Nuclear Reactions KinematicsPowerPoint PresentationSlide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Nuclear ReactionsKinematics€ d +6Li → α + α€ Td= 2MeV€ θ€ φ € r p α1 € r p d € r p α2 € pd= 2mdTdvd=2Tdmdr p cm=r p d+r p Lir V cm=r p dmd+ mLi € r p d*+r p Li*= 0Td*=p*( )22mdTLi*=p*( )22mLiT*= Td*+ TLi* € r v d=r v d*+r V cmr p d*=r p d− mdr V cmr p d*=r p d− mdr p dmd+ mLir p d*=r p dmLimd+ mLi= md'r v d€ d +6Li → α + α€ Td= 2MeV€ θ€ φ € r p α1 € r p d € r p α2€ Ti*mdc2 + mLic2E*mBec20.0€ Tf*m c2 + mc2€ d +6Li → α + α € pd= 2mdTd=2mdc2Tdc MeVc ⎛ ⎝ ⎜ ⎞ ⎠ ⎟vd=2Tdmd→ βd≡vdc=2Tdmdc2r p cm=r p d+r p Li MeV/c( )r V cm=r p dmd+ mLi→ βcm≡r V cmc=r p dcmdc2+ mLic2Work in units of MeV€ d +6Li → α + α€ Ei= EfMdc2+ MLic2+ Td= Mαc2+ Mαc2+ Tα1+ Tα2Calculate the Q-valueLet M-values = nuclear masses€ mZ ,Ac2= MNc2− Z mec2+ Bii=1Z∑m-values = isotopic massesConservation of electric charge  total me values before/after are equal; therefore subtract to zero.€ m1,2c2+ m3,6c2+ Td≈ m2,4c2+ m2,4c2+ Tα1+ Tα2Approximation…€ d +6Li → α + αCalculate the Q-value€ m1,2c2+ m3,6c2+ Td≈ m2,4c2+ m2,4c2+ Tα1+ Tα2€ Q = m1,2c2+ m3,6c2− m2,4c2− m2,4c2= Tα1+ Tα2− Tdinitial state final state€ Bii=1Z =1∑+ Bii=1Z = 3∑≈ Bii=1Z = 2∑+ Bii=1Z = 2∑Appproximation…€ Q = Δ1,2+ Δ3,6− Δ2,4− Δ2,4mass defects€ d +6Li → α + αCalculate the Q-value€ Q = m1,2c2+ m3,6c2− m2,4c2+ m2,4c2= Tα1+ Tα2− Td€ Q > 0 → m1,2c2+ m3,6c2> m2,4c2+ m2,4c2 ; Tα1+ Tα2− Td> 0€ Q < 0 → m1,2c2+ m3,6c2< m2,4c2+ m2,4c2 ; Tα1+ Tα2− Td< 0Q < 0 cannot occur spontaneously - threshold Td € Q = m1,2c2+ m3,6c2− m2,4c2+ m2,4c2= Tα1*+ Tα2*− Td*€ Q = Tα1*+ Tα2*− Td (min)*€ d +6Li → α + αCross section€ I x( )= I 0( )e−Ndσ x€ Y = I 0( )Ndσ xσ =YI 0( )Ndxσ E,θ( )=Y E,θ( )I 0( )Ndx€ I x( )≈ I 0( )1− Ndσ x...[ ]€ Ndx ≡areal target densityx€ I 0( )€ I x( )€ Ndtarget density (#/cm3)€ Y x( )= I 0( )− I x( )Y is“yield”€ d +6Li → α + αCross section€ I x( )= I 0( )e−Ndσ x€ Y x( )= I 0( )− I x( )€ Y = I 0( )Ndσ xσ =YI 0( )Ndxσ E,θ( )=Y E,θ( )I 0( )Ndx€ I x( )≈ I 0( )1− Ndσ x...[ ]x€ I 0( )€ I x( )detector€ dσ E,θ( )dΩ=1I 0( )NdxdY E,θ( )dΩdY E,θ( )dΩ≈ΔC E,θ( )ΔΩ€ Ω ≡4πR2R2= 4πΩ =total areaR2ΔΩ =detector areaR2€ d +6Li → α + αCross sectionx€ I 0( )€ I x( )detector* to *P1*P2*Pcm € r v α1=r v α1*+r V cm€ vα1(x )= vα1(x )*+ Vcmx*y*€ vα1(y )= vα1(y