22.101 Applied Nuclear Physics (Fall 2006)

Lecture 1 2 (10/25/06)

Empirica l Binding Energy Formu l a and Mass Parabolas

References :

W. E. Meyerhof, Elements of Nuclear Physics (McGraw-Hill, New York, 1967), Chap. 2.

P. Marm ier and E. Sheldon, Physics of Nuclei and Particles (Academ ic Press, New York, 1969), Vol. I, Chap. 2.

R. D. Evans, The Atomic Nucleus (M cGraw-Hill, New York, 1955).

The binding energy curve we have discusse d in the last lecture is an overall representation of how the stab ility of nuclides varies acro ss th e entire rang e of m a ss num b er A. The curve shown in Fig. 10.1 wa s based on experim e ntal data on atom ic m a sses. One way to analyze th is cu rve is to decompose the binding energy into various contributions from the interactions among th e nucleons. An em pirical form ula for the binding energy consistin g of contrib u tions representing vo lu m e , surface, Coulom b and other effects was first proposed by von W e izsä cker in 1935. Such a formula is usef ul because it n o t only allo ws one to calculate th e m a ss of a nucleus, thereb y elim inatin g the need for table of m a ss data, but also it lead s to qu alitative und erstand i ng o f the essential features of nuclear binding. More detailed theo ries exist, for exam ple, Bruecker et al., Physical Review 121 , 255 (1961), but they are beyond the scope of our study.

The em pirical m a ss for m ula we consider here w a s derived on the basis of the liquid crop model of the nucleus. T h e essential assum p tions are:

1. The nucleus is com posed of incom p re ssible m a tter, thus R ~ A 1/3 .

2. The nuclear force is the sam e a m ong neutrons and protons (excluding Coulom b interactions).

3. The nuclear force satu rates (m ean ing it is very short ranged).

The em pirical m a ss for m ula is usually given in term s of the binding energy,

B ( A , Z )  a v A  a s A

2 / 3

 a c

Z ( Z  1 )

A 1 / 3

 a a

( N  Z ) 2

A

  (11.1)

where the co efficients a are to b e determ ined (by f itting the mass data ), with subscr ipts v, s, c, and a referring to vo lum e , surface, Coul om b, and asymmetry respectively. The last term in (11.1) represents the pairing effects,

  a p /

even-even n u clei

= 0 even-odd, odd-even nuclei

=  a p /

odd-odd nuclei

where coefficient a p is also a f itting p a ram e ter. A set of value s f o r the f i ve coef f i cients in (11.1) is:

a v a s a c a a a p

16 18 0.72 23.5 11 Mev

Since the fitting to experim e ntal data is not perfect one can find seve ral slightly different sets of coefficien ts in the literature. The average accuracy of (11.1) is ab out 2 Mev except where strong shell effects are present. One can add a term , ~ 1 to 2 Mev, to (11.1) to represen t the shell effects, ex tra b i nding fo r nuclei with closed shells of neutrons or protons.

A sim p le wa y to interpret (11.1) is to rega rd the first term as a first approxim ation to the binding energy. T h at is to say, the binding energy is proportional to the volum e of the nucleus or the m a ss num b er A. This assum e s every nucleon is like every other nucleon. Of course this is an oversimplif ication, the rem a ining term s then can be regarded as corrections to th is first approxim a tion. That is w hy the term s representing surface, Coulom b and asymm e try com e in with negativ e sign s, each one s ubtracting from

the volum e effect. It is quite understandable that the surface term should vary with A 2/3 , or R 2 . The Coulom b term is also quite self-e vident considering that Z(Z - 1)/2 is the num b er of pairs that one can fo rm from Z protons, and the 1/A 1/3 factor com e s from the 1/R. The asymm e try term in (11.1) is le ss obvious, so we digress to derive it.

