Measurement of nuclear radiusSlide 2Diffraction scatteringSlide 4Slide 5Slide 6Slide 7Atomic X-raysSlide 9Slide 10Slide 11Slide 12Muonic X-raysCoulomb Energy DifferencesSlide 15Measurement of nuclear radius•Four methods outlined for charge matter radius:–Diffraction scattering–Atomic x-rays–Muonic x-rays–Mirror NuclidesMeasurement of nuclear radius•Three methods outlined for nuclear matter radius:–Rutherford scattering–Alpha particle decay-mesic x-raysDiffraction scattering•q = momentum transferαkikfαki-kfq € r k i=r k f≡ k → q = 2k sin(α /2)Diffraction scattering•Measure the scattering intensity as a function of  to infer the distribution of charge in the nucleus, € r k i=r k f≡ k → q = 2k sin(α /2)€ ρ′ r ( ) € Fr k i,r k f( )= ψf*V r( )∫ψidvF q( )= eir q •r r ∫V r( )dvV r( ) equation 3.4F q( )=4πqsin q ′r ( )∫ρe ′r ( ) ′r d ′rDiffraction scattering•Measure the scattering intensity as a function of  to infer the distribution of charge in the nucleus • is the inverse Fourier transform of • is known as the form factor for the scattering.•c.f. Figure 3.4; what is learned from this?€ F q( )=4πqsin q ′r ( )∫ρe ′r ( ) ′r d ′r € F q( )2€ ρe ′r ( )€ F q( )Diffraction scattering•Density of electric charge in the nucleus is ≈ constant € ρe ′r ( )≈ constantρe ′r ( )∝A4πR34πR3∝ AR = RoA1/ 3Diffraction scattering•The charge distribution does not have a sharp boundary–Edge of nucleus is diffuse - “skin”–Depth of the skin ≈ 2.3 f–RMS radius is calculated from the charge distribution and, neglecting the skin, it is easy to show € r2=35R2Atomic X-rays•Assume the nucleus is uniform charged sphere.•Potential V is obtained in two regions:–Inside the sphere–Outside the sphere€ ′ V r( )= −Ze24πεoR32−12rR ⎛ ⎝ ⎜ ⎞ ⎠ ⎟2 ⎧ ⎨ ⎪ ⎩ ⎪ ⎫ ⎬ ⎪ ⎭ ⎪ r ≤ R€ V r( )= −Ze24πεor r ≥ RAtomic X-rays•For an electron in a given state, its energy depends on -•Assume does not change appreciably if Vpt Vsphere•Then, E = Esphere - Ept•Assume can be giving (3.12) € V = ψn*Vψndv∫€ ψn€ ′ V = ψn* ′V ψndvr<R∫+ ψn*Vψndvr>R∫ € ψ1,1(1s), n=1, l =0€ ψnAtomic X-raysE between sphere and point nucleus for •Compare this E to measurement and we have R.•Problem!•We will need two measurements to get R --•Consider a 2p 1s transition for (Z,A) and (Z,A’) where A’ = (A-1) or (A+1) ; what x-ray does this give?€ ΔE1s=25Z4e24πεoR2ao3€ ψ1,1(1s)€ ΔE1s€ E1s(pt)€ E1s(sphere)€ EKαA( )− EKα ′A ( )= = E2 pA( )− E1sA( )[ ]− E2 p ′A ( )− E1s ′A ( )[ ]Atomic X-rays•Assume that the first term will be ≈ 0. Why? •Then, use E1s from (3.13) for each E1s term. Why? € EKαA( )− EKα ′A ( )= = E2 pA( )− E1sA( )[ ]− E2 p ′A ( )− E1s ′A ( )[ ] = E2 pA( )− E2 p ′A ( )[ ]− E1sA( )− E1s ′A ( )[ ]€ EKαA( )− EKα ′A ( )= = ΔE1s ′A ( )− ΔE1sA( )[ ] =25Z4e24πεo1ao3Ro2A2 / 3− ′A 2 / 3( )Atomic X-rays•This x-ray energy difference is called the “isotope shift”•We assumed that R = Ro A1/3. Is there any authentication?•How good does your spectrometer have to be to see the effect?•We assumed we could use hydrogen-line 1s wavefunctions Are these good enough to get good results?•Can you use optical transitions instead of x-ray transitions?€ EKαA( )− EKα ′A ( )Muonic X-rays•Compare this process with atomic (electronic) x-rays:–Similarities–Differences–Advantages–Disadvantages•What is ao ? •Pauli Exclusion principle for muons, electrons? € ψn,l,m= 2Zao ⎛ ⎝ ⎜ ⎞ ⎠ ⎟3/ 2e−Zrao n =1,l = 0,m = 0ao=4πεoh2me2En= −mZ2e432π2εo2h2n2Coulomb Energy Differences•Calaulate the Coulomb energy of the charge distribution directly Consider mirror nuclides: Measure EC; How? Assume R is same for both nuclides. Why? € EC=35Q24πεoRΔEC=35e24πεoRZ2− Z −1( )2[ ]ΔEC=35e24πεoR2Z −1( )€ Z =A +12;N =A −12Z =A −12;N =A +12€ Z =A +12→ A = 2Z −1( )€ ΔEC=35e24πεoRoA2 / 3Measurement of nuclear radius•Three methods outlined for nuclear matter radius:–Rutherford scattering–Alpha particle decay-mesic