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Calculate the number of atoms per unit cell of a metal having lattice parameter 2.9A0 and density 7.87 gm/cm3. Atomic weight of metal is 55.85, Avagadro number is 6.023 x 1023 /gm-mole.
1 Answer
written 3.3 years ago by |
Given:
$a=2.9A^\circ =2.9\times 10^{-8} cm$
$ρ=7.87$gm/cm3
$A =55.85$
$N=6.023\times 10^{23}$/gm-mole
Formula:
$a^3 ρ=\dfrac{nA}{N}$
Solution:
$a^3 ρ=\dfrac{nA}{N}$
$∴n=Na^3 \dfrac{ρ}{A}$
$n= \dfrac{6.023\times 10^{23}\times {(2.9\times 10^{-8})}^3\times 7.87}{55.85}$
$n=2$
Number of atoms per unit cell is 2.