# Talk:Semi-empirical mass formula

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## Form of the pairing term

I have the A dependence of the pairing term to be ${\displaystyle A^{-1/2}}$ in my undergraduate notes, instead of ${\displaystyle A^{-3/4}}$ . Is the 3/4 figure chosen from any particular experiment? Or it one of those things that is open to choice of definition? --Zapateria 15:49, 2 June 2006 (UTC)

Can anyone give a more detailed explanation how the ${\displaystyle A^{1/2}}$ comes from, both experimentally and theoretically? —The preceding unsigned comment was added by Shinbu3 (talkcontribs) 10:49, 20 March 2007 (UTC).
Roughly speaking, you can imagine the pairing as occurring between pairs of nucleons that are orbiting in the same orbit, but in opposite directions. Classically, they would pass by each other twice per orbit. Since nuclear forces have short ranges, they only have a chance to interact when they're passing by each other. In a heavier nucleus, the orbits are physically bigger, so the passings are less infrequent, so for this kind of naive classical reason, you expect it to be maybe A^-1/3. There may be more careful, explicitly quantum mechanical arguments that produce the exponents -1/2 and -3/4. In reality, you fit this kind of thing to a wide range of data, and the reasons for certain features of the resulting fit may be obscure.--207.233.84.39 02:10, 28 March 2007 (UTC)

## What units does EB have?

Is the output value of EB in eV?

According to my undergraduate notes, EB should be about 8MeV per nucleon, but as far as I can tell, that's using the "Wapstra" values from the table. 134.226.1.234 21:03, 4 March 2007 (UTC)

## merge

Liquid drop model is a redundant, lower-quality version of Semi-empirical mass formula. I fixed a few of the more egregious errors in Liquid drop model, but I think there's still zero useful content in it that isn't in Semi-empirical mass formula.--207.233.84.39 02:02, 28 March 2007 (UTC)

Merge per nom --h2g2bob (talk) 22:16, 17 May 2007 (UTC)

I agree, this Semi-empirical mass formula is an expanded version of the formula I added to the Binding energy article a few months ago. The Liquid drop model contains nothing that is not in these two articles. P.S. I added the Liverhant reference. 138.194.161.242 05:44, 5 June 2007 (UTC)

Merge due to reasons given above. Dan Gluck 07:00, 5 June 2007 (UTC)
Merge Liquid drop model is more than adequately explained here. Zapateria (talk) 15:22, 23 January 2008 (UTC)

## pairing energies

I'm a little surprised about the vast difference between the pairing parameters quoted from Wapstra and Rohlf. I think the Rohlf values are right. Is it possible that someone made a mistake in tabulating the Wapstra values, and didn't understand that there was some difference in the definitions or something? I realize that all this stuff depends somewhat on what data set you're fitting, but I can't believe it really varies by a factor of three. Both the external links agree with the Rohlf values.--207.233.84.39 02:02, 28 March 2007 (UTC)

After looking at the discussion above of the -1/2 and -3/4 exponents, my guess is that Wapstra used one exponent, and Rohlf the other. If that's the case, then it's probably an error to include the Wapstra values here as if they pertained to the same exponent.--207.233.84.39 02:12, 28 March 2007 (UTC)

Just to chip in: the unverified least-square values are the same that our lecturer at Oxford quoted for us. Not sure of the original source thought. —Preceding unsigned comment added by 86.131.92.51 (talk) 13:00, 5 December 2007 (UTC)

## Merge

I will soon start merging the article into Liquid drop model due to the consensus regarding the issue.Dan Gluck 14:33, 4 July 2007 (UTC)

## Hidden away

Why is this hidden away with redlinks at Weizsacker's formula and Weizsäcker's formula? Gene Nygaard (talk) 02:38, 7 January 2008 (UTC)

## What does aP do?

Might seem silly, but what is the constant aP (given as 12 MeV) used for? I can't find it in any other place in the article than in the table of empirical values of the constants. —Preceding unsigned comment added by 85.228.17.185 (talk) 01:15, 3 January 2009 (UTC)

## Coulomb Term

I don't understand the following sentence in the article: 'However, because electrostatic repulsion will only exist for more than one proton, Z^2 becomes Z(Z − 1).' Could you explain it please? 86.42.243.140 (talk) 17:47, 12 June 2010 (UTC)

Each proton has one unit of charge. Electrostatic repulsion occurs between two particles, each repelling the other. With just one proton there is, of course, no repulsion; with two protons there are two units (A repels B, B repels A). The number of interactions between Z protons is (Z(Z-1))/2, but since each interaction involves 2 units of charge, the denominator '2' is 'cancelled out', leaving Z(Z-1). I hope that this makes sense. I'm not an expert but this expalantion works for me. :) --TraceyR (talk) 12:16, 15 October 2010 (UTC)

## Original paper

Hello, I do not succeed to find the original paper publish by Weizsacker where he introduces this formula. Could you help me and add this paper as a reference? 193.48.109.64 (talk) 09:09, 14 September 2010 (UTC)

