Question

The shell of a hollow spherical nanoparticle has a uniform thickness of 3 nanometers (nm). The outer radius of the nanoparticle is 5 nm. The ratio of the volume of the shell to the volume of the hollow core is (Round off to one decimal place).

Hint:

Find the inner (core) radius first (outer radius minus shell thickness), then compare cubes of the two radii.

Answer 1

Correct Answer: 14.6

Step 1: Set up the two radii.
The nanoparticle is a sphere with a hollow core inside it, surrounded by a shell of uniform thickness.
The outer radius is given as \(R = 5\) nm and the shell thickness is \(t = 3\) nm.
The inner radius, which is the radius of the hollow core, is the outer radius minus the shell thickness:
\[ r = R - t = 5 - 3 = 2 \text{ nm} \]

Step 2: Write the volume of a sphere.
The volume of a sphere of radius \(a\) is
\[ V = \frac{4}{3}\pi a^3 \]
The full outer sphere (radius \(R = 5\) nm) has volume \(\frac{4}{3}\pi(5)^3\), and the hollow core (radius \(r = 2\) nm) has volume \(\frac{4}{3}\pi(2)^3\).

Step 3: Find the volume of the shell.
The shell is what is left of the outer sphere once the hollow core is taken out, so
\[ V_{shell} = \frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (R^3 - r^3) \]
\[ R^3 = 125, \quad r^3 = 8, \quad R^3 - r^3 = 117 \]

Step 4: Take the ratio of shell volume to core volume.
Since both volumes share the same \(\frac{4}{3}\pi\) factor, it cancels out in the ratio:
\[ \frac{V_{shell}}{V_{core}} = \frac{R^3 - r^3}{r^3} = \frac{117}{8} = 14.625 \]
Rounded to one decimal place, this is 14.6.

Final Answer:
The ratio of the shell volume to the hollow core volume is 14.6. \[ \boxed{\dfrac{V_{shell}}{V_{core}} \approx 14.6} \]