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clutch12

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## Homework Statement

One mole of iron (6 *10^23 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 *10-10 m. You have a long thin bar of iron, 2.9 m long, with a square cross section, 0.05 cm on a side.

You hang the rod vertically and attach a 29 kg mass to the bottom, and you observe that the bar becomes 1.65 cm longer. From these measurements, it is possible to determine the stiffness of one interatomic bond in iron.

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

## Homework Equations

## The Attempt at a Solution

A wire = 2.9 * 5 * 10^-4

= (1.45 * 10^-3 ) ^2

= 2.1025 * 10^-6

A1 atom = (2.28 *10 ^10)^2

= 5.1984 * 10^-20

N chains = Awire/ A1atom

= 2.1025 * 10^-6 / 5.1984 * 10^-20

= 4.044 * 10 ^13

Is this right?