Conservation problems and solutions for nuclear physics PDF

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Solutions to nuclear physics problems. Includes: conservation laws in nuclear reactions, binding energy, radioactivity, decay, potential barrier, decay constant, beta decay...

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P2241 Nuclear Physics Exercises for Ch. 4 (Nuclear Reactions) Solutions Conservation laws in nuclear reactions: 1. Fill in the blank for the particle(s) that conserve total charge and number of nucleons in the following nuclear reactions: (a) 197 ! 79Au + ! 126C → !206 85 At + ___ (b) 27 13 Al + p → ___ + n Solution Ztot,left = Ztot,right and Atot,left = Atot,right (a) left side: ZAu = 79, ZC = 6, Ztot,left = ZAu + ZC = 85, Atot.left = AAu + AC = 209 right side: ZAt = 85, Zblank = Ztot,left − ZAt = 0, Ablank = Atot,left − AAt = 3 the blank must be 3 neutrons (n + n + n) (b) left side: ZAl = 13, Zp = 1, Ztot,left = ZAl + Zp = 14, Atot.left = AAl + Ap = 28 right side: Zn = 0, Zblank = Ztot,left − Zn = 14, Ablank = Atot,left − An = 27 the blank must be 27Si

Energy release and change of binding energy in nuclear reactions: 2. For the reaction n + ! 147N → !146C + p! (a) Using the equations for the binding energy BN for 147 N and BC for 146C, together with the difference in binding energy as ∆B = BC − BN , show that the energy release in the reaction is given by Q = ∆B + mec2 , where me is the electron mass. (b) Use the semi-empirical mass formula to show for this reaction that: ! ΔB = a c(2Z N − 1)A−1/3 − 4a sym A−1(A − 2Z N + 1) + 2a p A−3/4 where ZN is the charge number of 14N Solution (a) ∆B = BC − BN = (ZCmH + NCmn − mC)c2 − (ZNmH + NNmn − mN)c2 with ZC = ZN − 1 and NC = NN + 1 and mH = mp + me ∆B = ((ZN − ZN − 1)mH + (NN − NN + 1)mn + mN − mC)c2 = (mN − mC − mp − me + mn)c2 Q = (mtot,i − mtot,f)c2 = (mN + mn − mC − mp)c2 = ∆B + mec2 (b) B ! = a v A − a s A2/3 − a c Z2 A−1/3 − a sym A−1(A − 2Z)2 ± a p A−3/4 with ZC = ZN − 1 and NC = NN + 1 and AC = AN = A and ∆B = BC − BN ΔB = (a v A − a v A) − (a s A2/3 − a s A2/3) − (a c ZC2A−1/3 − a c Z N2 A−1/3)

! −(a sym A−1(A − 2Z C)2 − a sym A−1(A − 2Z N )2) + a p A−3/4 + a p A−3/ ! = − a c[(Z N − 1)2 − ZN2 ]A−1/3 ! −(a sym A−1(A − 2(Z N − 1))2 − a sym A−1(A − 2Z N )2) + 2a p A−3/ ! = a c(2Z N − 1)A−1/3 − 4a sym A−1(A − 2Z N + 1) + 2a p A−3/4

Radioactivity: 3. The radioactive isotope 90Sr undergoes β− decay into 90Yr with a half-life of 28.8 years. We will compare with 85Kr which β− decays into 85Rb with a half-life of 10.76 years. (a) Explain why the decay constant λ gives the (negative) slope of ln(N) vs. time, where N is the number of parent (undecayed) nuclei. Also, calculate λ for both 90Sr and 85Kr. (b) Starting with an initial sample N0 = 100 of both 90Sr and 85Kr, use the empty graph below (or create one) to draw N vs. t for both 90Sr and 85Kr on the same semi-log graph. Solution (a) starting with a number of undecayed nuclei N0 at time t0, the number of undecayed nuclei N1 at a later time t1 is given by: N1 = N0⋅e−λ(t1 − t0) −λ = ln(N1 / N0) / (t1 − t0) = ( ln(N1) − ln(N0) ) / (t1 − t0) which is the slope of ln(N) vs. t the half-life T1/2 = t1 − t0 when N1 = (1/2)N0 , so λ = −ln((1/2)N0 / N0) / T1/2 = ln(2) / T1/2 λSr = ln(2) / T1/2,Sr = 0.0241 [yr−1] , λSr = ln(2) / T1/2,Kr = 0.0644 [yr−1]

Alpha (α) Decay: 4. For the α decay of 226Ra → 222Rn + 4He calculate the kinetic energy of 222Rn. The rest masses are: m(226Ra) = 226.0254 u, m(222Rn) = 222.0176 u, and m(4He) = 4.0026 u. What fraction of the total energy Q is the kinetic energy of 222Rn ?

