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Mirror Nuclei 1.2) Determination of the Asymmetry Term from Line of Maximum Β Stability 1.3) Estimate of the Pairing Term

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Mirror Nuclei 1.2) Determination of the Asymmetry Term from Line of Maximum Β Stability 1.3) Estimate of the Pairing Term Selected Topics in Physics a lecture course for 1st year students by W.B. von Schlippe Spring Semester 2007 Lecture 11 1.) Determination of parameters of the SEMF 2.) α decay 3.) Nuclear energy levels 4.) Spontaneous fission 1 1.) Determination of parameters of the SEMF 1.1) Determination of the Coulomb term from mirror nuclei 1.2) Determination of the asymmetry term from line of maximum β stability 1.3) Estimate of the pairing term 1.1) Determination of the Coulomb term from mirror nuclei In Lecture 9 we have derived a formula for the binding energy, based on the liquid drop model: 2 2 ZZ−−1 N Z 3 ()() BAZ(, )=− aA12 aA − a 31 − a4 +δ A 3 A where the terms are, in that order: volume term, surface term, Coulomb term, asymmetry term and pairing term. We can see that the volume and surface terms are identical for isobars. Therefore, if we compare the binding energies of two isobars, then these terms cancel. 2 If we choose these isobars to be odd-A mirror nuclides, then the asymmetry term also cancels. Mirror nuclei: Two nuclides are called mirror nuclei if they have equal mass numbers A, and if the number of protons Z in one of them is equal to the number of neutrons N in the other. Examples of mirror nuclei: 11 11 boron-11and carbon-11 : 56BC− 13 13 carbon-13 and nitrogen-13: 67CN− If we define the neutron excess of a nucleus by TNZ= − (also called isotopic number), then we have for odd-A mirror nuclei TZ= ±=1, and hence N±1 3 Thus these mirror nuclides lie next to the line N = Z in the Segre chart. Now we know from the Segre chart that the band of stability moves with increasing A away from the line N = Z towards neutron-rich nuclides, and unstable nuclides are never very far away from the band of stability. Therefore we must expect that mirror nuclides are found only up to mass numbers A around 60. All mirror nuclides with A >10 are shown in the following table 4 Mirror nuclides from A = 11 to A = 63: there are no mirror nuclides with A > 63 N - Z = +1 N - Z = -1 N - Z = +1 N - Z = -1 A Nuclide Z Delta Nuclide Z Delta A Nuclide Z Delta Nuclide Z Delta 11 B 5 8.668 C 6 10.650 39 K 19 -33.807 Ca 20 -27.274 13 C 6 3.125 N 7 5.345 41 Ca 20 -35.135 Sc 21 -28.642 15 N 7 0.101 O 8 2.856 43 Sc 21 -36.188 Ti 22 -29.321 17 O 8 -0.809 F 9 1.952 45 Ti 22 -39.006 V 23 -31.880 19 F 9 -1.487 Ne 10 1.751 47 V 23 -42.002 Cr 24 -34.560 21 Ne 10 -5.732 Na 11 -2.184 49 Cr 24 -45.331 Mn 25 -37.620 23 Na 11 -9.530 Mg 12 -5.474 51 Mn 25 -48.241 Fe 26 -40.220 25 Mg 12 -13.193 Al 13 -8.916 53 Fe 26 -50.945 Co 27 -42.640 27 Al 13 -17.197 Si 14 -12.384 55 Co 27 -54.028 Ni 28 -45.340 29 Si 14 -21.895 P 15 -16.953 57 Ni 28 -56.082 Cu 29 -47.310 31 P 15 -24.441 S 16 -19.045 59 Cu 29 -56.357 Zn 30 -47.260 33 S 16 -26.586 Cl 17 -21.003 61 Zn 30 -56.350 Ga 31 -47.090 35 Cl 17 -29.014 Ar 18 -23.047 63 Ga 31 -56.547 Ge 32 -46.900 37 Ar 18 -30.948 K 19 -24.800 Delta is the mass excess in MeV; the values are from the BNL Wallet Cards 2005, http://www.nndc.bnl.gov/wallet/ 5 Let us now compare the binding energies of mirror nuclides using the SEMF: we have 2 2 ZZ−−1 N Z 3 ()() BAZ(, )=− aA12 aA − a 31 − a4 +δ A 3 A or in terms of A and T = N – Z: 2 2 AT−−− AT 2 3 1 ( )( ) T BAT(,)=− aA12 aA − a 3 1 −a4 +δ 4 A 3 A hence Δ≡BBAT(, =+− 1) BAT (, =− 1) 1 ()()()()AA−13−− AA + 11 − =− a3 1 4 A 3 − 1 3 =−aA3 ()1 A 6 Thus, if we plot the measured differences of the binding energies as a function of (A-1)/A1/3, then we expect a straight line through the origin. This is shown in the next figure. Determination of the Coulomb coefficient from mirror nuclei 12 y = 0.6448x 10 8 6 4 Delta E_Coulomb (MeV) 2 0 0 5 10 15 20 (A-1)/A^(1/3) 7 The slope of the line, which is constrained to pass through the origin, is the