One mole of iron (6 x 10“ atoms) has a mass of 56 grams, and its density is 7.87

Question

One mole of iron (6 x 10“ atoms) has a mass of 56 grams, and its density is 7.87 grams per cubic centimeter, so the center-to-center distance between atoms is 2.28 x 10'" m. You have a long thin bar of iron, 2.7 m long, with a square cross section, 0.09 cm on a side. You hang the rod vertically and attach a 106 kg mass to the bottom, and you observe that the bar becomes 1.73 cm longer. From these measurements, it is possible to determine the stiffness of one interatomic bond in iron. 1) What is the spring stiffness of the entire wire, considered as a single macroscopic (large scale), very stiff spring? 11 N/m 2) How many side-by-side atomic chains (long springs) are there in this wire? This is the same as the number of atoms on the bottom surface of the iron wire. Note that the cross-sectional area of one iron atom is (2.28× 10-10)2 m2 Number of side-by-side long chains of atoms- Top of wire Bottom of wire 3) How many interatomic bonds are there in one atomic chain running the length of the wire? Number of bonds in total length- 4) What is the stiffness of a single interatomic "spring"?

Explanation / Answer

Elongation of bar, x = 0.0173 m

weight acting on the bar is

F = kx

mg = kx

k = mg/x

k = (106 kg x 9.81 m/s^2) / 0.0173 m

k = 60108 N/m

number of iron per iron bar is

n = Abar / Aatom

n = (0.09 x 10^-2)^2 / (2.28 x 10^-10)^2

n = 1.558 x 10^13

the number of bond in one bar is

N = L / dcc

N = 2.7 / (2.28 x 10^-10)

N = 1.184 x 10^10

Kspring = k*n / N

Kspring = (60108 N/m x  1.558 x 10^13 ) / 1.184 x 10^10

Kspring = 7.909 x 10^7 N/m

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