Assignment 5. Ian Rittersdorf Nuclear Engineering & Radiological Sciences

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1 Assignment 5 Ian Rittersdorf Nuclear Engineering & Radiological Sciences ianrit@umich.edu March 19, 2007

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3 All mass values taken from Krane unless otherwise noted. P5.1, 20% Krane, Problem 9.3, p Find Q for β, β +, and ε for 196 Au. 196 Au 196 Hg + β + ν For β -decay: Z Z+1 Q β = [m( A X) Zm e ] [m( A X ) (Z + 1)m e ] m e + B i B i i=1 i=1 We see the masses of the electrons cancels out. We neglect the difference in electron binding energies because this value is very small, thus For β + -decay: Q β = [m( A X) m( A X )]c 2 And for electron capture: Q β + = [m( A X) m( A X ) 2m e ]c 2 Q ε = [m( A X) m( A X )]c 2 B n Where B n is the binding energy of the n-shell captured electron. We use the following approximation for all B n calculations: Some important values: B n = 13.6eV Z 2 m( 196 Au) = u = MeV / c 2 m( 196 Hg) = u = MeV / c 2 2

4 From this we calculate Q β, Q β +, Q ε for 196 Au. Q β = [ ] MeV Q β = kev Q β + = [ ( )] MeV Q β + = kev Q ε = [ ] u MeV/u MeV Q ε = kev 3

5 P5.2, 20% Krane, Problem 9.9, p Recall the inverse β-decay equations: ν + p = n + e + ν + n = p + e Using those relations, and the β-decay equations, we can supply the missing components to the equations in the text. (a) ν + 3 He 3 H + e + (b) 6 He 6 Li + e + ν (c) e + 8 B 8 Be + ν (d) ν + 12 C 12 N + e (e) 40 K ν + e Ar (f) 40 K ν + e + 40 Ca 4

6 P5.3, 20% Krane, Problem 9.10, p Some important values: m p = MeV / c 2 m e = MeV / c 2 (a) What is the kinetic energy give to the proton in the decay of the neutron when the electron has negligibly small kinetic energy? We know that the neutron decay is We also realize that n p + e + ν Where Q = T p + T ν Q = (m n m p m e m ν )c 2 In the text, Krane calculates this value for us as Q = We realize that the proton will not be moving at any significant fraction of the speed of light, and thus we can treat it nonrelativistically. Thus, T p = 1 2 m pv 2 p T ν = m ν c 2 (γ 1) Since the momentum of the electron is negligible, the momentum of the proton must be opposite and equal to that of the neutrino. Thus, p p = p ν m p v p = E ν c Using these equations, we can solve for the kinetic energy of the proton. 5

7 We say Q = T p + T ν = 1 2 m pv 2 p + E ν Where E ν is the kinetic energy of the ν. Substituting in momentum Q = 1 2 m pv 2 p + p ν c Substitute in the momentum of p p for p ν because we know their momenta to be equal. Also, This can be rewritten as a quadratic Q = 1 2 m pv 2 p + p p c 0 = 1 2 m pv 2 p + m p v p c Q Using a computer, we can substitute the following converted values into this equation and solve for v p. m p = kg c = m/s Q = J I used maple and solved for a velocity of v p = m/s. Plugging this into T p = 1 2 m pv 2 p, we get J. Converting this back to ev gives T p = ev 6

8 P5.3, 20% Krane, Problem 9.10, p Some important values: m p = MeV / c 2 m e = MeV / c 2 (b) What is the kinetic energy give to the proton in the decay of the neutron when the neutrino has negligibly small kinetic energy? We know that the neutron decay is We also realize that n p + e + ν Where Q = T p + T e Q = (m n m p m e m ν )c 2 In the text, Krane calculates this value for us as Q = We realize that the proton will not be moving at any significant fraction of the speed of light, and thus we can treat it nonrelativistically. Thus, T p = 1 2 m pv 2 p T e = m e c 2 (γ 1) Since the momentum of the neutrino is negligible, the momentum of the proton must be opposite and equal to that of the electron. Thus, p p = p e m p v p = m e cβγ Using these equations, we can solve for the kinetic energy of the proton. We say 7

