Nuclear Masses and Binding Energy Lesson 3. Nuclear Masses Nuclear masses and atomic masses Because B electron (Z)is so small, it is neglected in most.
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Nuclear Masses and Binding Energy
Lesson 3
Nuclear Masses
• Nuclear masses and atomic masses
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mnuclc2 = Matomicc
2 −[Zmelectronc2 + Belectron (Z)]
Belectron (Z) =15.73Z 7 / 3eV
Because Belectron(Z)is so small, it is neglected in most situations.
Mass Changes in Beta Decay
- decay
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14C→14N + β − + ν e
Energy = [(m(14C) + 6melectron ) − (m(14N) + 6melectron ) − m(β −)]c 2
Energy = [M(14C) − M(14N)]c 2
+ decay
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64Cu→64Ni− + β + + ν e
Energy = [(m(64Cu) + 29melectron ) − (m(64Ni) + 28melectron ) − melectron − m(β +)]c 2
Energy = [M(64Cu) − M(64Ni) − 2melectron ]c 2
Mass Changes in Beta Decay
• EC decay
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207Bi+ + e−→207Pb+ ν e
Energy = [(m(207Bi) + 83melectron ) − (m(207Pb) + 82melectron )]c 2
Energy = [M(207Bi) − M(207Pb)]c 2
Conclusion: All calculations can be done with atomic masses
Nomenclature
• Sign convention:Q=(massesreactants-massesproducts)c2
Q has the opposite sign as HQ=+ exothermicQ=- endothermic
Nomenclature
• Total binding energy, Btot(A,Z)
Btot(A,Z)=[Z(M(1H))+(A-Z)M(n)-M(A,Z)]c2
• Binding energy per nucleonBave(A,Z)= Btot(A,Z)/A
• Mass excess ()M(A,Z)-ASee appendix of book for mass tables
Nomenclature
• Packing fraction(M-A)/A
• Separation energy, SSn=[M(A-1,Z)+M(n)-M(A,Z)]c2
Sp=[M(A-1,Z-1)+M(1H)-M(A,Z)]c2
Binding energy per nucleon
Separation energy systematics
Abundances
Semi-empirical mass equation
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Btot (A,Z) = avA − asA2 / 3 − ac
Z 2
A1/ 3− aa
(A − 2Z)2
A± δ
Terms
•Volume avA•Surface -asA2/3
•Coulomb -acZ2/A1/3
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ECoulomb =3
5
Z 2e2
R
R =1.2A1/ 3
ECoulomb = 0.72Z 2
A1/ 3
Asymmetry term
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−aa
(A − 2Z)2
A= −aa
(N − Z)2
A
To make AZ from Z=N=A/2, need to move q protons qin energy, thus the work involvedis q2=(N-Z)2/4. If we add that=1/A, we are done.
Pairing term
A Z N # stable
e e e 201
o e o 69
o o e 61
e o o 4
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δ =+11A−1/ 2K ee
δ = 0K oe,eo
δ = −11A−1/ 2K oo
Relative importance of terms
Values of coefficients
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av =15.56MeV
as =17.23MeV
ac = 0.7MeV
aa = 23.285MeV
Modern version of semi-empirical mass equation (Myers and
Swiatecki)
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Btot (A,Z) = c1A 1− kN − Z
A
⎛
⎝ ⎜
⎞
⎠ ⎟2 ⎡
⎣ ⎢
⎤
⎦ ⎥− c2A
2 / 3 1− kN − Z
A
⎛
⎝ ⎜
⎞
⎠ ⎟2 ⎡
⎣ ⎢
⎤
⎦ ⎥− c3
Z 2
A1/ 3+ c4
Z 2
A+ δ
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c1 =15.677MeV
c2 =18.56MeV
c3 = 0.717MeV
c4 =1.211MeV
k =1.79
δ =11A−1/ 2
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Mass parabolas and Valley of beta
stability
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M(Z,A) = Z • M(1H) + (A − Z)M(n) − Btot (Z,A)
Btot (Z,A) = avA − asA2 / 3 − ac
Z 2
A1/ 3− aa
(A − 2Z)2
A
aa
(A − 2Z)2
A= aa
A2 − 4AZ + 4Z 2
A= aa A − 4Z +
4Z 2
A
⎛
⎝ ⎜
⎞
⎠ ⎟
M = A M(n) − av +as
A1/ 3+ aa
⎡ ⎣ ⎢
⎤ ⎦ ⎥+ Z M(1H) − M(n) − 4Zaa[ ] + Z 2 ac
A1/ 3+
4aa
A
⎛
⎝ ⎜
⎞
⎠ ⎟
This is the equation of a parabola, a+bZ+cZ2
Where is the minimum of the parabolas?
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∂M∂Z
⎛
⎝ ⎜
⎞
⎠ ⎟A
= 0 = b+ 2cZA
ZA =−b
2c=M(1H) − M(n) − 4aa
2ac
A1/ 3+
4aa
A
⎛
⎝ ⎜
⎞
⎠ ⎟
ZA
A≈
1
2
81
80 + 0.6A2 / 3
Valley of Beta Stability
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