Bohr & Wheeler Fission Theory Calculation 4 March 2009

Transcription

1 Bohr & Wheeler Fission Theory Calculation 4 March 9 () Introduction The goal here is to reproduce the calculation of the limiting Z /A against spontaneous fission Z A lim a S. (.) a C as first done by Bohr & Wheeler in their famous paper The Mechanism of Nuclear Fission [Phys. Rev. 56, (939)]. a S and a C are the surface and Coulomb energy factors, respectively ~ 8 MeV and.7 MeV. The development given here draws heavily on M. S. Plesset s analysis in American Journal of Physics 9(), - (94). Plesset incorporates more terms in the analysis than are necessary to determine the limit and his notation is a little confusing, so what is given here is more minimalist than what he presented. The deformed nucleus is presumed to be modeled as a sphere of initial radius R o which is distorted: R o + α + α P ( cosθ) r θ [ ], (.) where P is a Legendre polynomial. The idea is that both perturbing coefficients α and α are small. Two coefficients are included on the rationale that the volume of the nucleus is assumed to be conserved (incompressible) even as it distorts, so that one can then solve for, say, α in terms of α. It turns out that the lowest-order contributions to the surface and Coulomb energies are both of order α. Thus, I carry no terms to any higher orders. θ r

2 There are several steps in the calculation: The volume integral section below. The surface area integral - section 3. The Coulomb electrostatic potential-energy integral section 4. Results of the area and Coulomb integrals are combined to give the total energy, upon insisting on volume conservation. Detailed woring of various integrals appear in appendices A-H.

3 () The Volume Integral The element of volume in spherical coordinates is r sinθ dr dθ dφ. Hence V π r(θ ) π r sinθ dr dθ dφ. (.) θ r φ Note the order of integration with r and θ. The upper limit of r is a function of θ : r( θ) R o [ + α + α P ]. (.) The r integral must be done first, then the θ integral. The φ integral gives π directly. Integrating over r: V π 3 π r 3 ( θ ) sinθ dθ. (.3) θ This integral proceeds as V π R 3 π o [ + α + α P ] 3 sinθ dθ 3. (.4) θ We will have numerous integrals of this form. In all such cases, it is convenient to mae a change of variable x cosθ, which gives V π R 3 o [ + α + α P ] 3 dx 3. (.5) It is also handy to have the normalization condition for Legendre polynomials: j δ P i P j dx i. (.6) i + j + Expanding out (.5): V π R 3 o + α 3 3 dx α α P dx α α P dx + α 3 P 3 dx. 3

4 The first integral gives, the second integral vanishes by (.6), the third integral gives /5 by (.6), and the last integral is dropped as I retain terms only to order α. Hence V 3 4 π R o + α α α. (.7) If volume is to be conserved, then the contents of the brace bracet in (.7) must equal unity: + 3α + 3α + α α α α. At this point I follow Plesset s argument. The idea here is to solve for α in terms of α. Both are small. Begin by dropping the presumably very small cross term α α as well as the α 3 term; what remains is α + α + 5 α, a quadratic in α. Solving gives α ± 4α 5 ~ ± ( α ) where I invoed a binomial expansion of the square root. To decide on which root to choose, if α and volume is to be conserved, we must have α as well. The upper sign is the correct choice, leaving α ~ 5 α. (.8) This result will prove valuable in computing the area and Coulomb energies.. 4

5 (3) The Surface Area Integral The deformed nucleus does not have a spherical profile, so we have to use the general arc-length for spherical coordinates: ds dr + r dθ + r sin θ dφ. (3.) ds θ r Consider a line of longitude at fixed φ: ds dr + r dθ dθ r + dr r dθ + dr. (3.) dθ r dθ So, the area of a ribbon of surface at polar angle θ is da π r sinθ ds π r sinθ + dr r dθ dθ. (3.3) Put x dr. Presuming x to be small, that is, that the nucleus is not greatly distorted, so r dθ that dr/dθ ~, then ( + x) / ~ + x 8 x (3.4) 5

6 Now, the nucleus is described by r( θ) R o [ + α + α P ]. This means that ( dr / dθ) α R o ( dp / dθ), so, to order α, we can stop at the x/ term in (3.4): da π r sinθ + dr r dθ +... dθ π sinθ r + dr dθ +... dθ. (3.5) So the surface area of the deformed nucleus, to this level of approximation, has two contributions: π A π r sinθ dθ + π dr sinθ dθ +... dθ. (3.6) These integrals are done in Appendices A and B: π r sinθ dθ R o + α π dr dθ These give, in (3.6), A ~ 4 π R o + α + 5 α. (3.7) sinθ dθ 6 5 R o α. (3.8) α (3.9) Substituting into this the result of volume conservation, α ~ α / 5, [(.8) from above] gives, to terms of order α, A ~ 4π R o + 5 α If Ω is the factor which converts surface area to equivalent energy, and invoing the usual approximation R o ~ a o A / 3, then we have the surface energy U S as + 5 α U S ~ a S A / (3.) 6

