[poonilization]

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.

[gritch]

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:

$\frac1{{d_{hkl}}^2} =\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}$  $(d_{hkl})^2≠\frac{a^2}{h^2}+\frac{b^2}{k^2}+\frac{c^2}{l^2}$

You can't simply invert both sides of the equation like this with addition of fractions. It'd be like saying:

$\frac11=\frac13+\frac13+\frac13$  $\frac11≠\frac31+\frac31+\frac31$

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.