Respuesta :
Answer:
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
