Cu has FCC structure. If the interplanar spacing d is 2.08 A" for the set of (1 1 1) planes. Find the density and diameter of Cu atom. Given atomic weight of Cu is 63.54.

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written 9.4 years ago by

teamques10 ★ 70k

• modified 9.4 years ago

Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 7M

Year: May 2015

engineering physics

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written 9.4 years ago by

teamques10 ★ 70k

Given:

Cu has FCC structure.

$d = 2.08 A \ \ or \ \ 2.08 × 10^{-10}$m

$M=63.54$

To find: density (ρ ) & diameter (d) of Cu=?

Solution: plane is (1 1 1)

$ d= \frac{a}{\sqrt{1^2 + 1^2 + 1^2}}$

$ a = 2.08 × 10 ^{-10} ×\sqrt{3}$

$a …

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