## Asymmetry term

The Asymmetry term discusses at length the importance of the (A - 2Z) value in terms of being a significant value without mentioning that both the (A - 2Z) value and the simpler (N - Z) value are both the value of the number of neutrons contained within the nucleus that is in excess of the Z number of protons. Thus if the assumption is made that each proton in the nucleus is "paired" with an associated neutron, the (A - 2Z) and/or (N - Z) becomes the "extra neutron number" characteristic of the subject nucleus. And, since the atomic stability of the various isotopes has been determined to be considerably influenced by variations in the contained neutron numbers, in such a manner that the most stable isotopes are grouped around an increasing central tendency number of excess neutrons, it appears possible that the stability of the specific isotope is more closely associated with the "excess neutron" (A - 2Z) or (N - Z) number than it is for the N number. However it is to note that the majority of the stable isotopes (approximately 210 of the 285) have a total even number of neutrons (are of the categories EE or OE).WFPM (talk) 15:35, 25 September 2010 (UTC)

Also in the Symmetry image, the importance of "pairing" would be better shown for the A = 16 example, if the EE8O16 had been shown as 8 pairs of Proton/Neutron pairs as compared to the alternative EE6C16 image with 6 Proton/Neutron pairs plus 4 "extra neutrons" as an example of reduced stability.WFPM (talk) 18:05, 25 September 2010 (UTC) In the article on Radiocarbon dating it is pointed out that whereas the element (Carbon) is stable as EE6C12 with zero or as EO6C13 with 1 extra neutron, it then becomes unstable B- emission unstable as EE6C14 with only 2 extra neutrons, thus indicating the instability caused by the existence of too many (Unpaired) excess neutrons in the nucleus.WFPM (talk) 01:04, 26 September 2010 (UTC)

Beta decay does not necessarily indicate excess binding energy. It does not just depend on the binding energy part of the formula; you have to factor the proton and neutron mass terms of the full formula. If you were just to look at the binding energy, you would not expect a free neutron to decay to a proton and not conversely, since both have zero binding energy.Morngnstar (talk) 21:31, 14 February 2013 (UTC)

## Exponent in pairing term

Does anyone have an explanation why the pairing energy term is written here to the power of -½? In Kenneth S. Krane's book "Introductory Nuclear Physics" which I've been told is generally refered to as the bible of nuclear physics, the exponent is -¾, however I do note that he says "the pairing term is 'usually' expressed as ap * A^-¾". RubberTyres (talk) 20:50, 24 April 2011 (UTC)

## Preferred form of the asymmetry term?

There are three forms of the numerator in the asymmetry term that have been used in the history of this article and all are valid: (A - 2Z), (A/2 - Z), and (N - Z). The first and the third are equivalent. The second is half of these. Any could be used, but using the second would require using an ${\displaystyle a_{A}}$ value four times as much (since the expression is squared). I reverted to the form that was used in the oldest version of this article, (A - 2Z). The section explaining the term is written from the perspective of this expression. It was also used in the oldest version of the article that presented values for the constants, so presumably it is consistent with those values. It is furthermore consistent with the formula derived for the most-bound Z, whereas (A/2 - Z) is not.Morngnstar (talk) 21:31, 14 February 2013 (UTC)

According to this, the first sentence of the description of the asymmetry term is wrong. You need, as stated above, two different values for aA, whether you express the asymmetry term by (A - 2 Z)^2 / A or by (A/2 - Z)^2 /A - which is somehow obvious. I therefore suggest (and implement) to write the second formula like this (4 * aA) (A/2 - Z)^2 /A, since it is not necassary to introduce a new constant aA*. 134.34.142.179 (talk) 12:14, 15 February 2013 (UTC)

## Most-bound Z

Some original research for you. The derivation of the formula for the "stable nucleus of atomic weight A" is apparently done from the binding energy formula. It should be done from the mass formula, which includes the terms for proton and neutron mass, since beta decay respects mass, not binding energy. Factoring that, you get the slightly modified formula

${\displaystyle Z\approx {1 \over 2}A{{1+{(m_{n}-m_{p})c^{2} \over 4a_{A}}} \over 1+A^{2/3}{a_{C} \over 4a_{A}}}}$

The new numerator can be easily evaluated by using published values for m_n and m_p in MeV/c^2, consistent with the units for the coefficients given here. m_n - m_p works out to 1.29 MeV and using the least squares a_A the numerator works out to 1.0139, so this makes about a 1.4% difference in the Z result you get; probably of the same order as the inaccuracy of the formula itself. Of course we're still neglecting the undifferentiable pairing term, which will also lead to choosing the next or previous Z in some cases. If anyone has a source for this derivation (none cited here), check whether they consider this factor.Morngnstar (talk) 19:30, 15 February 2013 (UTC)

## Pairing term exponent (once again)

Since at least 2006, in 2011, and more recently, the exponent keeps getting changed back and forth, in what appear to be WP:AGF edits from a variety of editors, including one-shot IP editors. To assure some stability and credibility in the article, could an experienced and knowledgeable editor explicitly discuss the alternative exponent choices? If a well-known textbook is simply wrong, this needs to be mentioned, to avoid needless "corrections" and reversions. If there is legitimate controversy or confusion, this should be discussed here, and possibly in the article itself. Of course, ample references to WP:RS are absolutely essential to clear up the confusion. Reify-tech (talk) 19:51, 1 June 2015 (UTC)