empirical coefficient a3 of the Coulomb term. The least squares fit gives a3 = 0.645MeV 1.2) Determination of the asymmetry term from line of maximum β stability In the previous lecture we have defined the position of maximum β stability: ∂MAZ( , ) = 0 ∂Z A= const Now, the mass M(A,Z) is given by 2 M ( AZ,,) =+−− ZmHn( A Z) m B( AZ) c hence 8 2213− mcnH−+ m c aA34 +4 a Zmin = −−13 1 28aA34+ aA thus the position of the minimum of the mass parabola is determined by the coefficients a3 (Coulomb term) and a4 (asymmetry term). Solving for a4 we get 22 −13 mcnH−+− m c a3m(12 Zin) A a4 = −1 84ZAmin − Empirically, fitting the masses of isobars with parabolas, we will get somewhat different values of a4. Then these values are averaged. From fits to only a few mass parabolas of mass numbers A = 15, 63, 65, 97, 101, 107 and 135 I have got the following result: a4 = 21.8± 1.3 MeV 9 1.3) Estimate of the pairing term The pairing term δ is nonzero only for even-A nuclides. In lecture 9 we have seen in the discussion of stability rules that of the 174 stable even-A nuclides, 166 are even-even and only 8 are odd-odd nuclides. Four of the odd-odd stable nuclides are light nuclides: deuterium, lithium, boron and nitrogen; they are absolutely stable and symmetric, i.e. N = Z. The remaining four odd-odd nuclides are unstable but have lifetimes greater then 109 years, i.e. greater than the age of the earth: 40 9 19 Ky1.3× 10 ears 50 17 23V 1.4× 10 years 138 11 57 La 1.1× 10 years 180m 15 73Ta >×1.2 10 years 10 We will be guided by these stability rules to estimate the pairing term δ To do this we rewrite the expression for the minimum of the mass parabola in terms of the neutron excess T: −13 aA3 (1)−−− A( mnH m) TAZmin =−2 min = −−13 1 aA34+ 4 aA and then use this to get the following result: 22 Δ≡UMATcMATc( ,,) − ( min ) ⎛⎞1 −−13 1 2 =+−⎜⎟aA34 aA() T T min ⎝⎠4 Then we apply this to even-A nuclides. The stability rules suggest that the mass parabolas of the odd-odd nuclides are shifted upwards from the mass parabolas of the even-even nuclides. We denote the separation between these parabolas by δ. 11 Mass parabolas for even-A nuclides: the energy U=Mc2 is plotted vs T (Figures from Blatt and Weisskopf; their notation differs from ours: they write δ/A instead of our δ, also 4uC = a3 and 4uτ = a4) 12 Shown on the left is the situation where there is one stable odd-odd nuclide at T = Tmin and no stable even-even nuclides. Here ⎛⎞1 −−13 1 δ <+⎜⎟aA34 aA ⎝⎠4 From the stability rules we know that this does not happen in nature. Therefore we conclude that ⎛⎞1 −−13 1 δ >+⎜⎟aA34 aA ⎝⎠4 On the right-hand figure there are three stable even-even nuclides, the middle one at T = Tmin. Their stable even-even neighbors are separated by 2 units in T, and hence their energy difference is ⎛⎞1 −−13 1 δUaAaA=+4⎜⎟34 ⎝⎠4 13 The unstable odd-odd nuclides lie above the minimum of their mass parabola by ΔU with T - Tmin = 1, and therefore above the minimum of the mass parabola of the even-even nuclides by δ + ΔU, thus ⎛⎞11−−13 1 ⎛⎞ −−13 1 δ ++>+⎜⎟aA34 aA4⎜⎟ aA 34 aA ⎝⎠44⎝⎠ hence ⎛⎞1 −−13 1 δ >+3⎜⎟aA34 aA ⎝⎠4 This situation occurs rarely in nature; we must therefore conclude, together with the previous result, that 11−−13 1 ⎛⎞−−13 1 aA34+<≤ aAδ 3⎜⎟ aA34 + aA 44⎝⎠ Now, we had δ = a5/A, and that gives us an estimate of a5: 1123 ⎛⎞23 aA34534+<≤ a a3⎜⎟ aA + a 44⎝⎠ 14 For a numerical estimate let us take an intermediate value of A = 100; then, with our values for a3 and a4 we get 32< a5 ≤ 96 but for small values of A the limits get tighter: if we put A = 1, then 22< a5 ≤ 65 We must conclude that this parameter can be determined only semi-quantitatively. But then we must remember that the basis of the SEMF, the liquid drop model of nuclear matter, is only the simplest model that gives at least reasonable results. More sophisticated models require quantum mechanics. These are outside the scope of the present lecture course. We will continue using the SEMF next to discuss the (in)stability of nuclei against α decay.
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