1.

The nearest neighbouring silver atoms in the silver crystals are 2.87 xx 10^(-10) m apart. What is the density of silver ? Silver crystallises in fcc form. [Atomic mass of Ag = 108]

Answer»

Solution :Nearest neighbours if fcc crystal `=(1)/(2)XX" FACE diagonal"=(1)/(2)(sqrt2a)`
`"or"2.87xx10^(-10)m=(a)/(sqrt2)`
`"or"a=2.87xx1.414xx10^(-10)=4.05xx10^(-10)m=4.05xx10^(-8)cm`
`"orVolume "=a^(3)=(4.05)^(3)xx10^(-24)`
Use the following equation to CALCULATE d
`d=(zxxM)/(a^(3)xxN_(A))=(4xx108)/((4.05)^(3)xx10^(-24)xx6.023xx10^(23))=(108)/(1.008xx10)=(108)/(10.08)="10.7 G cm"^(-3)`


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