22.101 Applied Nuclear Physics (Fall 2006)

Lecture 1 1 (10/23/06) Nuclear Binding Energy and Stability

References :

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

The stability of nuclei is of great interest because uns table nuclei undergo trans i tions that resu lt in th e em ission of particles and/ or electrom agnetic radiation (gamm a s). If the transition is spontaneous , it is called a rad i o activ e decay . If the transition is induced by the bom bardm e nt of particles or radiation, then it is called a nuclear reaction.

The m a ss of a nucleus is the decisive f actor governing its stab ility. Knowing the m a ss of a particular nucleus and those of th e neighboring nuclei, one can tell whether or not the nuc leus is s t able. Yet the r e lation be twee n m a ss and stability is co mplicated.

Increasing the m a ss of a stable nucleus by adding a nucleon can m a ke the resulting nucleus unstable, but this is not always tr ue. Starting with the sim p lest nucleus, the proton, we can add one neutron after anothe r. This would generate the series,

 

H  n  H 2 ( stable )  n  H 3 ( unstable )  He 3 ( stable )  n  He 4 ( stable )

Because He 4 is a double-m a gic nucleus, it is particular ly stable. If we continue to add a nucleon we find the resulting nucleus is unstable,

He 4  n  He 5  He 4 , with t 1/ 2 ~ 3 x 10 - 21 s

He 4  H  Li 5  He 4 , with t 1/ 2 ~ 10 - 22 s

One m a y ask: W ith He 4 so stable how is it possible to bu ild up the heav ier elem ents starting with neutrons and pr otons? (This question arises in the study of th e origin of

elem ents. See, for example, the Nob e l lecture of A. Penzias in Reviews of Modern Physics , 51 , 425 (1979).) The answer is that the following reactions can occur

He 4  He 4  Be 8

He 4  Be 8  C 12 ( stable )

Although Be 8 is unstable, its lifetime of ~ 3 x 10 -1 6 s is apparently long enough to enable the next reaction to proceed. Once C 12 is form ed, it can react with another He 4 to give O 16 , and in this way the heavy elem ents can be form ed.

Instead of the m a ss of a nucleus one can use the binding energy to express the sam e infor m ation. The binding energy concept is useful for discussing the calculation of nuclear m a s s es and of energy released or absorbed in nuclear reactions.

Binding Energy and Separation Energy

We define the binding energy of a nucleus with m a ss M(A,Z) as

B ( A , Z )  [ ZM  NM  M ( A , Z )] c 2 (10.1)

where M H is the hydrogen m a ss and M(A,Z) is the atom ic m a ss. Strictly speaking one should subtract out the binding energy of the electrons; however, the e rror in not doing so is quite sm all, so we will just igno re it. According to (10.1) the nuclear binding energy B(A,Z) is th e difference between the m a ss of the constituent nucleons, w h en they a r e f a r separated from each oth e r, and th e m a ss of th e nucleus, wh en they are brought tog e ther to for m the nucleus. Therefore, one can inte rpre t B( A,Z) as the work required to sepa rate the nucleus into the indiv i dual nucleo n s (separ ated far from each other), o r equivalently, as the en erg y releas ed d u ring th e ass e m b ly of the nucleus f r o m the constituents.

Taking the actual data on nuclear m a s s for various A and Z, one can calculate B(A,Z) and plot the results in the form shown in Fig. 10.1.

8

6

4

2

0 8 16 24 30 60

90 120 150 180 210 240

Mass Number (A)

F i g u r e b y M I T O C W . Adapted from Meyerhof.

Fig. 10.1 . Variation of average binding en ergy per nucleon with m ass num ber for naturally occurring nuclides (a nd Be8). Note scale change at A = 30. [from Meyerhof]

The m ost striking featu r e of the B/A cu rve is the approxim a te constancy at ~ 8 Mev per nucleon, except for the ve ry light nu cle i . It is in struc t ive to s ee what th is behavior implies. If the bindi ng energy of a pair of nucleons is a constant, say C, then for a nucleus with A nucleo n s, in which there ar e A(A-1)/2 distinct pairs of nucleons, the B/A would be ~ C(A-1)/2. Since this is not what one sees in Fig. 10.1, one can surm ise that a given nucleon is not bound equally to all the other nucleons; in other words, nuclear forces, being short-ranged, extend ove r only a few neighbors. The constancy of B/A i m plies a saturation effect in nuclear forces, the intera ction energy of a nucleon does not increase any further once it has acquired a certain num be r of neighbors. This number seem s to be about 4 or 5.

