The fraction of `Ca` atoms that lies on the surface of a cubic crystal that is `1.0 cm` in length is

Options

A. `1.11 xx 10^(-8)`
B. `2.22 xx 10^(-8)`
C. `1.11 xx 10^(-7)`
D. `2.22 xx 10^(-7)`

Answer

A

Explanation

For bcc , `Z_(eff) = 2`
`:.` Number of atoms `= 2 xx 9.17 xx 10^(22)`
Surface area of one face of a crystal `= (10^(-2)m)^(2)`
`=10^(-4)m^(2)`
Surface area of one face of a unit cell `= a^(2)`
Number of square faces visiable at one face of crustal
`= (10^(-4)m^(2))/((22.17 xx 10^(-12))^(2)m^(2)) = 2.03 xx 10^(15)`

Number of atom per face of unit cell `= 1`
Number of atoms `= 1 xx 2.03 xx 10^(15)`
Fraction of Ca atoms `= (2.03 xx 10^(15))/(2 xx 9.17 xx 10^(22))`
`= 1.11 xx 10^(-8)`