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November 21[edit]

Free electron density[edit]

Interesting Question: in a typical metal, how many conduction band electrons are present per atom?

My answer: This isn't a topic I know, but a quick search conduction electrons per atom turned up [1]. The "table" link for it provides "free electron number densities" as in the first two columns of the following. I've added atomic weight and density to get a per-atom calculation by what I *think* is a relatively foolproof procedure - though I arbitrarily used white tin's density, and would get a slightly different answer for gray. Then I added Fermi energy and related parameters, which I believe can be calculated from the free electron number density and atomic mass...

Element	N/V x10²⁸/m³ ( x 10²² cm³)	g/cc	N/g (x 10²² cm³)	g/mol	N/mol (x 10²² cm³)	N/atom (x 1)	Fermi energy (eV)	Fermi temperature x 10⁴ K	Fermi velocity x 10⁶ m/s
Cu	8.47	8.96	0.94	63.5	60.0	0.96	7.00	8.16	1.57
Ag	5.86	10.49	0.559	107.9	60.27	1.00	5.49	6.38	1.39
Au	5.90	19.30	0.306	197.0	60.22	1.00	5.53	6.42	1.40
Be	24.7	1.85	13.35	9.012	120.3	2.00	14.3	16.6	2.25
Mg	8.61	1.738	4.95	24.305	120.4	2.00	7.08	8.23	1.58
Ca	4.61	1.55	2.97	40.08	119.2	1.98	4.69	5.44	1.28
Sr	3.55	2.64	1.34	87.62	117.8	1.96	3.93	4.57	1.18
Ba	3.15	3.51	0.897	137.32	123.2	2.05	3.64	4.23	1.13
Nb	5.56	8.57	0.649	92.906	60.3	1.00	5.32	6.18	1.37
Fe	17.0	7.874	2.159	55.845	120.5	2.00	11.1	13.0	1.98
Mn (alpha)	16.5	7.21	2.288	54.938044	125.7	2.09	10.9	12.7	1.96
Zn	13.2	7.14	1.849	65.38	120.9	2.01	9.47	11.0	1.83
Cd	9.27	8.65	1.072	112.414	120.5	2.00	7.47	8.68	1.62
Hg (78 K)	8.65	13.534 (not 78 K)	0.639	200.592	128.2	2.13	7.13	8.29	1.58
Al	18.1	2.70	6.703	26.9815	180.9	3.00	11.7	13.6	2.03
Ga	15.4	5.91	2.606	69.723	181.7	3.02	10.4	12.1	1.92
In	11.5	7.31	1.573	114.818	180.6	3.00	8.63	10.0	1.74
Sn	14.8	7.365 (gray)	2.010	118.710	238.5	3.96	10.2	11.8	1.90
Pb	13.2	11.34	1.164	207.2	241.2	4.01	9.47	11.0	1.83

The relationship given for Fermi energy and free electron density at the link above is the harder part for me at the instant. It should reduce to

E_F = (1/8) h²/m [n*3/pi]^2/3

h = [[Planck's constant] in J s = kg m² s^-1 m = (presumably) electron mass in daltons x (1.660538921E-27 kg) n is the number from the first column x 10²⁸ m^-3, so to the 2/3 power it is that number to the 2/3 power x 4.641589E+18 m^-2.

This should get me a result in kg m² s^-2, i.e. in joules, which should be multiplied by 1 eV/(1.6021766208E-19 J) for a result in eV - looks like the units are right...

Looking at the table there's a monotonic relationship between N and Fermi energy (sort by column), so I'm thinking the "m" is actually the electron mass 9.10938215E-31 kg. Putting that in I get EV = 1.692532 N ^2/3

Then for copper I calculate 7.03 eV from N, very close to the value given, and for lead 9.45 eV ... this is looking like I might have it, haven't checked all though. Wnt (talk) 21:29, 22 November 2015 (UTC)[reply]

From this I should be able to expand the table to cover some values at the above link's table where N is not given, but Fermi energy is. Computing (EV / 1.692532) ** 3/2 to fill the first column,

