One mole of magnesium 6 times 1023 atoms has a mass of 24 gr

One mole of magnesium (6 times 10^23 atoms) has a mass of 24 grams, and its density is 1.74 grams per cubic centimeter, so the center-to-center distance between atoms is 2.84 times 10 ^-10 m. You have a long thin bar of magnesium, 2.7 m long, with a square cross section, 0.06 cm on a side. You hang the rod vertically and attach a 8 kg mass to the bottom, and you observe that the bar becomes 1.47 cm longer. From these measurements, it is possible to determine the stiffness of one interatomic bond in magnesium. What is the spring stiffness of the entire wire, considered as a single macroscopic (large scale), very stiff spring k_s = N/m 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 magnesium wire. Note that the cross-sectional area of one magnesium atom is (2.84 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 = What is the stiffness of a single interatomic \"spring\" k_s, i = N/m An interatomic bond in magnesium 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. Expert Answer

Solution

1.

Weight of the object is balanced by the spring force

mg = k_sx

k_s= mg/x =8*9.8/0.0147

k_s=5333.3 N/m

2)

Area at the bottom

A=S²=(0.06*10^-2)²=3.6*10^-7 m²

Number of side by side Long chains of atoms in magnesium are

n =3.6*10^-7/(2.84*10^-10)² =4.46*10¹² atoms

3)

Number of bonds in one chain are

N =L/d =2.7/(2.84*10^-10)

N=9.5*10⁹

^d)

Stiffness of single interatomic spring is

K=k_sn/N =5333.3*4.46*10¹²/(9.5*10⁹)

K=2.5*10⁶ N/m