measurement of nuclear radius
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Measurement of nuclear radius
• Four methods outlined for charge matter radius:– Diffraction scattering– Atomic x-rays– Muonic x-rays– Mirror Nuclides
Measurement of nuclear radius
• Three methods outlined for nuclear matter radius:– Rutherford scattering– Alpha particle decay -mesic x-rays
Diffraction scattering
• q = momentum transfer
α
ki
kf
α
ki
-kfq
€
rk i =
r k f ≡ k → q = 2k sin(α /2)
Diffraction scattering
• Measure the scattering intensity as a function of α to infer the distribution of charge in the nucleus,
€
rk i =
r k f ≡ k → q = 2k sin(α /2)
€
ρ ′ r ( )
€
Fr k i,
r k f( ) = ψ f
*V r( )∫ ψ i dv
F q( ) = e ir q •
r r ∫ V r( ) dv
V r( ) equation 3.4
F q( ) =4π
qsin q ′ r ( )∫ ρ e ′ r ( ) ′ r d ′ r
Diffraction scattering
• Measure the scattering intensity as a function of α 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?
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F q( ) =4π
qsin q ′ r ( )∫ ρe ′ r ( ) ′ r d ′ r
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F q( )2
€
ρe ′ r ( )
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F q( )
Diffraction scattering
• Density of electric charge in the nucleus is ≈ constant
€
ρe ′ r ( ) ≈ constant
ρe ′ r ( )∝A
4π R3
4π R3 ∝ A
R = Ro A1/3
Diffraction 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 =3
5R2
Atomic X-rays
• Assume the nucleus is uniform charged sphere.• Potential V is obtained in two regions:
– Inside the sphere
– Outside the sphere
€
′ V r( ) = −Ze2
4πε oR
3
2−
1
2
r
R
⎛
⎝ ⎜
⎞
⎠ ⎟2 ⎧
⎨ ⎪
⎩ ⎪
⎫ ⎬ ⎪
⎭ ⎪ r ≤ R
€
V r( ) = −Ze2
4πεor r ≥ R
Atomic 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 = ψ n*Vψ n dv∫
€
ψn
€
′ V = ψ n* ′ V ψ n dv
r<R∫ + ψn
*Vψ n dvr>R
∫
€
ψ1,1(1s), n=1, l =0
€
ψn
Atomic 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?
€
E1s =2
5
Z4e2
4πεo
R2
ao3
€
ψ1,1(1s)
€
E1s
€
E1s(pt)
€
E1s(sphere)
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EKα A( ) − EKα ′ A ( ) =
= E2 p A( ) − E1s A( )[ ] − 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? €
EKα A( ) − EKα ′ A ( ) =
= E2 p A( ) − E1s A( )[ ] − E2 p ′ A ( ) − E1s ′ A ( )[ ]
= E2 p A( ) − E2 p ′ A ( )[ ] − E1s A( ) − E1s ′ A ( )[ ]
€
EKα A( ) − EKα ′ A ( ) =
= ΔE1s ′ A ( ) − ΔE1s A( )[ ]
=2
5
Z 4e2
4π εo
1
ao3
Ro2 A2 /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?
€
EKα A( ) − 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 = 2Z
ao
⎛
⎝ ⎜
⎞
⎠ ⎟
3/2
e−
Zr
ao n =1,l = 0,m = 0
ao =4πεoh2
me2
En = −mZ 2e4
32π 2εo2h2n2
Coulomb 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 =3
5
Q2
4πεo R
ΔEC =3
5
e2
4πεo RZ 2 − Z −1( )
2[ ]
ΔEC =3
5
e2
4πεo R2Z −1( )
€
Z =A +1
2;N =
A −1
2
Z =A −1
2;N =
A +1
2
€
Z =A +1
2→ A = 2Z −1( )
€
EC =3
5
e2
4πεoRo
A2 /3
Measurement of nuclear radius
• Three methods outlined for nuclear matter radius:– Rutherford scattering– Alpha particle decay -mesic x-rays
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