One can understand the initial rapid increas e of B/A for the very light nuclei as the resu lt of the com p etition between volum e effects, which m a ke B incre a s e with A like A, and surface effects, which m a ke B decrease (in the sense of a correction ) with A lik e A 2/3 . The latter should be less im portant as A becom e s large, hence B/A increases (see the dis cussion of the sem i -em p iricial m ass f o rm ula in the next chap ter ) . At the o t h er end of the curve, the gradual decr ease of B/A for A > 100 can be understood as the effect of

Coulom b repulsion which becom e s more im porta nt as the number of protons in the nucleus increases.

As a quick application of the B/A curve we m a ke a rough estim ate of the energies release in fission and fusion reactions. Suppos e we have symm etric fission of a nucleus with A ~ 240 producing two fragm e nts, each A/2. The reactio n gives a fin a l state with B/A of about 8.5 Mev, which is about 1 Mev gr eater than the B/A of the initial state.

Thus the energy released per fission reacti on is ab out 240 Mev . (A m o re accurate estim ate gives 200 Mev.) For fusion reaction we take H 2  H 2  He 4 . The B/A values of H 2 and He 4 are 1.1 and 7.1 Mev/nucleon respec tive l y. The gain in B/A is 6 Mev/nucleo n , so the energy released per fusion event is ~ 24 Mev.

Binding Energy in Nuclear Reactions

The binding energy concept is also app licab le to a binary reaction where the initial state consists of a pa rticle i incident upon a target nucleus I and the final state consists of an outgoing particle f and a residu al nu cleus F, as indicated in the sketch,

W e write th e reac tion in the f o rm

i  I  f  F  Q (10.2)

where Q is an energy called the ‘Q-v alue of the reaction’. C o rrespond in g to (10.2) we have the def i nition

Q  [( M  M )  ( M  M )] c 2 (10.3)

where the m a sses are understood to be atom ic masses. Every nuclear reaction has a characteristic Q-value; the reaction is cal led exotherm ic (endotherm ic) for Q > 0 (<0)

where energ y is given of f (absorbed). Thus an endotherm ic reaction canno t take p l ace unless additional energy, called th e threshold, is supplied. O f ten the source of this energy is the kinetic energy of the incident particle . One can also exp r ess Q in term s of the kinetic energies of the reactan ts and produc ts of the reaction by invoking the conservation of total energy, which must hold for any reaction,

T  M c 2  T

 M I

c 2  T

 M f

c 2  T

F  M

c 2 (10.4)

Com b ining this with (10.3) gives

Q  T f  T F  ( T i  T I ) (10.5)

U s ually T I is neglig ible c o m p ared to T i because the targ et nu cleus follow s a Maxwellian distribution at the tem p erature of the targe t sam p le ( t ypica lly room tem p eratur e), w h ile for nuclear reactions the inciden t particle can hav e a kinetic energy in the Mev range.

Since the res t m a sses can be express e d in term s of binding energies, another expression for Q i s

Q  B ( f )  B ( F )  B ( i )  B ( I ) (10.6)

As an example, the nu clear m e dicin e tech nique called boron neut ron capture therapy (BNCT) is based on the reaction

B 10  n  Li 7  He 4  Q (10.7)

In this case, Q = B(Li)+ B( )-B(B) = 39.245+28.296-64.750 = 2.791 Mev. The reaction is exoth e rm ic, therefore it can be ind u ced by a th erm a l neturo n. In prac tic e, the s i m p lest way of calculating Q values is to use the re st m a sses of the reactants and products. F o r m a ny reactions of interest Q values are in the range 1 – 5 Mev. An im portan t excep tion is fission, w h ere Q ~ 170 – 210 Mev depending on what one considers to be the fission

products. N o tice that if one defines Q in term s of the kinetic energies, as in (10.5), it m a y appear that the value of Q would depend on whether one is in laboratory or center-of - m a ss coordinate sys t em . This is illu sory becaus e the equ i valent definitio n, (10.3), is clearly frame independent.