9 Substituting in momentum Q = T p + T e = 1 2 m pv 2 p + m e c 2 (γ 1) p 2 p Q = 1 + m e c 2 (γ 1) 2 m p Substitute in the momentum of p e for p p because we know their momenta to be equal. Thus, This can be rewritten as a quadratic Q = 1 p 2 e + m e c 2 (γ 1) 2 m p We know that 0 = 1 p 2 e + m e c 2 (γ 1) Q 2 m p γ = 1 1 β 2 β = v2 c 2 Using a computer, we can substitute values for γ and solve the following equation for β. Where 0 = 1 2 (m ecβ) 2 1 β 2 + m e c 2 1 ( m 1) Q p 1 β 2 m p = kg m e = kg c = m/s Q = J Solving for β gives β = c, meaning that our electron is travelling at times the speed of light. Plugging this β into the momentum equation allows us to calculate a value of v p = m/s. Plugging that into T p yields J. Converting this into ev gives 8

10 T p = ev 9

11 P5.4, 20% Krane, Problem 9.11, p The decay that we are looking at is 7 Be + e 7 Li + ν We know that Q = T Li + T ν The Q-value of this reaction is equal to the kinetic energy of the 7 Li and the KE (or just energy because we assume that the neutrino mass is negligible). Also, We assume that Q ε = [m( A X) m( A X )]c 2 B n For Be: B n = [13.6 Z 2 ] ev Other important values: B n = = 217 ev m( 7 Be) = u = MeV / c 2 m( 7 Li) = u = MeV / c 2 Thus, Q ε = MeV We treat the proton as a classical particle because it will not be moving at any significant fraction of the speed of light. We ignore initial momentum of the electron in our conservation 10

12 equations. We do this because we view the capture in the zero-momentum frame of the Be nucleus. The moment that the electron is captured, the nucleus is travelling the same speed as the electron in that frame. Thus, the net momentum there is zero. So we say and p Li = p ν m Li v Li = E ν c Q = T p + T ν = 1 2 m Liv 2 Li + E ν Substituting in momentum Q = 1 p 2 Li + p ν c 2 m p Substituting p Li for p ν gives p 2 ν Q = 1 + p ν c 2 m p We can substitute in the value for momentum and then write this in the quadratic form 0 = 1 2 m Liv 2 + m Li vc + Q and then solve. I used maple using the following values. m Li = kg c = m/s Q = J Thus, v = m/s Therefore the kinetic energy of the Li nucleus is 11

13 T Li = 1 2 m Liv 2 = J = ev And then E ν = m Li vc = = MeV Thus, T Li = ev E ν = MeV 12

14 1 P5.5, 20% Krane, Problem 9.14, p Classify the following decays according to degree of forbiddenness: (a) 89 Sr( 5 + ) 89 Y( 1 ) 2 2 I = = 2 π = yes This transition is a first-forbidden transition. (b) 36 Cl(2 + ) 36 Ar(0 + ) I = 2 0 = 2 π = no This transition is a second-forbidden transition. (c) 26 Al(5 + ) 26 Mg*(2 + ) I = 5 2 = 3 π = no This transition is a second-forbidden transition. (d) 26 Si(0 + ) 26 Al*(0 + ) 26 Mg(0 + ) I =

15 = 2 π = no These transitions are both allowed transitions. (c) 97 Zr( 1 + ) 97 Nb*( 1 ) 2 2 I = = 0 π = yes This transition is a first-forbidden transition. 14

16 2 P5.6, 20% Krane, Problem 9.20, p From the data given in the problem, here is the decay scheme that I constructed. As you can see from the figure, the mass difference between the two ground states is Or, if you prefer the mass in amu, MeV u 15

Radioactivity Solutions - Lecture 28B (PHY315)

Radioactivity Solutions - Lecture 28B (PHY315) Radioactivity s - Lecture 8B (PHY35) Problem solutions.strategy In beta-minus decay, the atomic number Z increases by while the mass number A remains constant. Use Eq. (9-). 4 For the parent 9 K Z 9, so