Element	N/V x10²⁸/m³ ( x 10²² cm³)	g/cc	N/g (x 10²² cm³)	g/mol	N/mol (x 10²² cm³)	N/atom (x 1)	Fermi energy (eV)	Fermi temperature x 10⁴ K	Fermi velocity x 10⁶ m/s
Li	4.68	0.534	8.76	6.94	60.8	1.01	4.74	5.51	1.29
Na	2.65	0.968	2.74	22.990	63.0	1.05	3.24	3.77	1.07
K	1.40	0.862	1.62	39.098	63.5	1.05	2.12	2.46	0.86
Rb	1.14	1.532	0.744	85.468	63.6	1.06	1.85	2.15	0.81
Cs	0.911	1.93	0.472	132.905	62.7	1.04	1.59	1.84	0.75
Tl	10.56	11.85	0.891	204.38	182.1	3.02	8.15	9.46	1.69
Bi	14.15	9.78	1.447	208.98	302.3	5.02	9.90	11.5	1.87
Sb	16.34	6.697	2.440	121.760	297.0	4.93	10.9	12.7	1.96

Not very surprisingly, all the alkali metals come out close to one conduction electron per atom. The others seem plausible. The number of free electrons is never more than the number in the outermost shell, but we see in the first table it can be less... and that part may indeed be complicated. Wnt (talk) 22:16, 22 November 2015 (UTC)[reply]

The link worked for me just now when I retested it - anyone else having trouble? I've been having some damn witchy internet performance today. As for the temperature, well - in the case of Hg the number given for N was at 78 K, and the density was at room temp, and I presume the disagreement between the two is to blame for the non-integer result. It looks like the Fermi temperature and free electron number per volume are interchangeable; either of these can be calculated based on the number per atom, the density of the material and the atomic weight. One thing I don't know at the moment is which quantities are obtained most accurately, and which is derived from which in practice by calculation. I cannot guarantee the number of conduction band electrons doesn't change with temperature, though I'm thinking that ought to be some kind of a phase transition at least. Wnt (talk) 04:07, 23 November 2015 (UTC)[reply]

Note: well, that last comment was stupid, since the whole point of the Fermi distribution is it changes with temperature... what I mean, though, is that I think of the number of holes and mobile electrons in a semiconductor as being comparatively small. To change the integer part of the N/atom value above, I think you'd have to have an empty spot in half of the atoms of the metal, which I'd think would be an unusually hot temperature. But I realize now I don't actually know this. Wnt (talk) 15:24, 23 November 2015 (UTC)[reply]

Banned user

The following discussion has been closed. Please do not modify it.

In a metal, electrical conduction is said to be provided by electrons flying about within the metal, not bound to any one atom or being an inter-atom bond. Only some electrons do this, I think, certainly not all electrons each atom could contribute. In TYPICAL pure metals, what is the ratio of the number of conduction electrons to the number of atoms? For COMMON alloys? 124.178.79.219 (talk) 02:53, 21 November 2015 (UTC)[reply]

You'll want to be researching concepts like Fermi level and Electronic band structure and Metallic bonding and band gap. Delocalized electrons in metals are located in the conduction band. --Jayron 32 04:51, 21 November 2015 (UTC)[reply]

And if I could make sense of those articles, I'd probably have a PhD and wouldn't need to ask such a simple question! So how many electrons does each atom TYPICALLY contribute to the conduction band? 1.122.54.74 (talk) 06:01, 21 November 2015 (UTC)[reply]

It depends on the metal in question, the temperature of the metal, and probably many other factors. It's not as simple as you seem to require to answer your question... --Jayron 32 06:08, 21 November 2015 (UTC)[reply]

According to Electrical_resistivity_and_conductivity#In_metals only the outer shell electrons are delocalised. The number therefore depends on which metal(s) you have and their electron structures. For example, copper has a single outer electron so the ratio is 1:1. I'm sure someone will be along to tell me it's not quite as simple as that.--Heron (talk) 09:33, 21 November 2015 (UTC)[reply]

Well Jayron, if it depends on the metal, in what way? If that is complex, what about some examples? If it depends on temperature, to what degree, TYPICALLY? What is the ratio at around normal temperatures? Is it linear with temperature? A power relationship? 1.122.126.61 (talk) 09:54, 21 November 2015 (UTC)[reply]

I thought you were having trouble understanding such things? Either this explanation would make sense to you, and you'll have your answer, or it won't and you'll come back here asking questions and expecting simple answers which we can't give because they don't exist. The simplest answer (and the very wrong one) is that each atom in a metal contributes one electron to the sea of electrons. The more complex answer (and more correct) answer is given by that paper I linked to, which discusses modeling delocalized electrons as a fermion gas to calculate their density in a given material. --Jayron 32 16:08, 21 November 2015 (UTC)[reply]

Jayron, this isn't helpful. The website you linked states in the first line that each atom contributes one conduction electron. What the rest of it says is beyond me - it certainly doesn't give any usable result such as "pure copper at room temperature has 0.7 electrons per atom available for electrical conduction." or whatever the ratio might happen to be. Surely, no matter how what level of math is required to work it out, the result is dependent on only two things - the material and the temperature.

