Calculating Distance Between Ions in Sodium Chloride Crystal Structure

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Homework Help Overview

The discussion revolves around calculating the distance between sodium and chloride ions in the crystal structure of sodium chloride, specifically focusing on a cube with an edge length of 0.281nm. The problem involves understanding the geometric relationships within the cube to find the distance between ions positioned at opposite corners.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using the Pythagorean theorem to calculate the distance, with one suggesting the diagonal of the cube can be derived from the edges. Questions arise about the visualization of the problem and the use of different variable representations in the equations.

Discussion Status

The discussion is ongoing, with participants exploring different geometric interpretations and questioning the clarity of the problem setup. Some guidance on using the Pythagorean theorem has been provided, but no consensus has been reached on the approach or solution.

Contextual Notes

There is mention of difficulty in visualizing the problem, particularly for those new to the concepts, which may affect their understanding of the geometric relationships involved.

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Homework Statement


A drawing shows sodium and chloride ions positioned at the corners of a cube that is part of the crystal structure of sodium chloride. The edge of the cube is 0.281nm in length. Find the distance between the sodium ion located at one corner of the cube and the chloride ion located on the diagonal at the opposite corner.


Homework Equations


I'm thinking pythagorean theorem and one or more of the trig functions perhaps?


The Attempt at a Solution


Well...I figured the diameter if the cube is drawn inside a circle is 2(0.281)=0.562nm.
Not sure what to do this...it's difficult for me to conceptualize.

(The answer in the book is 0.487nm)
 
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Just Pythagoras. If each edge of the cube has length x, then A diagonal of one side, say, the base, is given by d^2= x^2+ x^2= 2x^2 so that d= x\sqrt{2}. Now, that "face diagonal" together with a vertical edge gives you another right triangle having the diagonal of the cube from one corner to the opposite corner as hypotenuse. D^2= x^2+ d^2= x^2+ 2x^2= 3x^2. That diagonal has length D= x\sqrt{3}.
 
so is this two triangles?
 
maybe I'm confused because of the variables you are using...i'm use to seeing a,b,c with pythagoras equations..i know it doesn't make a difference..but for us newbies it kinda does sometimes make it easier to picture. I"m trying to visualize what this would all look like if flatttened out i guess.
 

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