A perfectly spherical iron ball bearing weighs 21.91 grams. Derive the diameter of the ball bearing assuming an iron atom has an effective radius of 0.124nm and iron is BCC at room temperature. The answer should be in cm with 2 decimals of accuracy.

Explanation:

We need to find the diameter of the ball bearing given its mass of 21.91 grams. We can use the density to find the volume knowing the mass.

They tell you the effective radius of the atom which has a body centered cubic unit cell (BCC). This will allow us to calculate the volume of the unit cell.

The mass of the unit cell can be determined since a BCC unit cell contains two atoms ( 1/8 of 8 corners and 1 in the center of the cube)

Having mass and volume of the unit cell, compute the density and solve the question.

Then we can compute the diameter of of the bearing in this question.

In a BCC unit cell, the length of the side of the cube is:

a = 4r/√3 where r is the effective radius given.

r = 0.124 nm x 1 x 10⁻⁷ cm / nm = 1.24 x 10⁻⁸ cm

( converted the radius to cm because we will be using the density in the typical  units of g/cm³ )

a = 4 x 1.24 x 10⁻⁸ cm / √3 = 2.86 x 10⁻⁸ cm

The volume of the cubic cell is a³:

v = (2.86 x 10⁻⁸ cm)³ = 2.35 x 10⁻²³ cm³

m unit cell = # atoms unit cell x mass of an iron atom

= 2 atoms x 55.85 g/mol x 1 mol/ 6.022 x 10²³ atom = 1.85 x 10⁻²² g

d = 1.85 x 10⁻²² g / 2.35 x 10⁻²³ cm³ = 7.88 g/cm³

Volume ball = m/d = 21.91 g / 7.88 g/cm³ = 2.78 cm³

The radius can be calculated from the volume of a sphere :

V = 4/3 π r³ ⇒ r = ∛(3 V/4 π)

r = ∛ ( 3 x 2.78 cm³ /4 π) =0.66 cm

Finally the diameter is:

d = 2r = 2 x 1.28 cm = 0.66 cm

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