{ } or. ( ) = 1 2 ψ n1. ( ) = ψ r2 n1,n2 (, r! 1 ), under exchange of particle label r! 1. ψ r1 n1,n2. ψ n. ψ ( x 1

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1 Practice Modern Physics II, W08, Set 3 Question : Symmetric (Boson) and Anti-symmetric (Fermions) Wavefunction A) Consider a system of two fermions Which of the following wavefunctions can describe the system? ψ r n,n (, r ) = ψ r n( )ψ r n ( )+ψ r n ( )ψ r n ( ) ψ n,n r, r ψ n,n r, r ( ) = ψ n ( ) = ψ n { } or r { ( )ψ r n ( ) ψ r n ( )ψ r n ( )} Justify your answer r { ( )ψ r n ( )+ψ r n ( )ψ r n ( )}is symmetric ( ) =ψ r n,n (, r ), under exchange of particle label r r, appropriate for ψ r n,n, r two identical bosons ψ r n,n (, r ) = ψ r { n( )ψ r n ( ) ψ r n ( )ψ r n ( )}is anti-symmetric ( ) = ψ r n,n (, r ), under exchange of particle label r r, appropriate for ψ r n,n, r two identical fermions B) Two identical particles in a box (in one dimension D) is described by the total wavefunction, ψ ( x,x ) = L sin π x sin 4π x L sin 4π x sin π x, where L is the length of the box and x and x are the positions of particle and, respectively What are the quantum states of the two particles? Are the two particles bosons or fermions? Explain your answer What is the total energy of the two identical particles? The mass of a particle is m = kg, and L = nm Hint: A particle in a D box of length, L, is described by the wavefunction ψ n = nπx sin, where n =,,3 is the quantum state of the particle For a L particle in the n state, its energy is E n = n π ml Obviously, it is a wavefunction for two identical fermions, one fermion is in a state, n =, the other in n = 4 The total energy is = E n= + E n=4 = π ml + 4 π = 0π ml ml C) Repeat part D) for the wavefunction, ψ ( x,x ) = L sin π x sin 3π x + L sin 3π x sin π x Obviously, it is a wavefunction for two identical bosons, one boson is in a state, n =, the other in n = 3 The total energy is

2 E = E + E = π total n= n=3 ml + 3 π = 5π ml ml D) repeat for the wavefunctionψ x,x ( ) = L sin π x L sin π x wavefunction for two identical bosons, both in the state, n = The total energy is = E n= + E n= = π ml + π = π ml ml E) Consider two electrons in a box of width L = 50 nm The quantized energy of n π 3 each electron is E n =, n =,,3, where m = 9 0 kg is the mass of ml the electron Calculate the ground state energy (lowest energy) and first excited energy (second lowest energy) of the two electrons system Express your answer 9 in ev ( ev = 6 0 J ) Since electrons are fermions: Ground state one in n =, the other in n =, = E n= + E n= = π ml + π = 5π ml ml First excited state one in n =, other in n = 3 = E n= + E n=3 = π ml + 3 π = 5π ml ml Question Magic Numbers and Hassium As discussed in the textbook, certain values of protons, Z and neutrons N are called magic since they are believed to correspond to nuclear stability These magic numbers are, 8, 0, 8, 50, 8, and 6 However, there are other magic numbers associated only with protons or neutrons For example, Hassium, Hs is one of the heaviest isotopes ever observed, and is stable for about 0s Its surprising stability is attributed by some to be due to the fact that 08 is a magic number for protons, and 6 is a magic number for neutron A) Despite this explain in no more than two sentences, the physical reasons why Hs should not be stable Usually stable isotopes have slightly more neutrons than protons, but Hassium has far too many neotrons than protons B) Use the nuclear liquid drop model (see equation sheet) to calculate the binding energy of 48 8 Ni Does the results support the magic number hypothesis? Justify in two sentences

3 Use Nuclear liquid drop model B( A Z X) = a v A a A A /3 3 Z ( Z )e a s 5 4πε 0 r a s = 3MeV, δ = Δ (even-even),δ = 0 (even-odd, odd-even),δ = Δ (odd-odd), Δ = 33MeV A 3/4, with r = r 0 A /3, r 0 = 0 5 m ( N Z ) + δ, a v = 58MeV, a A = 83MeV, A C) Hs has atomic mass of 70349u Use this to calculate the binding energy, and compare (no more than two sentences) with the results of part B) Use B = Nm n + ZM ( H ) M ( A Z X) c Question 3 β + andα decay, and electron capture A) Consider the nuclear reaction, 4 8 O 4 7 N + β + +ν e Calculate the disintegration energy of the decay using Q = M ( 4 8 O) M ( 4 7 N ) m e c In no more than two sentences, explain why it is necessary to included the mass of two electrons in the equation for Q In one sentence, justify, based on your answer, why the decay will or will not occur? What isν e? In no more than two sentences, explain how that the law of quantum mechanics requires its presence in the reaction u mc = 935MeV, m e c = 5keV = 05MeV, u mc = 935MeV, m e c = 5keV = 05MeV, m e = 05MeV 935MeV /u = u Q = M 8 ( 4 O) M 4 7 N ( ) m e c = u u u c 935MeV / c = 4MeV Since Q > 0, the decay can occur Since 4 7 N has one fewer electron than 4 8 O, and we must include the mass of β +, which is the same as the electron s mass, the product mass includes m e The decay 4 8 O 4 7 N + β + +ν e effectively converts one proton into a neutron, p n+ β + +ν e, where all the particles have spin s = ½ so that the total spin on the RHS is s = ½, and on the RHS it is can also be ½, and the total spin is conserved If ν e does not exists, the reaction will be p n+ β +, and the spin on RHS is s = 0 or, and the spin cannot be conserved B) Repeat part A), but for electron capture 4 8 O + β 4 7 N +ν e Why does the relation for Q of the electron capture not include m e? Q = M 8 ( 4 O) M ( 4 7 N ) c

