Given the facts that: - Silver (Ag) adopts an fcc unit cell, and cadmium (Cd) ad

Question

Given the facts that:

- Silver (Ag) adopts an fcc unit cell, and cadmium (Cd) adopts an hcp unit cell.

- The atomic weights are: Ag= 107.8682 g/mol, and Cd= 112.411 g/mol (Cd has one extra proton and three extra neutrons)

- the lattice parameter of Ag is a = 408.53 pm, and the lattice parameters of Cd are a = 297.94 pm and c = 561.86 pm.

Part a)

Calculate the theoritical densities of Cd and Ag (Vhcp = a*a*cos(30)*c)

Part b)

calculate the effective atomic radii of the two elements based upon the assumption that:

<110> Direction along which atoms touch in Ag

<11-20> Direction along which atoms touch in Cd

Part C)

In part b, you should have calculated the the density of Cd < the density of Ag, even though Cd is a heavier atom. Using this information in part c and an appropriate equation or theory, explain why Ag is denser than Cd even though Cd has a higher atomic mass. ( Hint: what does ideal mean for one crystal structure?)

Explanation / Answer

Mass of one Silve atom =107.8682/(6.023*1023)= 1.79X10-22 g/atom

mass = 4*107.86*/6.023*1023=7.16*10-22 gm

Volume of unit cell= a3= (408.53/10-10)3 = (4.08X10-8)3=6.818X10-23cm3

density = mass/ volume = 7.16X10-22/6.818X10-23 =10.4 g/cc

For cadmium

Density= no of atoms per unit cell* Avagadro constant/Volume

Volume of cell = 3*Sqrt(3)*a2*c/2= 3*sqrt(3)*(2.9794X10-8)2*561.86X10-8/2=1.159*10-22 cm3

mass =6( no of atoms/ unit cell)*112.41/6.023*1023=111.98X10-23 gms

Density= 1.1198X10-21/1.159X10-22=10.34 g/cc

b)

(4/3)*Pi*r3=1.159X10-22

4.19 r3= 1.159X10-22

r= (1.159X10-22/4.19)1/3=3.04X10-8cm

Density= 10.4 g/cc

c = 1.63a = 3.26R

R= c/3.26=561.86/3.26=172pm

C) cadmium has= one extra proton and hence cadmium is les dense than Ag.

For cadmium atomic radii=a/2 =297.94/2= 148.97pm

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