Understanding Nuclear Physics: The Liquid Drop Model and Nuclear Binding Energy, Slides of Advanced Physics

Download Understanding Nuclear Physics: The Liquid Drop Model and Nuclear Binding Energy and more Slides Advanced Physics in PDF only on Docsity! 3. Introductory Nuclear Physics – 1; The Liquid Drop Model docsity.com Each nucleus is a bound collection of N neutrons and Z protons. The mass number is A = N + Z, the atomic number is Z and the nucleus is written with the elemental symbol for Z AZ E.g. 12C, 13C, 14C are isotopes of carbon all with Z= 6 and neutron numbers N = 6, 7, 8 The neutrons and protons are bound together by the residual strong or color force docsity.com In addition there is a collection of bosons whose exchange mediates the four fundamental forces. , W+-, Z0, 8 gluons, graviton Only quarks and gluons experience the color force and quarks are never found in isolation docsity.com In the standard model …. Hadrons are collections of three quarks (baryons) or a quark plus an anti-quark (mesons). This way they are able to satisfy a condition of color neutrality. Since there are three colors of quarks, the only way to have neutrality is to have one of each color, or one plus an antiparticle of the same (anti-)color. Each gluon carries one color and one anti-color and there are eight possible combinations, hence 8 gluons. Why 8? The color force only affects quarks and gluons. gluons are linear combinations of red antired red antigreen red antiblue green antired green antigreen green antiblue blue antired blue antigreen blue antiblue http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/gluons.html red antirred + blue antiblue + green antigreen carries not color and thus is not counteed – non interacting docsity.com A red quark emits a red-antigreen gluon which is absorbed by a green quark making it red. The color force binds the quarks in the hadrons docsity.com Mesons are quark-antiquark pairs and thus carry net spin of 0 or 1. They are Bosons while the baryons are Fermions. They can thus serve as coupling particles. Since they are made of quarks they experience both strong and weak interactions. The lightest two mesons consist only of combinations of u, d, u, and d and are Name Made of Charge Mass*c2 τ (sec) π o uu − dd 2 0 135 MeV 8.4(-17) π ± ud, du ±1 139.6 2.6(-8) π ± → µ±+νµ π 0 →2γ , occasionally e+ + e- There are many more mesons. Exchange of these lightest mesons give rise to a force that is complicated, but attractive. But at a shorter range, many other mesons come into play, notably the rho meson (776 MeV), and the nuclear force becomes repulsive. docsity.com There are two ways of thinking of the strong force - as a residual color interaction (like a van der Waals force) or as the exchange of mesons. Classically the latter was used. Exchange of (pseudo)scalar mesons like the  meson give rise to an attractive force Exchange of vector mesons like the  and % give rise to a repulsive force. At very short distances one has to consider the exclusion principle for the quarks docsity.com The nuclear force at large distances is not just small, it is zero. Repulsive at short distances. Nuclear density nearly constant. 2 / E t Mc c Δ Δ ≈ ≈    docsity.com Since the nucleons are Fermions they obey FD statistics p0 = 3h3 8π n ⎛ ⎝⎜ ⎞ ⎠⎟ 1/3 where n = 0.17 2 fm−3 =8.5×1037 cm−3 is the density of n or p. Here h = 6.626×10−27 erg s . This implies a speed for the nucleons of about c / 4. and a peak Fermi energy, εF = p0 2 2M =39 MeV. The average Fermi energy is 3/5 of this εF = 23 MeV per nucleon As we shall see shortly, Coulomb energies are much smaller than this. To zeroth order the nucleus is a degenerate gas of nucleons confined by the (residual) strong force. docsity.com Nuclear density is a constant. Deformation is an indication of nuclear rotation R∝A1/3 docsity.com The nuclear crossing time for a typical nu- cleon, T ~ 10R/c ~ 107”? s sets the time scale for the shortest nuclear reactions. The electrical force between charged nu- cleons is 2 e€ 1.44 MeV Pool = = “Rlimy << &F for any one nucleon. However, since electri- cal energy rises as Z? and nuclear binding goes only as A, for large mass nuclei, the electrical force does become non-negligible. The nuclear force is independent of charge and is the same between neutrons and