Answer:
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
