of light in a given medium the particle emits Cerenkov light (photons) in a cone shape, which is picked up by the photomultiplier tubes.
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Additionally, electrons from upstream encounter lead nuclei and emit photons, leading to electromagnetic showering. Electromagnetic showering is the production of electron-positron pairs by a photon that induces more pairs and photons and so on. The Cerenkov light emitted by these electron positron pairs enters photomultiplier tubes. The photon, by way of the photoelectric effect, kicks an electron out of the valence shell of an alkali metal atom which is accelerated through a potential difference to the other end of the tube, producing an avalanche of electrons along the way. The number of electrons detected at the end of the tube is proportional to the number of photons produced by the electromagnetic showering, which in turn is proportional to the energy of the incident electron.
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Schematic of avalanche formation: e Ar Ar+ e Ar
e Ar
Ar+ Ar+
Ar+ e
e
Ar
e Ar
Ar
e Ar
Fig 9
The last station of the detector is the Muon Identification System: The goal of the muon identification system is to identify charged current interactions of muons in the target and use that information to point to the event vertex. Additionally it is used to find muons from tau decays, which occurs in 17% of all tau decays. This system has three layers of steel walls interleaved with proportional tubes totaling 4 m in length. The tubes are filled with an ArCO2 gas mixture that becomes ionized when a charged particle passes through. The freed electrons are accelerated to a wire running down the center of the tube that is held at a high potential with respect to the walls. This signal is amplified and processed, identifying and tracking muons. Dayalbagh Educational Institute
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Tauon is the heaviest of leptons, was discovered in e +eannihilation experiments in 1975
Fig 10: τ pair production in e+e- annihilation
Electron is a stable particle, while muon and tauon have a finite lifetime: τμ = 2.2 ✕ 10-6 s and ττ = 2.9 ✕ 10-13 s Muon decays in a purely leptonic mode: μ- → e- + νe + νμ
(3)
Tauon has a mass sufficient to produce even hadrons, but has leptonic decay modes as well: τ- → e- + νe + ντ
(4)
τ- → μ- + νμ + ντ
(5)
Fraction of a particular decay mode with respect to all possible decays is called branching ratio. Branching ratio of the process (4) is 17.81%, and of (5) - 17.37%. Note: lepton numbers are conserved in all reactions ever observed.
Important assumptions: 1) Weak interactions of leptons are identical, just like electromagnetic ones (“interactions universality”) 2)One can neglect final state lepton masses for many basic calculations Dayalbagh Educational Institute
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The decay rate of a muon is given by expression: (6)
Here GF is the Fermi constant. Substituting mμ with mτ one obtains decay rates of tauon leptonic decays, equal for both processes (4) and (5). It explains why branching ratios of these processes have very close values. Using the decay rate, the lifetime of a lepton is:
(7) Here 1 stands for μ or τ . Since muons have basically only one decay mode, B=1 in their case. Using experimental values of B and formula (6), one obtains the ratio of muon and tauon lifetimes:
This again is in a very good agreement with independent experimental measurements. Universality of lepton interactions is proved to big extent. That means that there is basically no difference between lepton generations, apart of the mass. Dayalbagh Educational Institute
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Problem 1 List all of the known leptons. How does µ+ decay? Considering this decay and the fact that ν µ + n →e− +p is found to be forbidden, Discuss possible lepton quantum number assignments that satisfy additive quantum number conservation laws. How could νµ produce a new charged “heavy lepton”?
Solution 1 Up to now 10 kinds of leptons have been found. These are e− , νe, µ−, νµ, τ− and their antiparticles e+,νe, µ+,νµ, τ+. ντ and ντ have been predicted theoretically, but not yet directly observed. µ+ decays according to µ+ →e+ +νe +νµ. It follows that
νe +µ + → e+ +νµ.
On the other hand the reaction νµ+n →e− +p is forbidden. From these two reactions we see that for allowed reactions involving leptons, if there is a lepton in the initial state there must be a corresponding lepton in the final state. Accordingly we can define an electron lepton number Le and a muon lepton number Lµ such that: Dayalbagh Educational Institute
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Le =1 for e− ,νe Lµ =1 for µ− ,νµ with the lepton numbers of the antiparticles having the opposite sign, and introduce an additional conservation rule that the electron lepton number and the µ lepton number be separately conserved in a reaction. It follows from a similar rule that to produce a charged heavy lepton, the reaction must conserve the corresponding lepton number. Then a new charged “heavy lepton” A+ can be produced in a reaction νµ +n →A+ +νA +µ− +X where νA is the neutrino corresponding to A+, X is a baryon. For example A+ = τ+, νA = ντ
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Problem 2 Use lepton universality and lepton-quark symmetry to estimate the branching fraction for the decay τ − → e −+νe+ντ. Ignore final states that are cabibbo suppressed relative to the lepton modes.
Solution 2 The leptonic decays of τ − are τ − → e −+νe+ντ.
(1)
τ − → μ −+νμ+ντ.
(2)
having equal probability from lepton – quark symmetry. Also the possible hadronic decays are of the form τ− → ντ + X where X can be du or sc so that X may have negative charge. However the latter possibility is ruled out because
mτ < ms + mc We are then left with only one possibility for X so that the hadronic decay will be τ− → ντ+ du
(3)
Reaction (3) has relative weightage of 3 because of three colours. Thus the branching fraction of (1) is predicted to be 1/5 or 0.2, a value which is in agreement with the experimental value of 0.18. Dayalbagh Educational Institute
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Problem 3 Which of the following reactions are allowed and which forbidden as under weak interactions? (a) νμ + p → μ + + n (b) νe + p → e − + π + + p (c) K + → π 0 + μ + + νμ (d) Λ → π + + e − + νe
Solution 3 (a) The reaction νμ + p → μ + + n is forbidden because lepton number is not conserved, left side has Lμ = −1 while right side has Lμ = +1 (b) The reaction νe + p → e − + π + + p is forbidden because charge is not conserved. (c) from the reaction
K + → π 0 + μ + + νμ It is allowed because B, Lμ, Q etc are conserved. Also the selection rule, ΔQ = ΔS = ±1 is obeyed. (d)
Λ → π + + e − + νe
is allowed for reason stated in (c).
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Problem 4 Given the mean life time of μ+ meson is 2.197 μs and its branching fraction for μ+ → e+ + νe + νμ is 100%, estimate the mean lifetime of τ+ if the branching fraction B for the decay τ+ → e + + νe + ντ is 17.7%. The masses of muon and τ-lepton are 105.658 and 1,784 MeV.
Solution 4
which is in excellent agreement with the experimental value of 0.3 × 10−12 s
Problem 5 Consider the semi leptonic weak decays (a) Σ− → n + e − + νe (b) Σ+ → n + e + + νe Explain why the reaction (a) is observed but (b) i s not.
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Solution 5: The reaction (a) can go via the mechanism of the diagram shown in Figure (a).
However, no diagram with single W exchange can be drawn for the reaction (b) which at the quark level implies the transformation uus → udd + e + + νe as in Fig (b)
and would require two separate quark transitions which involve the emission and absorption of two W bosons – a mechanism which is of higher order and therefore negligibly small. Dayalbagh Educational Institute
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The above conclusion can also be reached by invoking for t he selection rule for semi leptonic decays. Reaction (a) obeys the rule ΔS = ΔQ =+1 and is therefore allowed, While in reaction ( b) we have ΔS =+1, but ΔQ = −1, and therefore forbidden.
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