7 where a S 4 π Ωa o, nown to be ~ 8 MeV from fits to the semi-empirical mass formula. The areal energy increases upon perturbation of the nucleus from its initially spherical shape. 7

8 (4) The Coulomb Potential Integral r r θ r The charge distribution gets divided into elements of volume. The Coulomb self-potential is given by ρ dv dv 4πε o, (4.) V V r where ρ is the electrical charge density and r is the distance between the two elements of volume. There is one integral for volume elements labeled and one for those labeled. The volume elements are both of the form r sinθ dr dθ dφ, so (4.) is actually a sextuple integral: ρ 4πε o r r sinθ sinθ dr dr dθ dθ dφ dφ. (4.) r r r θ θ φ φ As in the volume and area integrals, the limits on the r s are functions of θ since r( θ) R o [ + α + α P ]. Again the r integrals must be done first, then the θ integrals. Great Care must be taen to eep trac of the and integrals. For the r in the denominator, use the identity: 8

9 r r + r r + r P ( cosθ ), r < r P ( cosθ ), r > r (4.3) where θ is the angle between the directions from the origin to volume elements and. Using dτ s to represent volume elements in the usual way, Eq. (4.) can be expressed more compactly as ρ dτ 8πε o dτ, (4.4) r where the () and () are to remind that each integral is actually a triple integral over the coordinates for volume elements and. Do the inner integral first; label it U. Brea it into two regimes, one for r to r (for which r < r ) and then from r to r ( θ ) (for which r > r ), and use (4.3): U dτ r r r + P dτ + r r ( θ ) r + P dτ. (4.5) r r Note carefully that P still means P ( cosθ ). These integrals are over the coordinates; the factors of r can be extracted but must remain within the sums. Write the volume elements as dτ r dr dω: U r r + + P dr dω r + r r P dr dω. (4.6) r ( θ ) r +3 The first integral over r is trivial and gives r case of, where a logarithmic form arises: ( + 3); the second requires a bit of care for the P U r dω r r ( ) ( θ )P dω r P ( ) dω + r ln r θ P dω. r (4.7) 9

10 Now, from the addition theorem for spherical harmonics, the P ( cosθ ) can be written in terms of Associated Legendre polynomials whose arguments are the cosines of the individual direction angles of the and volume elements: P ( cosθ ) ( m)! ( + m)! P m ( cosθ ) P m ( cosθ )exp[ ιm( φ φ )]. (4.8) m Imagine (4.8) substituted into (4.7). In the integrals over φ and φ, only m will give non-zero contributions. The Associated Legendre polynomials thus reduce to regular Legendre polynomials. Designate them as P () and P () to eep trac of which belongs to coordinates and ; thus, for example, P () designates the th-order Legendre polynomial for coordinate set. P U r P + 3 dω + r P ( ) r θ P dω P r P ( ) ( dω ) + r P ln r θ P dω r. (4.9) The first and third integrals in (4.9) will vanish except when ; on integrating over φ and θ they respectively give 4 π r / 3 and π r, for a total of π r / 3: U π 3 r + r P ( ) r ( ) ( θ )P dω + r P ln r θ P r dω. (4.) The second and third terms on the right side are done in Appendices C and D. The second term emerges as (to order α, as usual) r P ( ) r θ P dω π R o + α + π α R o r P R o ( ),, + α { }(4.) where {,,} designates the integral of a product of three Legendre polynomials over θ : {,,} P P dx. (4.) The third term in (4.) comes out as

11 r P ln r θ P r dω 4π 5 P r α + α π P r α {,,}. (4.3) ( + α ) Loo at the last terms on the right sides of each of (4.) and (4.3). The one that is missing from (4.), that for, is precisely that which appears in (4.3). Gathering (4.) and (4.3) into (4.) then gives: U π 3 r + π R o ( + α ) + 4π 5 P r α + π α + α R o r P R o ( ),, + α { }. (4.4) This expression goes bac into (4.4). Write the volume element for coordinates out explicitly: ρ π 8π ε o 3 + 4π 5 r ( θ ) r 4 dr dω + π R o + α α + α r ( θ ) r dr dω r ( θ ) P r 4 drdω ( + π α ) R o r ( θ ) {,,} R o ( + α ) r + P ( ) dr dω. (4.5) The four terms on the right side are done in Appendices E, F, G, and H. The results are (to order α ): π 3 r ( θ ) r 4 dr dω 4 π 5 R o 5 ( + α ) 5 + 4( + α ) 3 [ α ]. (4.6) π R o + α 4π 5 r ( θ ) 4π r dr dω α + α r ( θ ) P r 4 drdω ( 6π ) 3 R 5 o + α o ( 5 + α o) 3 α. (4.7) 3 α. (4.8) 5 R 5 o + α

12 ,, π α R { } o R o ( + α ) P ( ) r + dr dω 8 (4.6) (4.9) into (4.5) gives r ( θ ) 5 π α R 5 o + α 3. (4.9) ρ 4 π 5 R o 8π ε o 5 ( + α ) 5 + 4( + α ) 3 [ α ] π R 5 o + α ( 5 + α ) 3 α + 6π 5 R 5 o + α 3 α π α R 5 o + α. Rearrange ρ π 5 R o + α 8π ε o α 3 α , or ρ π 5 3 R o 8π ε o 5 + α α 75 + α, The charge density is given by ρ 3Z e 4π R o 3. so 9Z e 6π 6 π 5 R o 3 R o 8π ε o 5 + α α 75 + α, or 9Z e 3 8 π ε o R o 5 + α α 75 + α. Tae the 8 into the brace bracet, and tae a factor of 3 from the numerator of the prefactor into the brace bracet:

13 3Z e π ε o R o + α 5 + ( 5 + α ) 3 α. Extract a factor of into the denominator of the prefactor: 3Z e + α π ε o R o ( 5 + α ) 3 α. Now use the empirical nuclear size relationship R o a o A / 3, a o ~. fm: 3 e π ε o a o Z A / 3 + α ( 5 + α ) 3 α. The first prefactor is the Coulomb energy parameter a C ~.7 MeV: Z a C + α A / ( 5 + α ) 3 α. (4.) Now substitute the volume-conservation condition α ~ α / 5, [(.8) from above] into (4.): Z a C A / 3 5 α α 3 α Z ~ a C A / 3 5 α (4.) The Coulomb self-energy decreases upon perturbation of the nucleus from its initially spherical shape. 3

14 (5) Limiting Z /A Calculation From (3.) and (4.) we have the area and Coulomb energies of the distorted nucleus as + 5 α U S ~ a S A / (3.) Z ~ a C A / 3 5 α (4.) Fission will proceeds spontaneously if the total energy of the deformed nucleus is less than that of the initial spherical undeformed (α ) shape: ( U S + ) deformed ( U S + ) undeformed <. (5.) With (3.) and (4.) this corresponds to Z A > a S, (5.) a C as shown by Bohr & Wheeler. 4

15 Appendix A: π r ( θ) sinθ dθ Here we have, with r( θ) R o [ + α + α P ]; set x cosθ : A R o + α dx + ( + α )α P dx + α P dx. (A.) The orthogonality relation is j δ P i P j dx i. (A.) i + j + The first integral gives, the middle integral vanishes, and the third integral gives /5: A R o + α + 5 α. (A.3) 5

16 Appendix B: π dr sinθ dθ dθ Here we need to deal with the derivatives dr/dθ. Now, r( θ) R o [ + α + α P ]. (B.) But, the P s are functions of cosθ. Let x cosθ as usual: dp dθ dp dx dx dp sinθ dθ dx. (B.) So dr dθ R sinθ α o dp dx. (B.3) Hence B π dr R sinθ dθ α o dθ π dp dx sin 3 R θ dθ α o dp dx ( x ) dx (B.4) Now, P (3x -)/, so dp/dx 3x, hence B 9 α R o x ( x ) dx 6 5 α R o. (B.5) 6

17 Appendix C: r P ( ) r θ P dω Begin by substituting r( θ) R o [ + α + α P ]: C r P R ( ) o + α + α P [ ( ] P ) ( dω ). (C.) Within the square bracet, factor out ( + α ): C r P R + α ( ) o + α P ( ) P ( + α ) dω. (C.) For the square bracet, do a binomial expansion to order α ; also brea up the R o term: r C R P ( o + α ) ( ) + R o ( ) ( + α ) α ( ) ( ) P + α ( + α ) P ( +... ) P dω. (C.3) Consider the terms in the integrand. For the first, only will give a non-zero contribution, and the integral of P () over all (θ,φ) is 4π. The second would give a non-zero contribution for [the integral of P () P () ], but cannot have this value in the sum, so this term contributes nothing. Hence, to order α, C π R o + α + α r R o ( )P R o ( + α P ) P ( dω ). (C.4) Integrating over φ gives π ; the factor of cancels that in the denominator of the sum. For the θ integral, transform to x cosθ, and use the,, { } notation of the main text: C π R o + α + π α r R o R o ( )P + α {,,}. (C.5) 7

