How Does the Stiffness of an Interatomic Bond Affect Iron's Elasticity?

[clutch12]

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?

 

[nasu]

You need the area of the cross section of the wire. What you did calculate is the area of a lateral face. The cross section is a square with a side of 0.5 cm.

 

[clutch12]

so I am multiply 0.5 cm by 0.05 cm?

 

[denverdoc]

Area=length*width and L=W. So assuming your post is a typo, 0.5cm x 0.5cm But better yet convert to meters first so the units are the same as others in the problem.

note: .5cm=0.005 meters

 

[nasu]

Sorry if I confused you. The side of the square is 0.05 cm, according to your problem. I did not read all the zeroes.
So the cross section area is (0.05cm)^2= [5x10^(-4)m]^2

 

[clutch12]

oh so basically i don't need to multiply it by 2.9 then right?

 

[ideasrule]

No, you don't. You're trying to determine how many chains of atoms end on the end of the rod, not how many chains end on the side of the rod.