Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: poonilization on January 14, 2015, 03:02:13 AM

So here's the question: http://i.gyazo.com/fdaea1eb54bdd3debf0b649ec4b5f807.png
i just need someone to check if i've done the first part correct. I seem to be getting different answers to my friends and i'm not sure where i've gone wrong.
So heres my calculations.
The first compound is body centred so h+k+l = even number so the first 3 hkl values will be (110), (101), (011)
http://i.gyazo.com/ba3ce1f425ea2d3b824b38eefb4062b7.jpg
My friends have managed to get a value of 5.30 Angstroms for hkl (110), i'm not sure who's correct. Can someone confirm the correct answer using hkl (110) and i should be able to do the rest.

I'm not overly familiar with the (h,k,l) calculations. I'm actually just learning them myself but everything looks good to me. You however have made a math error:
[itex]\frac1{{d_{hkl}}^2} =\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}[/itex] :rarrow: [itex](d_{hkl})^2≠\frac{a^2}{h^2}+\frac{b^2}{k^2}+\frac{c^2}{l^2}[/itex]
You can't simply invert both sides of the equation like this with addition of fractions. It'd be like saying:
[itex]\frac11=\frac13+\frac13+\frac13[/itex] :rarrow: [itex]\frac11≠\frac31+\frac31+\frac31[/itex]
If you correct this error you'll get 5.30Å in agreement with your friend. Hope that helped but given the delay I suspect you've already figured the issues out by now.