Question

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, 3.0 m long, with a square cross section, 0.15 cm on a side.

You hang the rod vertically and attach a 142 kg mass to the bottom, and you observe that the bar becomes 1.55 cm longer. From these measurements, it is possible to determine the stiffness of one interatomic bond in titanium.

$04-017-chains_cross_section.jpg$

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 titanium wire. Note that the cross-sectional area of one titanium atom is (2.60 10^-10)² m².
Number of side-by-side long chains of atoms =

3) 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,i = N/m

An interatomic bond in titanium 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.

Answer 1

Giver 6x70 ( 6x10" allora, m=489 deste - 4.549 cm, 26xiom 7:37, 0-15 cm-Sides, 140 cm lais Spring stiffness mg = kn k=mg/n = 642) (98)/65570) = (89780.645 Nm (2) Area of one atom (260)² = 646x10-20 Potal area - (815x10²) ² (1.5x 10) = 2.25 x10-6 No. of atom 295x100 - 3328x10" alon ] 6-76/70-30 (3) Total length = 3m tength I one bond: 2610-10 No. 2 interatone bonds: 3 / 126x10 9 = 11154 xdo] ( Wikter One long chen I likain = k/3-328410 R-for individuel bond - 1/833880)< (1010) (89780-645 ) x (1.154x100) 3.398X102 [ 31.13 N/m