pro- tons. However it does depend on spin and orientation. The triplet state (tt) ' of two “Tpcieally the tight eam le thew state - tow tf states ont 6 Ramee wumtbdnsten of thr t00 *) aoe nuclear force is spin dependent docsity.com Unless weak interactions are involved (to be discussed later), the total energy released or absorbed in a given reaction Q = BE(speciesout) − BE(species in)∑∑ This is measured in MeV where 1 MeV = 1.6022 x 10-6 erg. docsity.com Semi-Empirical Mass Formulae •  A phenomenological understanding of nuclear binding energies as function of A, Z and N. •  Assumptions: –  Nuclear density is constant. –  We can model effect of short range attraction due to strong interaction by a liquid drop model. –  Coulomb corrections can be computed using electro magnetism (even at these small scales) –  Nucleons are fermions at T=0 in separate wells (Fermi gas model " asymmetry term) –  QM holds at these small scales " pairing term –  Nuclear force does not depend on isospin docsity.com surface area ~ n2/3 Liquid Drop Model •  Phenomenological model to understand binding energies. •  Consider a liquid drop –  Ignore gravity and assume no rotation –  Intermolecular force repulsive at short distances, attractive at intermediate distances and negligible at large distances " constant density. –  n=number of molecules, T=surface tension, BE=binding energy E=total energy of the drop,
,=free constants E=-
n + 4R2T  BE=
n-n2/3 •  Analogy with nucleus –  Nucleus has constant density –  From nucleon-nucleon scattering experiments we know: •  Nuclear force has short range repulsion and is attractive at intermediate distances. –  Assume charge independence of nuclear force, neutrons and protons have same strong interactions "check with experiment (Mirror Nuclei!) docsity.com Mirror Nuclei •  Compare binding energies of mirror nuclei (nuclei with n#"p). Eg 73Li and 74Be. •  If the assumption of isospin independence holds the mass difference should be due to n/p mass difference and Coulomb energy alone. •  From the previous page ΔEcoulomb(Z ,Z −1) = 3 5 e2 R [Z(Z −1) − (Z −1)(Z − 2)] = 3 5 e2 R 2(Z −1) 2/3( , 1)CE Z Z AΔ − ∝Z ~ A / 2 ; R = R0A 1/3>>1 to find that •  Now lets measure mirror nuclei masses, assume that the model holds and derive ECoulomb from the measurement. •  This should show an A2/3 dependence docsity.com nn and pp interaction same (apart from Coulomb) “Charge symmetry” ΔECoul ∝ A 2/3 docsity.com 27 More charge symmetry Energy Levels of two mirror nuclei for a number of excited states. Corrected for n/p mass difference and Coulomb Energy Ecorrected docsity.com For a degenerate gas p ∝n1/3 e.g. pΔx  h n = N V ~ (V / (Δx) 3) V ~ p 3 h3 page 15 docsity.com (1+λ)n ≈1+ nλ+ n(n −1) 2! λ2 + ... ' (1 ) 2 ' (1 ) 2 AZ AN λ λ = − = + ' 'N Z A λ −= The proportionality constant is about 28 MeV docsity.com So far we have BE = a A − b A2/3 − c N − Z( )2 A − d Z 2 A1/3 docsity.com The pairing and shell corrections are purely empirical quantum mechanical corrections (for now) to the liquid drop model docsity.com Putting it all together: z BE = aA - ae - BR — a, FNP _ ga) - (A) where terms 1 through 4 are the volume, surface, Coulomb, and symmetry energies respectively, 6(A) is the pairing correction, and S(A) is the shell correction. Without the last two terms which are strictly quan- tum mechanical, this is known as the “liquid drop model” or the Bethe-Weizsiicker semi- empirical mass formula. Empirically, and crudely from fitting to known binding en- ergies (deShalit and Feshbach, Theoretical Nuclear Physics, p. 126), we have ay = 15.68 MeV, aj = 18.56 MeV, a3 = 0.717 MeV, and ay = 28.1 MeV. Also 6(A) = +34/A3/4_ 0, -34/.45/4 MeV for odd-Z, odd- N; odd-A; and even-Z, even-N nuclei respec- tively. S(A) is complicated. docsity.com Each of the expressions here has been divided by A to give BE/A docsity.com N = A-Z N-Z = A-2Z Given A, what is the most tightly bound Z? Volume and surface energy are constant. Only the Coulomb and pairing terms contained Z explicitly docsity.com a3 A 2/3 <<4a4 symmetry dominates xxxxxxxxxxxxxx xxxx A Zs 20 9.6 40 18.6 60 27.3 docsity.com N − Z( )stable = A − 2Z( )stable = A 1− 4a4 a3A 2/3 + 4a4 ⎛ ⎝⎜ ⎞ ⎠⎟ = A a3 A 2/3 a3 A 2/3 + 4a4 ⎛ ⎝⎜ ⎞ ⎠⎟ ≈ a3 4a4 A5/3 = 0.0064 A5/3 a3 =0.717 a4 = 28.1 so a3 A 2/3 <<4a4 Evans 3.4 docsity.com