Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. At what radius (in nm) of sphere does the volume of a shell equal that of the interior sphere? Assume the shell thickness to be t = 1 nm.

Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. On log-log axes, plot the ratio of the shell volume to the interior sphere volume for an interior sphere radius of 10 microns to 1 nm. Assume the shell thickness to be t = 1 nm.

Question

Physics

Brice

18 July, 03:12

Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. At what radius (in nm) of sphere does the volume of a shell equal that of the interior sphere? Assume the shell thickness to be t = 1 nm.

Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. On log-log axes, plot the ratio of the shell volume to the interior sphere volume for an interior sphere radius of 10 microns to 1 nm. Assume the shell thickness to be t = 1 nm.

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Answers (1)

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Garner · Answer 1

3.85 nm

Explanation:

The volume of a sphere is:

V = 4/3 * π * r^3

The volume of a shell is the volume of the big sphere minus the volume of the small sphere

Vs = 4/3 * π * r2^3 - 4/3 * π * r1^3

Vs = 4/3 * π * (r2^3 - r1^3)

If the difference between the radii is 1 nm

r2 = r1 + 1 nm

Vs = 4/3 * π * ((r1 + 1) ^3 - r1^3)

Vs = 4/3 * π * ((r1 + 1) ^3 - r1^3)

Vs = 4/3 * π * (r1^3 + 3*r1^2 + 3*r1 + 1 - r1^3)

Vs = 4/3 * π * (3*r1^2 + 3*r1 + 1)

The volme of the shell is equall to the volume of the inner shell:

4/3 * π * (r1^3) = 4/3 * π * (3*r1^2 + 3*r1 + 1)

r1^3 = 3*r1^2 + 3*r1 + 1

0 = - r1^3 + 3*r1^2 + 3*r1 + 1

Solving this equation electronically:

r1 = 3.85 nm