Answer:
a) [tex]1.1823\times 10^{-29}m^3[/tex] is the average volume required for each iron atom.
b) The distance between the centers of adjacent atoms is [tex]2.2781\times 10^{-8} m[/tex].
Explanation:
a) Mass of an iron = 7.84
Volume of iron = [tex]1 cm^3[/tex]
Let the x atoms of iron weights 7.84 g and volume of [tex]1 cm^3[/tex].
Mass of an iron atom = [tex]9.27\times 10^{-26} kg=9.27\times 10^{-23} g[/tex]
[tex]x\times 9.27\times 10^{-23} g=7.84 g[/tex]
[tex]x =8.4573\times 10^{22} atoms[/tex]
Volume of x atoms of iron = [tex]1 cm^3[/tex]
Volume of an iron atom = v
[tex]x\times v=1 cm^3[/tex]
[tex]v=\frac{1 cm^3}{8.4573\times 10^{22} }=1.1823\times 10^{-23}cm^3[/tex]
[tex]1 cm^3= 10^{-6} m^3[/tex]
[tex]v=1.1823\times 10^{-29}m^3[/tex]
[tex]1.1823\times 10^{-29}m^3[/tex] is the average volume required for each iron atom.
b) Volume of an iron atom = v
Volume of the cube = [tex]s^3[/tex]
Length of iron cube -= s
[tex]1.1823\times 10^{-29}m^3=s^3[/tex]
[tex]s=2.2781\times 10^{-8} m[/tex]
According to question we are treating iron atom as cube. So all the cubic iron atoms will have a same side.
[tex]s=2.2781\times 10^{-8} m[/tex]
The distance between the centers of adjacent atoms:
Horizontal distance of the center from the one face of the cubic iron atom:
=[tex]\frac{s}{2}[/tex]
Horizontal distance of the center from the one face of the another cubic iron atom:
=[tex]\frac{s}{2}[/tex]
[tex]\frac{s}{2}+\frac{s}{2}=s[/tex]
The distance between the centers of adjacent atoms is [tex]2.2781\times 10^{-8} m[/tex].
