Asked by Chemgam

(a) What is the distance between second nearest neighbor atoms in vanadium? Express your answer in cm.

(b) What is the interplanar spacing between (110) planes in vanadium? Express your answer in cm.

(c) At what angle must (110) planes of vanadium be exposed to a collimated beam of X-rays in order to obtain diffraction by monochromatic chromium K-alpha radiation? Express your answer in degrees.
Answered by Anonymous
a) 3.03⋅10−8
Answered by lena
something wrong ..
Answered by Anonymous
b) 2.14*10^-8
Answered by c
pls. c
Answered by qpt
a)3.03*10^-8
b) 2.14*10^-8

any ideas for c)?
Answered by chemboy
c) 33.2
Answered by Anonymous
3.2e-8
Answered by Chemgam
Thank you
a)3.03*10^-8
b) 2.14*10^-8
c) 33.2
Answered by Anonymous
a is qrong can you please teach me how to do this
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