The number of atoms in `100 g an fcc` crystal with density `d = 10 g//cm^(3)` and the edge equal to 100 pm is equal to

Options

A. `3xx10^(25)`
B. `5xx10^(24)`
C. `1xx10^(25)`
D. `2xx10^(25)`

Answer

2

Explanation

`a=200 p m =200xx10^(-10)cm=2xx10^(-8)cm`
volume `=(2xx10^(-8))^(3)cm^(3)`
No. of atoms `=(ZxxA)/(d xx a^(3))=(4xx100)/(10xx(2xx10^(-8))^(3))=5xx10^(24)`