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Homework Help: Stiffness of a single interatomic spring ?

  1. Feb 25, 2012 #1
    Stiffness of a single interatomic "spring"?

    One mole of iron (6 x 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 x 10^-10 m. You have a long thin bar of iron, 2.8 m long, with a square cross section, 0.07 cm on a side.

    You hang the rod vertically and attach a 50 kg mass to the bottom, and you observe that the bar becomes 1.4 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?
    ks = 35000 N/m [Correct]

    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 x 10^-10)^2 m^2.
    Number of side-by-side long chains of atoms = 9.43e12 [CORRECT]

    3) How many interatomic bonds are there in one atomic chain running the length of the wire?
    Number of bonds in total length = 1.23e10 [CORRECT]

    4) What is the stiffness of a single interatomic "spring"?
    ks,i = [Units in N/m]

    I cannot figure how to get the 4th question. Any hints or steps would be useful. Thanks!
  2. jcsd
  3. Feb 25, 2012 #2


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    Re: Stiffness of a single interatomic "spring"?

    When a spring of length L is pulled with force F and the spring stretches by ΔL:

    F=ks ΔL.

    Now the spring is cut in half, the ends are connected at point C to a small bead. (See picture.) When equilibrium, the right string pulls the bead with the same force F applied at the right end, as the tension is the same along the whole spring. The left spring pulls the bead with the same force, and the bead pulls the left spring also by force F, according to Newton Third Law. So both half-springs are stretched by force F. But the half-springs stretch only by half of the original ΔL. F=ks' ΔL/2. So the stiffness factor of a half spring is ks' = 2F/ΔL, twice of the original. You can extend the same argument to any identical strings in series.

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    Last edited: Feb 25, 2012
  4. Feb 25, 2012 #3
    Re: Stiffness of a single interatomic "spring"?

    My ksΔL.
    ks is 35000N/m * 0.014m = 490N = F

    2*[490N] / 0.014m = ks'
    ks' = 70000N/m which is twice the original like you said.
    I still don't know how to get stiffness of a single interatomic "spring". Am I confused about what number to use? I believe the answer is a 2 digit N/m value and I don't know what I'm doing wrong.

  5. Feb 25, 2012 #4


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    Re: Stiffness of a single interatomic "spring"?

    You have calculated both the number of the side-by side atomic chains in the wire, and also the number of interatomic bonds in an atomic chain along the length of wire. One bond corresponds to an elementary spring.
    So the wire consist of parallel chains, and a chain consist of the elementary bonds-the elementary springs.

    If you know the spring constant of the wire and the number of side-by side chains, you get that of a single chain.

    (When the whole wire is pulled by force F, the chains equally share that force. The whole wire stretches ΔL, and all the chains stretch by the same length. From that you find out how is stiffness constant of a single atomic chain related to that of the whole wire. )

    One chain is made of N elementary springs, and you know already, how the stiffness constant of the whole chain is related to that of an elementary spring.

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