physics

One mole of titanium (6 1023 atoms) has a mass of 48 grams, and its density is 4.54 grams per cubic centimeter, so the center-to-center distance between atoms is 2.60 10-10 m. You have a long thin bar of titanium, 2.4 m long, with a square cross section, 0.11 cm on a side.

You hang the rod vertically and attach a 76 kg mass to the bottom, and you observe that the bar becomes 1.23 cm longer. From these measurements, it is possible to determine the stiffness of one interatomic bond in titanium.

What is the stiffness of a single interatomic "spring"?

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asked by hannah
  1. The force applied to the bar is 34*9,8 = 333.2 N. The extension of the wire ∆L = 0.0145 m, so the bar stiffness is F/∆L = 2.30*10^4 N/m

    The number of atoms in one layer of cross section is area of the bar divided by the area of one atom =0.0012²/(2.84*10^-10)² = 1.785*10^13.

    No no of bonds along the length is 2.5/(2.84*10^-10) = 8.80*10^9

    The applied force is divided among all the bonds along the length, and then divided among all the atoms in a cross-section layer The force applied to each atom in a layer is then 333.2/(1.785*10^13*8.80*10^9) = 2.12*10^-21 N

    The strain (fractional length increase, ∆L/L) on the bar is 0.0145/2.5 = 0.0058. This will also be the strain between the atom layers. The bond extension ∆L is then 2.84*10^-10 * 0.0058 = 1.6472*10^-12 m. The bond stiffness F/∆L = (2.12*10^-21)/(1.6472*10^-12) = 1.29*10^-9 N/m

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    posted by BOB

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