Question

A hanging magnesium wire with diameter 2.4 mm (2.4 × 10^-3 m) is initially 0.60 m long. When a 100 kg mass is hung from it, the wire stretches an amount 3.25 mm. A mole of magnesium has a mass of 24 grams, and its density is 1.74 g/cm³. Find the approximate value of the effective spring stiffness of the interatomic force.

Answer 1

Let, D -Diameter of wire L Initial length of wire x - Extended length m -Attached mass ρ-Density of magnesium d -Bond length Given: D- 2.4 x 10-3 m L-0.6 m m 100 kg x 3.25 mm - 3.25 x 10-3 m Cmm Mass of 1 mole of magnesium-24 g 9

k, = Yd Where, Y -Young's Modulus d -Bond length ki -Effective spring stiffness of interatomic forces Where Y is given by, Stress strain/ Here, A -Area of cross section of wire, Therefore,

(100 kg) (9.8 2.4 x 10-32 (0.6 m m2 (3.25 x 10-3 m) 4 x 1010 kg m s =4x1010- To Calculate Bond length If we consider tight binding model, then bond length is equal to diameter of 1 atom, and is given by Where VMg - Volume occupied by one atom, 24 mo (1.74 3) (6.022 x 1023 mol--)

2.840 × 10-8 cm = 2.840 x 10-10 m From (1) = (4 × 1010 (2.840 × 10-10 m) = 11.36 Hence effective spring stiffness of interatomic forces-k, 11.36 N/n