Body Centred Cubic Structure?

Question

At a certain temperature and pressure an element has the simple body-centred cubic unit cell, depicted below. The corresponding density is 11.098 g/cm^3 and the atomic radius is 1.780 Å. Wat is the atomic mass for this element (in amu).

1 Å = 1.0*10^-8 cm

thank you.

Answer 1

I noticed your question was unanswered for an hour so, I guess this question is my burden.

Applying the rules of Pythagorean Theorem;

16r^2 = l^2 + (√ 21)^2=3l^2
16r^2 = 3l^2

r=[(√ 3)/4]l
Using this equation, you can first solve for the Length of unit cell (l).

r=[(√ 3)/4]l ---> r = 0.433l ---> r/0.433 = l

l = 1.780Å x (1.0x10^-8cm/1Å) = 1.78x10-8cm

Once the length is found, you can cube it (length x height x width) to find the unit cell's volume.

v = (1.78x10-8cm)^3 = 5.64 x 10-24cm^3

As you know cm^3 is interchangable with ml therefore;

v = 5.64 x 10-24ml

To find the mass of unit cell, use the density given:

Mass Unit Cell = 5.64 x 10-24ml x (11.098g/ml) = 6.26 x 10^-23g

Since in Centered Cubic Structure, there are total of two atoms, Divide the mass of unit cell by 2 to determine the mass of each atom.

6.26 x 10^-23g/2atoms = 3.13 x 10^-23g/atom

Now, convert this mass to amu:

3.13 x 10^-23g (6.022amu/1.00g) = 18.85amu.

I hope this helps.