# 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 $A^{-1/2}$ in my undergraduate notes, instead of $A^{-3/4}$ . 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 $A^{1/2}$ 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 $a_{A}$ 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

$Z \approx {1\over 2} A {{1 + {(m_n - m_p)c^2\over 4 a_A}} \over 1 + A^{2/3} {a_C\over 4 a_A}}$

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)