(i) What is the total number of voids in cubic close-packed lattice? (ii) Metallic gold (Au) crystallises in face-centred cubic lattice. What is the number of unit cells in 2.0g of gold? [Au = 197] or, (i) What is a p-type semiconductor? (ii) A cubic crystal is made up of elements A and B. B is located at the corners of the unit cell and A is at the body-centre. What will be the probable formula of the compound?

(i) What is the total number of voids in cubic close-packed lattice?

1.	(i) What is the total number of voids in cubic close-packed lattice? (ii) Metallic gold (Au) crystallises in face-centred cubic lattice. What is the number of unit cells in 2.0g of gold? [Au = 197] or, (i) What is a p-type semiconductor? (ii) A cubic crystal is made up of elements A and B. B is located at the corners of the unit cell and A is at the body-centre. What will be the probable formula of the compound?
Answer» Solution :(i) In CUBIC close-packed lattice, the total number of voids equals to 12 per unit cell. (II) 197 G Au contains `6.022xx10^(23)` atoms. `THEREFORE" "2.0g` Au contains, `(2xx6.022xx10^(23))/197=6.144xx10^(21)` atoms In fcc lattice, there are 4 atoms per unit cell. `therefore6.114xx10^(21)` atoms consist `(6.114xx10^(21))/4=1.528xx10^(21)` unit cells. Therefore, `2.0`g Au consists `1.528xx10^(21)` unit cells of fcc lattice. (ii) B atoms are located at the corners of cubic unit cell. Therefore, number of B atom per unit cell = `8xx1/8=1` A atoms are located at the body-centre. Therefore, number of A atoms per unit cell = 1 Thus, the formula of the COMPOUND wil be AB.