If somebody asked you for the electrical resistivity of typical metals, would you point them to a web page that lists typical values for copper, steel and the like and perhaps state that it increases about 0.4% for each degree temperature (or whatever it does), which would be helpful, or would you give them some book on Ph.D level physics that covers the theory of why, and tell them if they cannot understand that then they are not entitled to ask the question? 1.122.126.61 (talk) 16:44, 21 November 2015 (UTC)[reply]

If you wanted to know the electrical resistivity of typical metals, you should have asked the first time. Here is a table of them. --Jayron 32 20:02, 21 November 2015 (UTC)[reply]

I don't, and did not, want to know electrical resistivity. I want to know: in some pure metals, what is the TYPICAL ratio of the number of conduction electrons to the number of atoms? At a convenient temperature(s) if varies significantly. And for a few COMMON alloys would be good.

If you don't have any idea, leave it for someone else, don't be a smartarse. If you can do the math, perhaps you could explain it and/or draw some conclusions. 124.178.132.112 (talk) 01:34, 22 November 2015 (UTC)[reply]

By 'conduction electrons', I'm assuming you mean 'free electrons'. That is, the number of those which circulate freely when no potential difference (voltage) exists to drive electrons along and form a current. As I understand it, when a current is made to flow, electrons that are weakly held in orbit detach and join the free electrons in forming the current. Is my understanding of your terminology correct? Akld guy (talk) 01:54, 22 November 2015 (UTC)[reply]

Yes, your understanding of my terminology is indeed correct. I mean the electrons that fly about within the metal like molecules do in a gas, due to having kinetic energy.124.178.132.112 (talk) 02:30, 22 November 2015 (UTC)[reply]

(edit conflict)this has some additional calculations, and has a simpler method of approximating the free electron density. The equation is basically N=(Avogadro's number*density)/molar mass. So, for copper, the value would be(6.02E23 * 8.96)/63.55 = 8.45E22 electrons per cubic centimeter. --Jayron 32 02:10, 22 November 2015 (UTC)[reply]

Also found, on the same page here is a place to enter data in yourself to calculate free electron density, if all you know is the Fermi level of the material. this page lists the fermi level of copper as 7.00 electron volts, by plugging that in to the calculation give 0.841E29 electrons/cubic meter, which is 8.41E22 electrons per cubic centimeter, which is damned close to the approximation given above using only the molar mass and density. --Jayron 32 02:26, 22 November 2015 (UTC)[reply]

Well, Jayron, the number of ATOMS per unit volume is also Avogrado x density / molar mass. Even I with just a junior high education knows that. So, according to your post just above, the ratio of free flying electrons to atoms is 1:1. But in your ealier post, you said "The simplest answer (AND THE VERY WRONG ONE) is that each atom in a metal contributes ONE electron to the sea of electrons." You also said it depends on temperature. Please go away Jayron. I'm hoping someone who understands this topic will answer. 124.178.132.112 (talk) 03:00, 22 November 2015 (UTC)[reply]

Thank you. I wanted to make sure that your terminology was understood for those who might be inclined to answer. Nobody can answer the question off the top of his/her head, since it depends on the metal, its purity, and its temperature. There may be other factors. Speaking as a former telecommunications engineer and with more than 50 years of experience as an electronics enthusiast, I can state that your question has no practical application whatsoever. You are almost certainly trying to get unpaid research assistance in the writing of a thesis or textbook. Others may wish to do unpaid research on your behalf. I do not. Akld guy (talk) 04:14, 22 November 2015 (UTC)[reply]

Writing a textbook or thesis? I wish! Not with my junior high education and career in cabinet-making. I'm just trying to get some understanding of solid state physics. If you can't help, you are under no obligation to do so. Just leave it for others. 124.178.132.112 (talk) 04:45, 22 November 2015 (UTC)[reply]

Cabinet-making? I always thought you were an engineer. At least, isn't that what you told us in the past? Nil Einne (talk) 05:40, 22 November 2015 (UTC)[reply]

Is this a help desk or a nut house? What past, exactly? I've asked a couple of questions here in the last few months. Haven't mentioned engineering in any of them. 121.221.73.213 (talk) 07:13, 22 November 2015 (UTC)[reply]

I was referring to the stuff in 2012 - 2013. Also this isn't a help desk, it's a reference desk (note, not an answer desk either). If you want the help desk, try WP:Help Desk but that isn't the place for questions like this. Nil Einne (talk) 08:47, 22 November 2015 (UTC)[reply]

