Question

A hypothetical metal has an orthorhombic unit cell for which the a, b, and c lattice parameters are 0.413 nm, 0.665 nm, and 0.876 nm, respectively. (a) If there are 8 atoms per unit cell and the atomic packing factor is 0.536, determine the atomic radius (in nm). (b) If the density is 3.99 g/cm3, calculate the metal's atomic weight (in g/mol).

Answer 1

Answer:

atomic radius  R = 0.157 nm

metal atomic weight = 72.27 g/mol

Explanation:

given data

parameters a =  0.413 nm

parameters b = 0.665 nm

parameters c =  0.876 nm

atomic packing factor = 0.536

density = 3.99 g/cm³

to find out

atomic radius and  atomic weight

solution

we apply  here atomic packing factor (x) that is

atomic packing factor (x) = $\frac{volume(sphere)}{volume(unit\ cell)}$  ..................1

put here value we get

atomic packing factor = $\frac{8*(4/3)*\pi R^3}{3*a*b*c}$

R = $(\frac{3(x)(abc)}{32\pi })^{1/3}$

R =  $(\frac{3(0.536)(0.413*0.665*0.876)}{32\pi })^{1/3}$

atomic radius  R = 0.157 nm

and

now we get here metal atomic weight that is

metal atomic weight = $\frac{\rho (abc)(N_A)}{no\ of\ atom}$   ....................2

metal atomic weight = $\frac{3.99 (0.413*0.665*0.876)(6.023*10^{23})}{8}$  

metal atomic weight = 72.27 g/mol