Metal `M` of radius `50 nm` is crystallized in `fcc` type and made cubical crystal such that face of unit cells aligned with face of cubical crystal. If the total number of metal atoms of `M` at all faces of cubical crystal is `6 xx 10^(30)`, then the area of one face of cubical crystal is `A xx 10^(16) m^(2)`. Find the value of `A`.

Answer

2

Explanation

Consider one face of unit cell as show below.

Number of atoms on one face
`= 4("corners) xx 1/8("per corner share") + 1 ("face centre")`
`xx 1/2 ("face centre share")`
`= 1/2 + 1/2= 1//"per face"`
Given number of atoms on all faces `= 6 xx 10^(30)`
Given number of atoms on one face `= 1/6 xx 10^(30)`
`= 10^(30)` atoms
Number of unit cells at one face of crystal
`= (6 xx 10^(30))/(6) = 10^(30)`
So, number of unit cells at the edge of crystal `= sqrt(10^(30))`
`= 10^(15)`
Now, edge length of unit cell `= 4/(sqrt(2)) xx 50 nm`
Edge length of cubical crystal `= 4/(sqrt(2)) xx 50 xx 10^(15) nm`
So, area of face of crystal `= (4/(sqrt(2)) xx 50 xx 10^(15))^(2)nm^(2)`
`= 16/2 xx 25 xx 10^(2) xx 10^(30)`
`= 2 xx 10^(34) nm^(2)`
`= 2 xx 10^(-18+34) m^(2)`
`= 2 xx 10^(16)m^(2)`
`:. A xx 10^(16)m^(2) = 2 xx 10^(16)m^(2)`
`A = 2`