Question

One mole of iron (6 × 1023 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, 2.7 m long, with a square cross section, 0.07 cm on a side. You hang the rod vertically and attach a 100 kg mass to the bottom, and you observe that the bar becomes 2.70 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= 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 iron wire. Note that the cross-sectional area of one iron atom is (2.28× 10-10)2 m2. Number of side-by-side long chains of atoms =

Answer 1

Answer:

a) k$_{s}$ = 3.6 N/m ; b) 9.43 * 10¹²; c) 1.18 * 10¹¹; d) 75.08 N/m

Step-by-step explanation:

Length of iron bar = L = 2.7m;

side length of cross section = a = 0.07cm = 0.0007m;

x = 2.7 cm = 0.027m;

m = 100kg;

ρ = 7.87gm/cm³;

da = 2.28 * 10⁻¹⁰m;

a)

Fnet = F - mg

where Fnet = 0

So,

F = mg where F=k$_{s}$x

k$_{s}$ = mg/x = 100*9.8/0.027

k$_{s}$ = 3.6 N/m

b)

Nchain = Aw/Aa =a²/(da)²

                           = (o.ooo7)²/(2.28 * 10⁻¹⁰)²

                           = 9.43 * 10¹²

c)

Nbond = L/da = 2.7/2.28 * 10⁻¹⁰

                       = 1.18 * 10¹¹

d)

Spring stiffness of wire = ksi = (Nbondk$_{s}$)/Nchain

                                      = [(1.18* 10¹¹)(6*10⁴)]/(9.43 * 10¹²)

                                      = 75.08 N/m