One mole of nickel (6 times 10^23 atoms) has a mass of 59 grams, and its density
ID: 1537589 • Letter: O
Question
One mole of nickel (6 times 10^23 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 times 10^-10 m. You have a long thin bar of nickel, 2.4 m long, with a square cross section, 0.08 cm on a side. You hang the rod vertically and attach a 67 kg mass to the bottom, and you observe that the bar becomes 1.23 cm longer. From these measurements, it is possible to determine the stiffness of one interatomic bond in nickel. What is the spring stiffness of the entire wire, considered as a single macroscopic (large stale), very stiff spring? 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 nickel wire. Note that the cross-sectional area of one nickel atom is (2.23 times 10^-10)^2 m^2. How many interatomic bonds are there in one atomic chain running the length of the wire? What is the stiffness of a single interatomic "spring"?Explanation / Answer
(a)
The force applied to the bar is 67*9.8 = 656.6 N.
The extension of the wire L = 0.0123 m, so
the bar stiffness is F/L = 53382.11 N/m
(b)
The number of atoms in one layer of cross section is area of the bar divided by the area of one atom =0.0008²/(2.23*10-10)² = 1.287 x 1013
(c)
Number of bonds along the length is 2.4/(2.23*10-10) = 1.076 x 1010
(d)
The applied force is divided among all the bonds along the length, and then divided among all the atoms in a cross-section layer The force applied to each atom in a layer is then 656.6/(1.287 x 1013 *1.076 x 1010 ) = 4.741x10-21 N
The strain (L/L) on the bar is 0.0123/2.4 = 0.005125
This will also be the strain between the atom layers. The bond extension L is then 2.23x10-10 * 0.005125
L = 1.142875 x10-12 m.
The bond stiffness F/L = (4.741x10-21)/(1.142875x10-12)
Bond stiffness = 4.148 x 10-9 N/m
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