One mole of nickel (6 times 10^23 atoms) has a mass of 59 grams, and its density

Question

One mole of nickel (6 times 10^23 atoms) has a mass of 59 grams, and its density is 8.9 grams per cubic centimeter, so the center-to-center distance between atoms is 2.23 times 10^-10 m. You have a long thin bar of nickel, 2.5 m long, with a square cross section, 0.05 cm on a side. You hang the rod vertically and attach a 27 kg mass to the bottom, and you observe that the bar becomes 1.32 cm longer From these measurements, it is possible to determine the stiffness of one interatomic bond in nickel. 1) What is the spring stiffness of the entire wire, considered as a single macroscopic (large scale), very stiff spring? K_s = 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 nickel wire. Note that the cross-sectional area of one nickel atom is (2.23 times 10^-10)^2 m^2. Number of side-by-side long chains of atoms 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"? K_s = N/m An interatomic bond in nickel is stiffer than a slinky, but less stiff than a pogo stick. The stiffness of a single interatomic bond is very much smaller than the stiffness of the entire wire.

Explanation / Answer

1)

The force applied to the bar is 27*9.81 = 264.87 N.

The extension of the wire L = 0.0132 m,

so the bar stiffness is F/L = 20065.90 N/m

2)

The number of atoms in one layer of cross section is area of the bar divided by the area of one atom

= 0.0005^2 / (2.23 x 10^-10)^2 = 5.02 x 10^12.

3)

No no of bonds along the length is 2.5 / (2.23 x 10^-10) = 1.12 x 10^10

4)

k = (20065.9 x 1.12 x 10^10) / (5.02 x 10^12)

k = 44.8 N/m

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