Reify-tech, I've consulted some amassed textbooks and it seems that Krane's 'Introductory Nuclear Physics' lists the pairing term as an exponent of (3/4). However, in this case they've used a different set of fit coefficients (pairing coefficient = 34 MeV). Regardless, calculating these values for a given heavy nucleus ~A = 50 will still yield a fairly close answer. My course textbook - Martin's "Nuclear and Particle Physics" actually lists the pairing term as a positive for the even even and negative for odd even, with essentially the same coefficients as least squares, alongside an exponent of (1/2). Tipler's "Fundamentals of Physics" has basically the same form listed as the page, along with Hodgon's "Introductory Nuclear Physics". I'm inclined to believe that the exponent of the term is completely dependent on the coefficients derived from the fitting model. Martin's textbook discusses this, and mentions that the actual function of the pairing term is "derived by fitting data" - not just the coefficient. 09:17, 4 November 2015 (UTC) Mithrils

## Pairing term: confusion explained

As noted by some contributors above, various authors use different formulae to describe the pairing term. However, when a different formula is used, the constants in the formula will change too. This is the reason that quite contradictory values for the pairing constant aP can be found in the literature. In quoting an aP value one should always specify the power of A used in the pairing term.

The basis for the pairing term is the observation that odd-odd nuclei are less stable, and even-even nuclei are more stable than comparable even-odd or odd-even nuclei. This is dramatically demonstrated by the fact that only four (4!) out of the ~280 nuclei that exist in nature are odd-odd. The increased stability of even-even nuclei is due to what is called the pairing effect: for protons it is energetically favorable to arrange in pairs of opposite angular momentum. The extra binding energy of such a pair of protons, as compared to the particles when apart, is about 1 MeV. This 1 MeV extra binding energy makes the nucleus more stable. Exactly the same applies for a pair of neutrons: its binding energy too is about 1 MeV larger (i.e., stronger binding) than two separate neutrons.

Compared to odd-even or even-odd nuclei, the total binding energy of an even-even nucleus is about 1 MeV larger, and the total binding energy of an odd-odd nucleus is about 1 MeV less. This amounts to a pairing term in the total binding energy EB, of plus or minus 1 MeV (±1 MeV; plus for even-even, and minus for odd-odd).

Now in fact the pairing energy is not a constant: the heavier the nucleus, the smaller the pairing energy. In analogy to the asymmetry term, which goes as A-1, one might expect a pairing term δ0 (:) A-1. In reality the pairing term falls off slower. To achieve good numerical agreement between formula and experimental data many authors choose a A-3/4 dependance for the term, others choose an A-1/2 term. (In fact one might take the exponent of A to be another free parameter in a least squares fit.) There seems to be no sound physical derivation for any particular choice.

When a different power of A is chosen, this goes with a different value of the coefficient aP; typical values are aP = ±12 MeV if δ0 (:) A-1/2 [Rohlf], and aP = ±34 MeV if δ0 (:) A-3/4 [Wapstra]. The following table shows that either choice (δ0 (:) A-1/2, or (:) A-3/4) describes the experimental values δ0(exp) quite well.

${\displaystyle A}$ ${\displaystyle \delta _{0}(exp)}$ ${\displaystyle 12MeV \over A^{1/2}}$ ${\displaystyle 34MeV \over A^{3/4}}$
16 ± 2895. keV ± 3000. keV ± 4250. keV
24 2361. 2449. 3136.
32 1747. 2121. 2527.
40 1603. 1897. 2138.
60 1347. 1549. 1577.
80 1365. 1342. 1271.
100 1120. 1200. 1075.
120 1239. 1095. 938.
160 983. 949. 756.
200 872. 849. 639.
240 663. 775. 558.

(Experimental binding energy values are from G. Audi et al., 'The AME2012 Atomic Mass Evaluation' (Chinese Physics C 36/12 (Dec 2012) 1287-1602); I intend yet to provide a graphic illustrating this table)

(I have inserted the promised diagram in the article text = file Pairing term nuclear physics.gif) Hans van Deukeren (talk) 10:47, 10 September 2016 (UTC)

Incidentally, one may note that in the case of the Coulomb term, as to the constant aC there is some ambiguity too, because some authors use a Coulomb term with Z2, while others use the improved term Z·(Z-1). Thus, depending on the choice, the value aC term will also vary. However, because the difference between Z2 and Z·(Z-1) is generally small, this seldom leads to problems.

Hans van Deukeren (talk) 10:47, 25 August 2016 (UTC)

## Pairing term times or divided by A to a power

In the "Pairing term" section there is a statement with a constant divided by A to a power, but then it refers to a factor of A to a power where the power is negative the power is listed as -3/4 or -1/2. So it should be a constant times A to a (negative) power. Can someone make this consistent? Thank you. RJFJR (talk) 19:04, 20 January 2017 (UTC)