An element E crystallizes in body centred cubic structure. If the edge length of the cell is 1.469 x 10^–10 m and the density is 19.3 g cm^–3, calculate the atomic mass of this element. Also calculate the radius of this element.

Solution

Solution:
We have given that
Z= 2 since it is bcc structure
edge length, a = 1.469 x 10^-10
Density, d= 19.3 g cm^-3Mass = ?

d = \frac{2 \times 3}{a^{3} \times N_{A}} 19.3 = \frac{2 \times M}{(1.469 \times 10^{- 8})^{3} \times 6.023 \times 10^{23}} or M = \frac{19.3 \times 3.17 \times 10^{- 34} \times 6.023 \times 10^{23}}{2} = \frac{19.3 \times 3.17 \times 6.023 \times 10}{2} = 18.4 g {mol}^{- 1} .

mass of element is 18.4g mol^-1

(ii) Cell edge,

a = 1.469 \times 10^{- 10} m

Body diagonal = 4R

Then 4R =

\sqrt{39}

= \sqrt{3} \times 1.469 \times 10^{- 10} m

R = \frac{\sqrt{3} \times 1.469 \times 10^{- 10}}{4} m = 0.64 \times 10^{- 10} m = 6.4 \times 10^{- 11} m .

Radius of element is 6.4 x 10^-11m.

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