Answer:
(a)
Stiffness of one inter-atomic bond in iron = 5.14*10^-9 N/m
(b)
Spring stiffness of entire wire = 45.37 N/m
Explanation:
Given data:
No of atoms = = 1 mole of iron = 6*10^23
Mass = 56 g
Density = 7.87 g / cm^3
Original Length = 2 m
Hanged mass = 231 kg
Change in length = 1.01 cm
Solution:
Force applied to bar = 231 * 9.8
= 2263.8 N
Change in length = ΔL = 1.01 cm = 0.0101 m
Bar stiffness = F / ΔL = 2263.8 / 0.0101
Bar stiffness = 2.24*10^5 N/m
Cross sectional area of bar= 0.0015*0.0015
= 2.25*10^-6 m^2
No. of atoms in one layer of cross-sectional area
= Cross-Sectional area of bar / Area of one atom
No. of atoms = 2.25*10^-6 / (2.28*10^-10)^2
No. of atoms in one layer of cross-sectional area = 4.33*10^13
Length of Bar = 2 m
No. of bonds along bar length
= Bar length / Center to center distance between atoms
No. of bonds = 2 / 2.28*10^-10
No. of bonds = 8.77*10^9
Force applied to each atom
= Force / ( No. of atoms * No. of bonds )
Force applied to each bond = 2263.8 / ( 4.33*10^13 * 8.77*10^9)
Force applied to each bond = 5.96*10^-21 N
Strain on bar on bar = ΔL / L = 0.0101 / 2
= 0.0051
Strain between atom layers will be same as above.
Bond extension (elongation) = 2.28*10^-10 * 0.0051
= 1.16*10^-12 m
Bond stiffness = Force applied to each bond / Bond extension
Bond stiffness = 5.96*10^-21 / 1.16*10^-12
Bond stiffness = 5.14*10^-9 N/m
Stiffness of one inter-atomic bond in iron = 5.14*10^-9 N/m
Part (b)
Spring stiffness of entire wire = (Bar stiffness * No. of bonds) / No. of atoms
Spring stiffness = (2.24*10^5 * 8.77*10^9) / 4.33*10^13
Spring stiffness = 45.37 N/m