What we would lik e to estim ate is th e energy d i fference b e tw een an actual nucleus, where N > Z, and an ideal nucleus, wh ere N = Z = A/2. This is then the ene r gy to transform a symm etric nucleus, in the sense of N = Z, to an asymm e tric one, N > Z. For fixed N and Z, the num b er of protons th at we need to transfor m into neutrons is , with N=(A/2) + and Z = (A/2) - . Thus, = (N – Z)/2. Now consider a set of energy levels for the neu t rons and an other se t for the pro t ons, each one filled to a certain level. To tra n sf orm protons into neutrons the prot ons in question have to go into unoccupied energy levels above the last ne utron. This m eans the am ount of energy involved is (the num ber of nucleons that ha ve to be transform e d) tim e s   (energy change for each nucleon to be transform e d) =  2   ( N  Z ) 2  / 4 , where  is the spacing between energy levels (assum e d to be the sam e f o r all th e leve ls) . To estim ate

 , we note that  ~ E F /A, where E F is the Ferm i energy (see Figs. 9.7 and 9.8) which is

known to be independent of A. Thus  ~ 1/A, and we have the expression for the asymm e try term in (11.1).

The m a gnitudes of the various contri butions to the binding energy curve are depicted in Fig. 11.1. The initial rise of B/A w ith A is seen to be due to the decreasing im portance of the surface contribution as A increases. The Coulom b repulsion effect grows in importance with A, causing a m a xi m u m in B/A at A ~ 60, and a subsequent decrease of B/A at larg er A. Except for th e extrem e ends of the m a ss number range the sem i -e m p irical m a ss formula generally can give binding energies accura te to with in 1 %

V olume Ener gy

Surface Ener gy

Coulomb Ener gy

Net Binding Ener gy

Asymmetry Ener gy

14

12

10

8

6

4

2

0

0 30 60 90 120

150 180 210

240 270

Mass Number (A)

Figure b y MIT OCW . Adapted f rom Evans.

Fig. 11.1 . Relative contributions to the binding energy per nucleon showing the im portance of the various term s in the sem i -em p i r ical W eizsäcker form ula. [from Evans]

of the experim e ntal values [Evans, p. 382]. This m eans that atom ic m a sses can be calculated correctly to roughly 1 part in 10 4 . However, there are conspicuous discrepancies in the neighborhood of m a gic nucle i. Attem pts have been made to take into account the nuclear she ll effects by generalizing the m a ss for m ula. In addition to what we have already m e ntioned, one can c onsider another term representing nuclear defor m ation [see Marm ier and Sheldon, pp. 39, for references].

One can use the m a ss form ula to determ ine the constant r o in the expression for the nuclear radius, R = r o A 1/3 . The radius appears in the coefficients a v and a s . In this way one obtains r o = 1.24 x 10 -1 3 cm .

Mass Parab olas and Sta bility Line

The m ass form ula can be rearra nged to give the m a ss M(A,Z),

M ( A , Z ) c 2  A  M c 2  a  a  a / A 1 / 3   xZ  yZ 2   (11.2)

4

Notice th at (11.2) is not an exact rearrange m e nt of (11.1), certain sm all term s having been neglected. W h at is im por tant about (11.2) is that it s hows that with A held constant the variation of M(A,Z) with Z is given by a parabola, as sketched below. The m i ni mum

of this parab o la occu rs at the atom ic num b er, which we label as Z A , of the stable nucleus f o r the given A. This the r ef ore rep r es ents a way o f determ ining the stable nuclide s .

W e can analyze (11.2) f u rther by co nsidering  M /  Z

Z A

 0 . This gives

Z   x / 2 y  A / 2 (11.5)

A 1  a 

1   c  A 2 / 3

4  a a 

Notice th at if we had considered on ly the volum e, surface an d Coulom b term s in B(A,Z), then we would have found inst ead of (11.5) the expression

( M  M

) c 2 A 1 / 3

Z  n H ~ 0 . 9 A 1 / 3 (11.6)

A 2 a

c

This is a v e ry different result b ecaus e for a stab le nucleus with Z A = 20 the corresponding m a ss number given by (11.6) would be ~ 9, 000, which is clearly unrealistic. Fitting (11.5) to the experi m e ntal data gives a c / 4 a a = 0.0078, or a a ~ 20 – 23 Mev. W e see theref ore the deviation of the stability line from N = Z = A/2 is the result of Coulom b effects, which favor Z A < A/2, becoming relatively m o re im portan t than th e asym m e try effects, which favor Z A = A/2.

W e can ask what happen s when a nuclide is unstable because it is proton-rich.