Separation Energy

Recall the definition of binding energy, ( 10.1), involves an initial state w h ere all the nucleons are rem oved far from each other. One can define another b i nding energ y where the in itial s t ate is one where o n ly one nucleon is separated off. The energy required to separate particle a from a nucleus is called the separation energy S a . This is also th e energy released, or energy av ailab l e f o r re action, when particle a is captur e d.

This concep t is usua lly a pplied to a neutron, proton, deuteron, or -particle. The energ y balance in g e nera l is

S  [ M ( A ' , Z ' )  M ( A  A ' , Z  Z ' )  M ( A , Z )] c 2 (10.8)

where par tic le a is tre a te d as a ‘nuc le us’ with a t o m ic num b er Z’ and m a ss num b er A’. For a neutron,

S  [ M  M ( A  1 , Z )  M ( A , Z )] c 2

= B ( A , Z ) – B ( A - 1 , Z ) ( 1 0 . 9 )

S n is som e tim es called the binding energy of the last neu t ron .

Clearly S n w ill v a ry f r om one nucleus to anothe r. In the r a nge of A where B/A is roughly constant we can estim ate from the B/A curve that S n ~ S p ~ 8 Mev. This is a rough figure, for the heavy nuclei S n is m o re like 5 – 6 Mev. It turns out that when a nucleus M(A-1,Z) absorbs a ne utron, there is ~ 1 Mev (or m o re, can be up to 4 Mev) difference between the neutron absorbed bei ng an even neutron or an odd neutron (see Fig. 10.2). This difference is the reason that U 235 can undergo fission with therm a l neutrons, w h ereas U 23 8 can f i ssion o n ly with f a st neutrons (E > 1 Mev).

10

5

0 1 18

120 122

124 126 128 130

Neutr on Number (N)

F i g u r e b y M I T O C W . Adapted from Meyerhof.

Fig. 10.2. Variation of the neutron separation en ergies of lead isotopes w ith neutron num b er of the absorbing nucleus. [from Meyerhof]

Generally speaking the following system atic beha vior is observed in neutron and proton separation energies,

S n (even N) > S n (odd N) for a given Z

S p (even Z) > S p (odd Z) for a given N

This effect is attribu t ed to the pairing prope rty of nuclear forces – the ex is tence of extra binding between pair of identical nucleons in the sam e state which have total angu lar mom e nta pointing in opposite di rections. This is also th e reas on for the ex ception al stability of the -particle. Because of pa iring the even-even (even Z, even N) nuclei are more stable than the even-odd and odd-even nuc lei, which in turn ar e m o re stable then the odd-odd nuclei.

Abundance Systemtics o f Stable Nuclides

One can con struc t a s t ability char t by plotting the neutron num ber N versus the atom ic number Z of all the stab le nu clides. The results, shown in Fig. 10.3, show that N

~ Z for low A, but N > Z at high A.

7

Odd A

A

N = Z

t

= constan

Even A

A

N = Z

t

= constan

100

60

20

20 40 60 80

20 40

60 80

Atomic Number (Z)

Atomic Number (Z)

F i g u r e b y M I T O C W . Adapted from Meyerhof.

Fig. 10.3. Neutron and proton num bers of stable nuclides which are odd (left) and even isobars (right). Arrows i ndicate the m a gic num b ers of 20, 28, 50, 82, and 126. Also shown are odd-odd isobars with A = 2, 6, 10, and 14. [from Meyerhof]

Again, one can readily unders tand th at in he avy nuclei the C oulom b repulsion will favor a neutron-proton distribution with m ore neutrons than protons. It is a little more involved to explain why there should be an equal distribution for the light nuclides (see the f o llowing discussion on the sem i -em p irical m ass f o rm ula). W e will sim p ly note that to have m ore neutrons than protons m eans that the nucleus has to be in a higher energy state, and is therefore less stable. This symmetry effect is m os t pronounced at low A and becom e s less im portant at high A. In connection with Fig. 10.3 we note:

(i) In the case of odd A, onl y one stab le isobar exists, except A = 113, 123.

(ii) In the case of even A, only even-even nuclides exist, except A = 2, 6, 10, 14.

Still anothe r way to sum m arize the tr end of stable nuclide s is shown in the f o llowing

Systematics of Stability T r ends in Nuclei

F i g u r e b y M I T O C W . Adapted from Meyerhof.