One mole of nickel (6 1023 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 10-10 m. You have a long thin bar of nickel, 2.1 m long, with a square cross section, 0.08 cm on a side. You hang the rod vertically and attach a 69 kg mass to the bottom, and you observe that the bar becomes 1.11 cm longer. From these measurements, it is possible to determine the stiffness of one interatomic bond in nickel.

6.58*10^-7 N/m

Step-by-step explanation:

The applied force  to the nickel bar = 69 x 9.8 = 676.2 N.

The change in length of the bar ∆L = 0.0008 m,

stiffness k = F/∆L = 8.45*10^5 N/m

Number of atoms per 1 layer of cross section

= area of the square cross section of the nickel bar / area of one atom =0.0008²/(2.23*10^-10)²

= 1.287  x 10^13

No of bonds per 1 nickel length is 2.1/(2.23*10^-10) = 9.41*10^9

The force applied per each nickel atom in a layer = 676.2/(1.287*10^13*9.41*10^9) = 5.58*10^-20 N

The strain (fractional  change in length, ∆L/L) on the bar =  0.0008/2.1 = 0.00038.

This is also defines the strain exacted between the nickel atom layers.

Bond extension ∆L = 2.23*10^-10 * 0.00038 = 8.474*10^-14 m.

stiffness of one inter atomic bond in nickel F/∆L

= (5.58*10^-20)/(8.474*10^-14) = 6.58*10^-7 N/m

the answer is n

chase is right because he already been to the swings and he know how far it will be to get there

step-by-step explanation: