Gold, which has a density of 19.32 g/cm3, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber. (a) If a sample of gold with a mass of 2.075 g, is pressed into a leaf of 3.842 μm thickness, what is the area of the leaf? (b) If, instead, the gold is drawn out into a cylindrical fiber of radius 2.900 μm, what is the length of the fiber?
Part A.
Given that density of gold, rho = 19.32 gm/cm^3
Now given that gold leaf has, m = 2.075 gm
thickness of leaf, L = 3.842 m = 3.842*10^-4 cm
Now we know that:
mass = density*Volume
Volume of leaf = Area*thickness
So,
m = rho*V = rho*A*L
A = m/(rho*L)
A = (2.075 gm/cm^3)/((19.32 gm/cm^3)*(3.842*10^-4 cm))
A = 2.075/(19.32*3.842*10^-4) cm^2
A = 279.55 cm^2 = 279.55*10^-4 m^2
A = Area of the leaf = 0.028 m^2
Part B.
Now when gold is drawn into cylindrical fiber of radius 2.900 m, then
m = rho*A*L
A = cross-sectional area of cylinder = pi*r^2
L = m/(rho*A) = m/(rho*pi*r^2)
So,
L = 2.075 gm/((19.32 g/cm^3)*pi*(2.900*10^-4)^2 cm^2)
L = 2.075/(19.32*pi*(2.900*10^-4)^2) cm
L = 406504.26 cm = 406504.26*10^-2 m
L = 4065.0 m = length of fiber
Gold, which has a density of 19.32 g/cm3, is the most ductile metal and can be...
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