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Question

The number of atoms in 100 g of an FCC crystal with density (d) equal to 10 g cm3 and cell edge of 200 pm is equal to :

A
3×1025
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B
5×1024
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C
1×1025
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D
2×1025
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Solution

The correct option is B 5×1024
Density of the unit cell,
d=(No.ofatomsinitsunitcell)(Massofeachatom)(a31030)
d=ZMNA(a31030).
here, Z=4,d=10gm/cm3,a=200pm
by putting values,
d=4100/N(20031030)=10
So, N=51024

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