measurement of nuclear radius

Measurement of nuclear radius

• Four methods outlined for charge matter radius:– Diffraction scattering– Atomic x-rays– Muonic x-rays– Mirror Nuclides


• 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

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rk i =

r k f ≡ k → q = 2k sin(α /2)


• Measure the scattering intensity as a function of α to infer the distribution of charge in the nucleus,

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rk i =

r k f ≡ k → q = 2k sin(α /2)

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ρ ′ r ( )

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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


• 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

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ρe ′ r ( )

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F q( )


• Density of electric charge in the nucleus is ≈ constant

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ρe ′ r ( ) ≈ constant

ρe ′ r ( )∝A

4π R3

4π R3 ∝ A

R = Ro A1/3


• 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

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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

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′ 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)

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V = ψ n*Vψ n dv∫

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ψn

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′ V = ψ n* ′ V ψ n dv

r<R∫ + ψn

*Vψ n dvr>R

∫

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ψ1,1(1s), n=1, l =0

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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?

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E1s =2

5

Z4e2

4πεo

R2

ao3

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ψ1,1(1s)

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E1s

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E1s(pt)

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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 ( )[ ]

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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?

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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?

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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?

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EC =3

5

Q2

4πεo R

ΔEC =3

5

e2

4πεo RZ 2 − Z −1( )

2[ ]

ΔEC =3

5

e2

4πεo R2Z −1( )

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Z =A +1

2;N =

A −1

2

Z =A −1

2;N =

A +1

2

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Z =A +1

2→ A = 2Z −1( )

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EC =3

5

e2

4πεoRo

A2 /3


• Three methods outlined for nuclear matter radius:– Rutherford scattering– Alpha particle decay -mesic x-rays

measurement of nuclear radius

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