Iron has a density of7.87g/cm3, and the mass of an iron atom isrole="math" localid="1654573436593" 9.27×1026kg. If the atoms are spherical and tightly packed, (a) what is the volume of an iron atom, and (b) what is the distance between the centers of adjacent atoms?

Short Answer

Expert verified

(a). The volume of an iron atom is 1.18×1029m3.

(b). The distance between the centers of adjacent atoms is 2.82×1010m.

Step by step solution

01

Given data

The density of iron,p=7.87g/cm3

The mass of iron atom,m=9.27×1026kg

02

Understanding the density of material

The density of a material (iron) is defined as the mass per unit volume. The distance between centers of adjacent atoms is equal to twice the radius of an atom.

The expression for density is given as:

p=mv … (i)

Here, pis the density, mis the mass and vis the volume.

03

(a) Determination of the volume of iron atom

Convert the density of iron into .

p=7.87gcm3×1kg1000g×100cm1m3

Using equation (i), the volume of an iron atom is,

localid="1660812050656" v=mp=9.27×1026kg7870kg/m3=1.18×1029m3

Thus, the volume of iron atom is 1.18×1029m3.

04

(b) Determination of the distance between the centers of adjacent atoms

The volume of atom is given as follows:

v=4πR33

Here, Ris the radius of the atom.

Solving for R and substituting the values,

R=3V4π=3×1.18×1029m34m=1.14×1010m

Now, the distance between the centers of two adjacent atoms is,

d=2R=21.14×1010m=2.82×1010m

Thus, the distance between the centers of two adjacent atoms is,2.82×1010m .

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