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Face centred cubic crystal lattice of copper has density of 8*966" g. cm"^(-3). Calculate the volume of the unit cell. Given molar mass of copper is 63*5"g. mol"^(-1) and Avogadro number N_(A) is 6*022xx10^(23)"mol^(-1). |
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Answer» Solution :Density of the unit CELL `(d)=("Maxx of the unit cell"(M))/("Volume of the unit cell"(V))`. . . (i) Let number of atoms in the unit cell be Z and mass of each atom m then `M=Zxxm` Mass of each atom `(m)=("Atomic mass")/("Avogadro's no.")` Thus from equation (i) `d=(ZXX"Atomic mass")/("Avogadro's no." xx V)` . . . .(ii) For FACE centered cubic lattice (FCC), the TOTAL no. of atoms in the unit cell `(Z)=(1)/(8)xx8+(1)/(2)xx6` `=1+3=4` So, `8*966"g cm"^(-3)=(4xx63*5"g mol"^(-1))/(6*022xx10^(23)"mol^(-1)xxV)` or, Volume of the unit cell `(V)=(254)/(53*99xx10^(23))cm^(3)` `=4*7xx10^(-23)cm^(3)` Hence, the volume of the unit cell is `4*7xx10^(-23)cm^(3)`. |
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