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The number of atoms in 100 g of an fcc c...

The number of atoms in `100 g` of an fcc crystal with density `= 10.0g cm^(-3)` and cell edge equal to `200` `"pm"` is equal to

A

`3times10^(25)`

B

`5times10^(24)`

C

`1times10^(25)`

D

`2times10^(25)`

Text Solution

Verified by Experts

The correct Answer is:
B
We know, `d=(ZM)/(N_(A)timesa^(3))`
or`N_(A)=(ZM)/(d timesa^(3))=(4times100)/(10times(2times10^(-8))^(3))=5times10^(24)`
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