18 Appendix D: r P ln r θ P dω r Expand the logarithm: D r P ln[ r ( θ) ]P ( dω ) ln r dω P. (D.) R o [ + α + α P ] in for r (θ) in the first integral, as was done in Appendix C above: The second integral vanishes. Put r θ and factor out + α D r P ln R o ( + α ) + α P ( + α ) P dω. (D.) Again expand the logarithm; the integral with ln [ R o ( + α )] vanishes lie the second integral in (D.) above, leaving D r P ln + α P P ( + α ) dω. (D.3) Do a binomial expansion on the logarithm, again eeping terms to order α : ln( + x) ~ x x +... (D.4) hence D r P α ( + α ) P ( dω ) r P α P + α 3 dω. (D.5) The first integral in (D.5) evaluates to 4π/5, and the second to π {,,} as in Appendix C above. Hence D 4 π r P α 5 ( + α ) π r P α {,,}. (D.6) ( + α ) 8

19 Appendix E: π 3 r ( θ ) r 4 dr dω Integrate over r. Integrating over φ gives a factor of π. Also invoe the usual transformation x cosθ : E π 3 r ( θ ) r 4 dr dω 4π 5 r 5 ( θ ) dx. (E.) Put r θ α : R o + α + α P E 4π R o 5 [ ]. In expanding this to the fifth power, eep only terms to order 5 [( + α ) 5 + 5( + α ) 4 α P + ( + α ) 3 α P +...] dx. (E.) The middle term in the integrand maes no contribution as the integral of P () over x (-, ) is zero. The other two terms are standard: E 4π 5 R o ( + α ) 5 + 4( + α ) 3 [ α ] 5. (E.3) 9

20 Appendix F: π R o + α r ( θ ) r dr dω This proceeds very much lie Appendix E above. Integrate over r, invoe a factor of π from integrating over φ, and invoe the usual transformation x cosθ : F π R o + α r ( θ ) 4π r dr dω r 3 θ 3 R o + α dx. (F.) Put r θ α : R o + α + α P F 4π 3 R 5 o + α [ ]. In expanding this to the third power, eep only terms to order [ + 3( + α )α P +...]dx. (F.) ( + α ) 3 + 3( + α ) α P The middle term in the integrand maes no contribution as the integral of P () over x (-, ) is zero. The other two terms are standard: F 4π 3 R 5 o + α ( 5 + α ) 3 α. (F.3)

21 Appendix G: 4 π 5 α + α r ( θ ) P r 4 dr dω This is similar to E and F above. Integrate over r, invoe a factor of π from integrating over φ, and invoe the usual transformation x cosθ : G 8π 5 α + α P r 5 ( θ) dx (. (G.) ) Put r θ α : R o + α + α P [ ]. In expanding this to the fifth power, eep only terms to order G 8π 5 5 α R o + α P [ 5 + 5( + α ) 4 α P + ( + α ) 3 α P +...] dx + α. (G.) Note the presence of the factor of P () in front of the square bracet in the integrand. This results in the first term within the square bracet maing no contribution as the integral of P () over x (-, ) is zero. We can ignore the last term in the square bracet as it will lead to a term of order α 3 when combined with the factor of α outside the integral. Only the second term gives a surviving contribution: 3 α G 8π 5 R 5 o + α o P ( dx 6π ) 3 α. (G.3) 5 R 5 o + α

22 ,, Appendix H: π α R { } o R o ( + α ) P ( ) r + dr dω Integrate over r, invoe a factor of π from integrating over φ, and invoe the usual transformation x cosθ : H π α R o ( ),, { } ( + 3) R o + α r ( θ ) r +3 ( θ ) P dx. (H.) We want to eep terms to order α, but there already is a factor of α in front of the integral. Hence, in expanding out r +3 ( θ), we need only eep the zero-th order term, that is, r +3 θ ~ R +3 o ( + α ) +3, so we have H π α R o 5 ( ),, + 3 { }( + α ) 3 P dx. (H.) The only term for which the integral gives a non-zero value is ; the value of the integral is : H 4 3 π α R 5 o ( + α ) 3 {,,}. (H.3) The value of {,,} is /5: H 8 5 π α R 5 o ( + α ) 3. (H.4)