A hanging wire made of an alloy of nickel with diameter 0.07 cm is initially 2.7 m long. When a 43 kg mass is hung from it, the wire stretches an amount 1.48 cm. A mole of nickel has a mass of 59 grams, and its density is 8.9 g/cm3. Based on these experimental measurements, what is Young's modulus for this alloy of nickel? Y = N/m2 As you've done before, from the mass of one mole and the density you can find the length of the interatomic bond (diameter of one atom). This is 2.22 10-10 m for nickel. As shown in the textbook, the micro quantity ks, i (the stiffness of one interatomic bond) can be related to the macro property Y: ks, i = N/m

Explanation:

Given that d = 0.07cm = 0.0007m, l = 2.7m, extension = 1.48cm = 0.0148m, mass = 43kgforce = 43 x 9.81 = 421.83N, Area = Pi x d^2/4 = 3.142 x 0.0007^2/4 = 3.849x10^-7m2Calculate the stress = Force/Area = 421.83N / 3.849x10^-7m2

= 1095961236N/m^2

calculate the strain = extension/original length = 0.0148 / 2.7 = 0.005481Young modulus of elasticity = Stress/Strain = 1095961236N/m^2 / 0.005481 = 1.99 x 10^11Pa = Young's modulus for this alloy of nickelGiven the inter-atomic bond length = 2.22 10-10 mTo Calculate (the stiffness of one inter-atomic bond = Ks,i =Y x dKs,i = 1.99 x 10^11 x 2.22 10-10Ks,i = 4.42 x 10 = 44.18N/m ; the stiffness of one inter-atomic bond