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The cell edge of a fcc crystal is 100 pm...

The cell edge of a fcc crystal is 100 pm and its density is `10.0 g cm^(-3)` . The number of atoms in 100 g of this crystal is `:`

A

`1 xx 10^(25)`

B

`2 xx 10^(25)`

C

` 3 xx 10^(25)`

D

`4 xx 10^(25)`

Text Solution

Verified by Experts

The correct Answer is:
D
Volume of units cell `= ( 100 xx 10^(-10) cm)^(3)`
`= 10^(-24) cm^(3)`
Density `= 10.0 g cm^(-3)`
Mass of unit cell `= 10^(-24) xx 10`
`= 10 ^(-23) g `
No. of unit cells in 100 g `= ( 100)/( 10^(-23)) = 10^(25)`
Since the lattice is fcc and each unit cell has 4 atoms per unit cell.
No. of atoms `= 4 xx 10^(25)`
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