A hanging iron wire with diameter 1.9 mm (1.9 × 10-3 m) is initially 0.75 m long. When a 68 kg mass is hung from it, the wire stretches an amount 0.881 mm. A mole of iron has a mass of 56 grams, and its density is 7.87 g/cm3. Find the approximate value of the effective spring stiffness of the interatomic bond. N/m

The approximate value of the effective spring stiffness of the interatomic bond 4.209×10^(-18) N/m.

Explanation:

To solve the question we first calculate the spring stiffness thus

F = k×e

Where F = force in Newtons

e = extension in meters

k = spring stiffness

The force = force of gravity = mass×acceleration

= 68 kg × 9.81 m/s^2= 667.08 N

Extension = 0.881 mm = 8.8×10^(-4) m

Therefore 667.08 N = k × 8.8×10^(-4) m

Or k = 7.573×10^5 N/m

Mass of one mole of iron Fe = 56 grams,

Density of one mole of iron = 7.87 g/cm3.

Therefor volume of one mole of iron = (Mass of one mole of iron)/(Density of iron) = ( 56 g)/( 7.87 g/cm3.) = 7.116 cm ^3 = 7.116 cm ^3 ×(1 m^3)/(1×10^6 cm^3)= 7.116×10^(-6) m^3

Since the diameter of wire = 1.9×10^(-3) m then the area of the cross section of the wire = D^2×pi/4 = Diameter^2×pi/4 = pi × ((1.9×10^(-3) m)^2)/4 = 2.84×10^(-6) m^2

Therefore one mole of iron will be present in

(7.116×10^(-6) m^3)/( 2.84×10^(-6) m^2) length of wire or 2.51 m length of wire

However, length of wire = 0.75 m which is equivalent to (0.75 m)/ (2.51 m/mole) = 0.2988 moles

By Avogadro's law

0.2988 moles contains

0.2988 × 6.02 × 10^23 atoms = 1.799 × 10^23 atoms

The approximate value of spring stiffness = (7.573×10^5 N/m)/( 1.799 × 10^23 atoms) = 4.209×10^(-18) N/m

The approximate interatomic bond spring stiffness = 4.209×10^(-18) N/m

the correct answer would be d. hope this !