Help desk, reference desk, answer desk, whatever. I have no idea what you are on about re 2012 or 2013. I didn't discover this page until about 6 months ago. This is the Science Desk, and my question is a science question. People ask questions because they want help. Clearly this site was set up so people could get ANSWERS to questions. There is no help in all in this thread, except from Heron. Certainly not in whatever mysterious thing happened in 2012. 121.221.73.213 (talk) 10:12, 22 November 2015 (UTC)[reply]

Actually this was set up primarily as a reference desk, which will often hopefully help people get answers to questions, but ultimately it's intended to act as a reference desk and not a place for people to get answers to questions. Anyway if you don't understand why this is different from the WP:help desk, and in particular, why it's important that this is a reference desk and not an answer desk, you're probably at the wrong place. Try Yahoo Answers or something. As for your claim about not being WickWack, perhaps you're telling the truth. Your attitude here makes me doubt that though, but either way you're not likely to get much help here, or anywhere else, when you yell at everyone who answers. Nil Einne (talk) 14:51, 22 November 2015 (UTC)[reply]

What claim? Where? Who or what WickWack is, I have no idea. I made no claim to be not WickWack - until now I had never heard of him/her. I do so claim now though. So your response is quite odd. As for the rest of it, this whole page has lots of questions, which have attracted many ANSWERS. Answers I have recieved in response to other questions have been very helpful. But the responses here have been essentially "Read this advanced Ph.D level text. If you can't understand that, you are not entitled to ask your question." And "this is of no practical application." Who says? Why should it matter? If you think these are useful approaches, you deserve to be "yelled at", because they are unhelpful. They add no value. And whatever happened in 2012 has nothing to do with my question.

The solution to this is obvious: If you & Jayron see a question on a topic you don't understand, don't respond. Don't waste your time. Leave it. Someone else who does understand the topic or has a good hint may well then respond. 121.221.73.213 (talk) 15:59, 22 November 2015 (UTC)[reply]

Your response is nonsense. If you are claiming you weren't here in 2012-2013 then you are denying you are WickWack. The same way if I say I was not the Prime Minister of Australia in 2000 I am denying I am John Howard. I haven't seen an PhD level text referenced here and I'm pretty sure no one who has a PhD in Physics or chemistry has taken part in this discussion so again so again your comment is utter nonsense. Also you are gravely mistaken.. WickWack is still banned from the RD so if you are WickWack, you are banned and all your comments can be deleted, so it is highly relevant to this question. If I wasn't uncertain you are WickWack your comments would have been deleted long ago. Although, as stated, your attitude is highly reminiscent of WickWack. So if it continues, don't be surprised if the assumption is made you are WickWack and any comments or questions are treated accordingly. And yes it is very simple, if you think people who offered a lot of help deserve to be yelled at, you don't belong on the RD whoever you are. And from what I can tell, everyone who took part in this discussion except me, who never claimed to understood your question as it wasn't relevant, so again your statement is nonsense. BTW no one said you don't need to know the answer, in this discussion. Are you thinking of a discussion somewhere else or still having problems understanding what people have said? Nil Einne (talk) 07:06, 23 November 2015 (UTC)[reply]

Is there more than two caves in an human skull?[edit]

I know two caves: nasal cave and orbital cave. Is there more? 92.249.70.153 (talk) 03:19, 21 November 2015 (UTC)[reply]

Do you mean cavity rather than cave? The skull has a nasal cavity, orbital cavity and sinus cavity. --Jayron 32 04:48, 21 November 2015 (UTC)[reply]

I found also middle ear cavity as well as oral cavity and carnial cavity. Is there a difference between sinuses to the cavities? 92.249.70.153 (talk) 06:27, 21 November 2015 (UTC)[reply]

There's also the cranial cavity - containing the brain. Roger (Dodger67) (talk) 16:39, 22 November 2015 (UTC)[reply]

Fruit Preservation[edit]

As a result of chemical or physical reactions when heating/cooking foods: 1.will it make certain fruits(example plums,durian,banana,apple) to change from sour to sweet or vice –versa? 2.will it help to make foods that has gone slightly bad to be safer for eating? Please also state the reasons for your answers.

This sounds an awful lot like homework. Why not try answering the questions yourself and then coming back here if you get stuck? You could even post your answer here and we'll likely vet it for you. In particular, think about what makes something sweet or sour and about how food is made more safe for eating. Is cooking used to accomplish either of these things? 99.235.223.170 (talk) 19:59, 21 November 2015 (UTC)[reply]