Sum

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).

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#### 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`

Concept: Calculations Involving Unit Cell Dimensions

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