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Question
Niobium crystallises as body centred cube (BCC) and has density of 8.55 Kg / dm-3 . Calculate the attomic radius of niobium.
(Given : Atomic mass of niobium = 93).
Solution
Crystal structure of Niobium = bcc
Density (d) = 8.55 kg dm-3
Atomic mass of Niobium = 93
`d=(nxxM)/(VxxN_A)`
No. of atoms per unit cell (n) in bcc = 2
Total volume of unit cell = a3
`d=(2xx93)/(a^3 xx 6.022 xx 10^23`
`a^3=(2xx93)/(8.55xx6.022xx10^23)=36.12xx10^-24`
`a=root(3)(36.12xx10^-24)`
a= 3.305 x 10-8
In BCC unit cell,
`r=(sqrt3xxa)/4=(sqrt3xx3.305xx10^-8)/4=1.431xx10^-8`
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