1. Apr 4, 2004

Claire84

Okay, I'm hoping someone here can explain this homework qustion I've got bcause we don't seem to have anything in our notes that covers it, so any help would be great- thanks!

Potassium fluoride (K+F-) is an ionic crytal having density 2.48x10^3 kg/m^3

Assuming that the diameters of the K+ and F- ions are approx equal, then make an estimate of the inyer-nuclear separation between the ions in the crystal. The info we're provided with are the relative atomicmasses of the K and F atoms, what one atomic mass uni equals and what 1eV equals. I'm not wanting an answer here, jsut how I should go about it (hence me not filling in all the details fo the quantities there). I was thinking that using the eqts from the Mie potential we could work it out since the attravtive force and repulsive force would be the same at this equilibrium separation bu it doesn't seem clear as to why I'd have to use all these values at the end of the question. Please help!

2. Apr 5, 2004

Kurdt

Staff Emeritus
Consider how many atoms of which approximately half are potassium and half are fluorine in a cubic meter of material and you should be able to make an estimate of the inter-nuclear size.

3. Apr 5, 2004

Claire84

Okay, so if I take the density which is mass per unit volume and the divide it by the mass of one ion I can the the number of ions per unit volume. Here I'd add the 2 relative atomic mases together, multiply them by 1.66x10^-27, and divide the density by them, and then mutiply this by 2 to get the number of ions per unit volume. Then take the cube root of this to work out the number of ions per unit length and then divided 1 by this to get the inter-nuclear separation? My answer is roughly x10^-10. Is this correct? And do we make the assumption here to neglect the diameters of the ions or what?

4. Apr 5, 2004

Kurdt

Staff Emeritus
Seeing as there's no information given on the atomic diameters in the question its reasonable to assume that you can neglect them. you're answer sounds ok its is generally accepted that the distance between atoms in a crystal is of the order of your calculated value. Just think of X-ray crystalography. The crystal is used as it has a similar width of spacing as the wavelength of the X-rays which are of the order of one or two Angstroms.

5. Apr 25, 2004

Claire84

So to obtain thiss experimentally, would you scatter x-rays off crystals? I've got here that that gives the spacing between the atoms, but as you've said (I think), that's to be neglected it's to the same order? Or am I barking up the wrong tree? Would you then use Bragg's law of 2d=msintheta to work out the internuclear separation? Thanks. :)

6. Apr 26, 2004

Claire84

In the initial calcualtion I mentioned, what assumptions would be made? I thought about maybe the structure should be regular all the way through, but is that just something that is actually the case anyway? Thank you.