1000 Solved Problems in Modern Physics

464 8 Nuclear Physics – II Eα= QMLi Mα+MLi = 2. 79 × 7. 018 4. 004 + 7. 018 = 1 .78 MeV ELi=Q−Eα=^2.^79 −^1.^78 =^1 .01 MeV 8.51 ...

8.3 Solutions 465 8.54 Inelastic scattering is like an endoergic reaction except that the identity of particles is unchanged. Q= ...

466 8 Nuclear Physics – II 8.58 E(He)+Eγ= 5 .3 MeV (energy conservation) (1) PHe= √ 2 M(He)E(He)=Pγ=Eγ/c (momentum conservation) ...

8.3 Solutions 467 Fig. 8.11Energy levels 8.61 For the nuclear reaction (^3) H+p→ (^3) He+n− 0 .7637 MeV (1) (^3) H− (^3) He=n−p− ...

468 8 Nuclear Physics – II When the target and the projectile are interchanged, the same excitation energyW∗produced withTais gi ...

8.3 Solutions 469 8.68 dσ dΩ = I I 0 NdΩ =A+Bcos^2 θ I 0 =^8 ×^1012 /m^2 −s N=number of target atoms intercepting the beam = N 0 ...

470 8 Nuclear Physics – II Ignoring the statistical factor g, at resonance σt= 7 , 000 × 10 −^24 cm^2 = λ^2 π Γs Γ (4) Butλ= 0. ...

8.3 Solutions 471 For the forward reaction, energy available in the CMS is E∗=Q+Td∗+TN∗=Q+ TdmN mN+md = 13. 57 + 20 × 14 14 + 2 ...

472 8 Nuclear Physics – II where the first term on the right hand side denotes the absorption rate of neu- trons in^23 Na, and i ...

8.3 Solutions 473 Writing the Laplacian for spherical geometry (1) becomes d^2 φ dr^2 + 2 r dφ dr −K^2 φ=0(2) Equation (2) is ea ...

474 8 Nuclear Physics – II The counting rate, R=Nσνn(r) (10) As the absorption obeys the 1/vlaw, the productσ.v=constant. We the ...

8.3 Solutions 475 Area of the unit cellA=πr 12 Now, Σam Σau = Nm Nu × σam σau = ρm/Am ρu/Au × σam σau = 1. 62 / 12 18. 7 / 238 × ...

476 8 Nuclear Physics – II The actual radius will be shorter byd= 0. 71 λtrwhered=extrapolated distance. 8.81 Letnfissions take ...

8.3 Solutions 477 Number of^235 Uatoms/cm^3 , N= N 0 ρ A × 0. 7 100 = 6 × 1023 × 19 238 × 0. 7 100 = 3. 353 × 1020 Σa=σaN= 550 × ...

478 8 Nuclear Physics – II Writing sinθ∗dθ∗=−d(cosθ∗), and using (3), (4) becomes 〈 ln E 1 E 2 〉 = ∫E^1 0 ln ( E 1 E 2 ) dE 2 E ...

8.3 Solutions 479 We can now calculatep, the resonance escape probability. Σs/N 0 = Σs(U)+Σs(m) N 0 =σs(U)+ Nm Nu σa(M) = 8. 3 + ...

480 8 Nuclear Physics – II At the surface the neutron density corresponding tor= 9 .6 cm and mean neutron velocity 2. 2 × 105 cm ...

8.3 Solutions 481 Fig. 8.14 The net current, J=− λtr 3 ∂φ ∂x =A 2 K λtr 3 e−kx Asx→ 0 ,J= Q 2 =A 2 K λtr 3 whenceA 2 = 3 Q 2 Kλt ...

482 8 Nuclear Physics – II Σa(graphite)= 1 λa(C) = 1 2 , 700 = 3. 7 × 10 −^4 cm−^1 Σa=Σa(C)+Σa(B)=Σa(C)+σa(B)N = 3. 7 × 10 −^4 + ...

8.3 Solutions 483 8.3.15 Fusion ........................................... 8.93 The minimum energy of neutrino is zero whendand ...