Question
A metallic element has a density of , a lattice constant of
and an atomic weight
. Calculate the number of atoms per unit cell of this element and predict its lattice crystal structure.
Answer :
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Skip in 10
We are asked to calculate the number of atoms per unit cell of a metallic element and predict its lattice structure. Let’s solve it step by step. Given data: * Density, ( \rho = 2.70 \text{ g/cm}^3 ) * Lattice constant, ( a = 4.0495 , \text{Å} = 4.0495 \times 10^{-8} , \text{cm} ) * Atomic weight, ( M = 26.9815 \text{ g/mol} ) * Avogadro's number, ( N_A = 6.022 \times 10^{23} , \text{atoms/mol} ) The number of atoms per unit cell ( n ) is related to the density via: [ \rho = \frac{n M}{N_A a^3} ] Rewriting for ( n ): [ n = \frac{\rho N_A a^3}{M} ] Step 1: Convert (
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