##
This **preview** shows page *1-2-3-4-5*
out of 16 **pages**.

*View Full Document*

End of preview. Want to read all 16 pages?

Upload your study docs or become a GradeBuddy member to access this document.

View Full Document**Unformatted text preview:**

Measurements of thenuclear radiusMeasurement of nuclear radius• Four methods outlined for charge matterradius:– Diffraction scattering– Atomic x-rays– Muonic x-rays– Mirror NuclidesMeasurement of nuclear radius• Three methods outlined for nuclear matterradius:– Rutherford scattering– Alpha particle decay– p-mesic x-raysDiffraction scattering• q = momentum transferakikfaki-kfq † r k i=r k f≡ k Æ q = 2k sin(a/2)Diffraction scattering• Measure the scattering intensity as a function of a to inferthe distribution of charge in the nucleus, † r k i=r k f≡ k Æ q = 2k sin(a/2)† r¢ r ( ) † Fr k i,r k f( )=yf*V r( )ÚyidvF q( )= eir q •r r ÚV r( )dvV r( ) equation 3.4F q( )=4pqsin q¢ r ( )Úre¢ r ( )¢ r d¢ rDiffraction scattering• Measure the scattering intensity as a function of ato 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( )=4pqsin q¢ r ( )Úre¢ r ( )¢ r d¢ r † F q( )2† re¢ r ( )† F q( )Diffraction scattering• Density of electric charge in the nucleus is ≈ constant† re¢ r ( )ª constantre¢ r ( )µA4pR34pR3µ 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( )= -Ze24peoR32-12rRÊ Ë Á ˆ ¯ ˜ 2Ï Ì Ô Ó Ô ¸ ˝ Ô ˛ Ô r £ R† V r( )= -Ze24peor r ≥ RAtomic X-rays• For an electron in a given state, its energy depends on -• Assume does not change appreciably if VptÆ Vsphere• Then, DE = Esphere - Ept• Assume can be giving (3.12)† V =yn*VyndvÚ† yn† ¢ V =yn*¢ V yndvr<RÚ+yn*Vyndvr>RÚ † y1,1(1s), n=1, l=0† ynAtomic X-rays•DE between sphere and point nucleus for• Compare this DE to measurement and we have R.• Problem!• We will need two measurements to get R --• Consider a 2p Æ1s transition for (Z,A) and (Z,A’) where A’ = (A-1) or (A+1) ; what x-ray does this give?† DE1s=25Z4e24peoR2ao3† y1,1(1s)† DE1s† E1s( pt)† E1s(sphere)† EKaA( )- EKa¢ A ( )= = E2 pA( )- E1sA( )[ ]- E2 p¢ A ( )- E1s¢ A ( )[ ]Atomic X-rays• Assume that the first term will be ≈ 0. Why?• Then, use DE1s from (3.13) for each E1s term. Why?† EKaA( )- EKa¢ A ( )= = E2 pA( )- E1sA( )[ ]- E2 p¢ A ( )- E1s¢ A ( )[ ] = E2 pA( )- E2 p¢ A ( )[ ]- E1sA( )- E1s¢ A ( )[ ]† EKaA( )- EKa¢ A ( )= = DE1s¢ A ( )- DE1sA( )[ ] =25Z4e24peo1ao3Ro2A2 / 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 thesegood enough to get good results?• Can you use optical transitions instead of x-ray transitions?† EKaA( )- EKa¢ 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? † yn,l ,m= 2ZaoÊ Ë Á ˆ ¯ ˜ 3/ 2e-Zrao n = 1,l = 0,m = 0ao=4peoh2m e2En= -mZ2e432p2eo2h2n2Coulomb Energy Differences• Calaulate the Coulomb energy of the charge distribution directly Consider mirror nuclides: Measure DEC; How? Assume R is same for both nuclides. Why?† EC=35Q24peoRDEC=35e24peoRZ2- Z -1( )2[ ]DEC=35e24peoR2Z -1( )† Z =A + 12;N =A -12Z =A -12;N =A + 12† Z =A + 12Æ A = 2Z -1( )† DEC=35e24peoRoA2 / 3Measurement of nuclear radius• Three methods outlined for nuclear matterradius:– Rutherford scattering– Alpha particle decay– p-mesic

View Full Document