Solution first, calculate the Q value given by Q = (mX − mY − mα)c2 = KY + Kα and use the conversion 1 u = 931.5 MeV/c2 Q = (226.0254 − 222.0176 − 4.0026)c2(931.5 [MeV/c2]) = 4.84 [MeV] since this energy is non-relativistic for 222Rn and α we can use the Newtonian conservation of momentum and the Q formula: ! pY = mYvY = pα = mαvα so vα = (mY / mα)vY Q = KY + Kα = (1/2)mYvY2 + (1/2)mαvα2 = (1/2)mYvY2 + (1/2)mα((mY / mα)vY)2 = (1/2)mYvY2(1 + (mY/mα)) = KY(1 + (mY/mα)) KY = Q / (1 + (mY/mα)) for 222Rn: K = 4.84 [MeV] / (1 + (222.0176 [u] / 4.0026 [u]) = 0.0857 [MeV] this is a fraction K / Q = 0.0177 5. The α decay of 232U → 228Th + 4He has a total energy release Q = 5.46 MeV. The potential energy inside the parent nucleus has a depth of V0 ≈ 35 MeV. (a) Calculate the frequency f at which the α particle collides with the potential barrier at the edge of the parent nucleus (r = a), given by a ≈ r0A1/3, where r0 = 1.2 fm, and A = 232. (b) Calculate the probability P for the α particle penetrating through the potential barrier. (c) Calculate the half-life T1/2 using the estimated decay constant λ. Solution (a) the frequency f = v / a the radius a = r0A1/3 = (1.2 [fm])(232)1/3 = 7.37 [fm] calculate the velocity v from the kinetic energy inside the parent nucleus: K = Q + V0 ≈ 40.5 [MeV], which is just in the relativistic regime for an α particle the approach from Exercise Sheet 3 problem 2 can be used to calculate the velocity: K = (γ − 1)mc2 , E = K + ER = γmc2 , ER = mc2 , thus γ = (K + ER) / ER = (40.5 + 3727.3 [MeV]) / (3727.3 [MeV]) = 1.0109 β = v/c = (γ2 − 1)1/2 / γ = 0.146 , v = βc = (0.146)(3×108 [m/s]) = 4.39×107 [m/s] f = (4.39×107 [m/s]) / (7.37×10−15 [m]) = 5.95×1021 [s−1] (b) the probability P is given by: P = e −2k 2(1/2)(b−a) where k2 = ((2m/ħ2)(1/2)(B − Q))1/2 ħc = 197.3 [MeV fm] and m = 3728.4 [MeV/c2] e2 z(Z − z )e 2 z(Z − z )e 2 with ! B = and ! b = and ! = 1.44 [eV nm] 4π ϵ0 a 4π ϵ0Q 4π ϵ0 α particle: z = 2, Th: Z = 92, Ra: (Z − z) = 90 B = 2(90)(1.44×10−15 [MeV m]) / (7.37×10−15 [m]) = 35.2 [MeV] k2 = (3728.4 [MeV] ⋅ (1/197.3)2 [MeV−2 fm−2] ⋅ (35.2 − 5.46 [MeV]))1/2 = 1.69 [fm−1] b = 2(90)(1.44×10−15 [MeV m]) / (5.46 [MeV]) = 47.5 [fm]

P = exp(−1.69 [fm−1] ⋅ (47.5 − 7.37 [fm])) = exp(−67.8) = 3.52×10−30 (c) the decay constant λ = f⋅P = 5.95×1021 [s−1] ⋅ 3.52×10−30 = 2.09×10−8 [s−1] T1/2 = ln(2) / λ = ln(2) ⋅ (1 / 2.09×10−8) [s] = 3.31×107 [s] = 3.31×107 [s] ⋅ (1 [yr] / 3.154×107 [s]) = 1.05 [yr]

Beta (β) Decay: 6. For the β+ decay of 127N (a) Write out the reaction. (b) Calculate the largest neutrino energy possible from this reaction when ! 127N is at rest. The atomic mass of 127 N is m(12N) = 12.01861 u. A positron’s mass is m(e) = 0.00055 u. Solution (a) the reaction is 127N → 126C + e+ + ν (b) the largest neutrino energy in this reaction is when the e+ kinetic energy is Ke ≈ 0 m(12C) = 12.0 u as defined, and the reaction energy for this β+ decay is: Q = [m(12N) − m(12C) − 2m(e)]c2 = [12.01861 − 12.0 − 0.0011]c2(931.5 [MeV/c2]) = 16.3 MeV since Q = Kf − Ki with Ki = 0 and the maximum Kν is when Ke ≈ 0 this gives Q = Kν ≈ Eν since the neutrino’s rest mass is so small, thus Eν = 16.3 MeV....