The answ er is tha t a nuc leus w ith too m a ny protons for stability can em it a positron (positive e l e c tron e + or   ) and thus convert a proton into a neutron. In this process a neutrino ( ) is also em itte d. A n exam ple of a positron dec a y is

F 16  O 16      (11.7)

By the sam e token if a nucleus has too m a ny neutrons, th en it can em it an electron (e - or

  ) and an antineutrino  , converting a neutron into a prot on. An electron decay for the isobar A = 1 6 is

N 16  O 16      (11.8)

A com p eting process with positron d ecay is el ectron captu re (EC). In this process an inner shell atom ic electron is cap tured by the nucleus so th e n u clear charg e is redu ced f r om Z to Z – 1. (N ote: O r bita l ele c trons can spe nd a f r action of their tim e insid e the nucleus.) T h e atom as a whole would rem a in neutra l but it is lef t in an ex cited sta t e because a v a cancy has been create d in one of its in ner shells.

As far as ato m ic m a ss balance is con cer ned, the requirem e nt for each p r ocess to be energe tic ally a llow e d is:

M(A, Z+1) > M(A,Z) + 2 m e   - decay (11.9)

M( A,Z) > M( A, Z+1)   - decay (11.10)

M(A, Z+1) > M(A,Z) EC (11.11)

where M(A,Z) now denotes atom ic mass. Noti ce that EC is a less stringent condition for the nucleus to decrease its atom ic number. If the energy differen ce be tw e e n initia l an d f i nal sta t es is less than tw ice the e l e c tron rest m a ss (1.02 M e v), the transition can take place v i a EC while it would be energ e tically forb idden via po sitron d ecay. The reason for the appearance of the electr on rest m a ss in (11.9) m a y be explained by looking at an energy balance in term s of nuc lear m a ss M’(A ,Z), w h ich is rela ted to th e atom ic m a ss by M(A,Z) = Zm e + M’(A,Z), if we ignore the binding en ergy of the electrons in the atom. For   -decay the energy balance is

M ' ( A , Z )  M ' ( A , Z  1 )  m e   (11.12)

whi c h we can r e wr i t e as

Zm e  M ' ( A , Z )  ( Z  1 ) m e  M ' ( A , Z  1 )  2 m e   (11.13)

The LH S is just M ( A , Z) w h ile the R H S is at le as t M(A , Z-1) + 2m e w ith the neutr i no having a variable energy. Thus one obtains (11.9). Another way to look at this condition is tha t is that in addition to the positro n em itted, th e daughte r n u clide also h a s to e j ect a n electron (fro m an outer s h ell) in ord e r to preserv e charge neu t rality.

Having discussed how a nucleus can change its atom ic num b er Z while preserving its m a ss num ber A , w e can pred ict w h at trans i tions w ill oc c u r as an uns table nuc lide moves along the m a ss parabola toward the point o f stability. S i nce th e pair ing te rm  vanishes for odd-A isobars, one has a single m a ss parabola in this case in contrast to two m a ss parabolas for the even-A isobars. One m i ght then expect that when A is odd there can be only one stable isobar. This is gene rally true with two ex ceptions, at A = 113 and

123. In these two cases the discrepancies ar ise from s m all m a s s differences which cause

one of the isobars in each case to have excepti onally long half life. In the case of even A there can be stable even-even isobars (three is the largest number found). Since the odd- odd isobars lie on the upper m a ss parabola, one w ould expect there s hould be no stable odd-odd nuclides. Yet there are several exceptions, H 2 , Li 6 , B 10 and N 14 . One explana tion is tha t the r e ar e rapid variations of the bi nding energy for the very light nuclides due to nuclear structure effects th at are not taken into account in the sem - em pirical m a ss form ula. For certain odd-odd nuc lides bo th co nditions for   and   decays are s a tisfied, and indeed both decays do o ccur in the s a m e nucleus. Exam ples of odd- and even-A m a ss parabolas ar e shown in Fig. 11.2.

A

b

?

Even A A = 102

Odd Z, odd N

Even Z, even N

?

 -

 +

 -

 -

 - EC

 +

T e I

Ba

Nb Mo T c Ru Rh Pd Ag Cd

52

Z

53

54 55 56 57 58

41 42

Z

43

44 45 46 47 48

Z A = 55.7

Z A = 44.7

Ce

La

Cs

Xe

 -

 +

 -

?

Odd A A = 135



Figure b y MIT OCW . Adapted f rom Meyerhof.

Fig. 11.2. Mass parabolas for odd and even isobars. Stable and radi oactive nuclides are denoted by closed and open circle s respectively. [from Meye rhof]