4 C) Calculate the disintegration energy of the decay 6 8 O A Z C + α Determine the value of A and Z for A Z C In one sentence, justify, based on your answer, why the decay will or will not occur? 6 8O 6 C + α Q = M ( 6 8 O) M ( 6 C) M ( 4 He) c = [ u 0u u]c 935MeV / c = 76MeV Since Q < 0, this alpha decay should not occur It is well known that 6 8 O is a stable element that does not decay Question 4 Elementary Particle Physics A) State one (or two) reason why the following two reactions are not allowed: π e + γ and π + e + e + + µ + For π e + γ, spin is not conserved since for LHS π has spin 0, but for RHS e has spin ½ and γ has spin giving a RHS spin of ½ or 3/ Also electron Lepton number is not conserved, since L e = 0 on LHS, but L e = on RHS For π + e + e + + µ + spin is not conserved since for LHS π + has spin 0, but for RHS e, e +, and µ + are spin ½ particles giving a RHS spin of ½ or 3/ Also muon Lepton number is not conserved, since L µ = 0 on LHS, but L µ = on RHS B) Complete the following reactions: µ + p n +? ; n + p Σ 0 + n +? Briefly justify your answers? µ + p n + ν µ, muon neutrino ν µ has muon lepton number L µ = needed to balance the L µ = of µ n + p Σ 0 + n + K +, K + balances the charge of the proton on LHS, and also has strangeness number + that cancels the - strangeness of Σ 0 to equate with the zero strangeness of LHS Also since K + is a meson, the baryon number of on LHS is conserved C) Determine the quark composition of Λ, π and π + See equation sheet for information on Λ Λ has baryon number of so it must be composed of three quarks It has zero charmed and strangeness number, hence it must have one strange quark, s, which has a charge of e /3 For the charge to add up to zero, its composition must be uds NOTE: top and bottom or charm quarks are ruled out since Λ does not have top, bottom or charm numbers π is a meson (baryon number 0) so it must be made of a quark/anti-quark pair It has no strangeness or charm To have charge e, it must be composed of ud π + is ud since it is the anti-particle of π

5 D) Quarks are spin ½ particles The Ξ 0 baryon has a quark composition ssu Is this composition consistent with the Pauli exclusion principle? Why or why not? All quarks are spin ½ fermions, so it appears that the two identical s quarks (fermions) disobey the Pauli exclusion principle, since they are in the same quantum state However, one spin of an s quark can be in the spin up state, and the other in the spin down state, so they are in distinct quantum state, which are allowed by the Pauli exclusion principle In addition, the existence quantum color states means that in the ssu composition the two s quarks may be in the same spin state (up and up, or down and down, as long as they are in in different color states Question 5) Cyclotron The figure below shows a particle undergoing a uniform circular motion in a cyclotron of radius r = 55m by a magnetic field of B = 3T It was then determined that the particle has a kinetic energy of K = 000MeV A) Using the appropriate equation, find the relativistic momentum, p, of the particle HINT: Assume that the particle has a charge of magnitude q = e, and the small cross means that the magnetic field B points into the page, while the charge particle follows a clockwise path First using the right hand rule we can see that, since the B field is into the page, so if the velocity is up then a positive charge should follow a counterclockwise path Since it follows a clockwise path it must be a negative charge Using the appropriate equation, find the relativistic momentum, p, of the particle r = p / qb q = rqb = 55m ( ) 3T ( ) C ( ) = kg i m i s B) Starting from K = E mc and E = p c + mc ( ) / K energy can be expressed as mc = p c K ( ), show that the rest mass K = E mc = p c + ( mc ) mc K + mc = p c + ( mc ) Squared both sides K + mc K + K mc ( ) = p c + ( mc ) ( ) + m c 4 = p c + m c 4 K + K ( mc ) = p c Solving for mc gives ( ) / K QED mc = p c K C) Use the results of part A and B to find the rest mass energy of the particle in MeV Inspect the nuclear physics data on the equation sheet to infer the identity of this particle NOTE: Select the particle in the table 44 with the closest mass, and charge

6 kg i m ( mc = p 44 0 c K ) 8 m s s ( J ) = K J mc = J J i ev = ev or 49 MeV The closest particle is the negative pi meson,π with 40 MeV and charge e This particle is a negative pi mesonπ