Calculating Nanoparticle Volume Ratio

The problem asks for the ratio of the volume of the shell to the volume of the hollow core of a spherical nanoparticle.

Given Parameters

• Shell thickness, $t = 3$ nm
• Outer radius, $R_{outer} = 5$ nm

Calculating Radii

The inner radius ($R_{inner}$) is found by subtracting the shell thickness from the outer radius:

$R_{inner} = R_{outer} - t = 5 \text{ nm} - 3 \text{ nm} = 2 \text{ nm}$

Calculating Volumes

The volume of a sphere is given by the formula $V = \frac{4}{3}\pi R^3$.

1. Volume of the Hollow Core ($V_{core}$):

Using the inner radius $R_{inner} = 2$ nm:

$V_{core} = \frac{4}{3}\pi R_{inner}^3 = \frac{4}{3}\pi (2 \text{ nm})^3 = \frac{4}{3}\pi (8 \text{ nm}^3) = \frac{32}{3}\pi \text{ nm}^3$

2. Volume of the Shell ($V_{shell}$):

The shell volume is the difference between the volume of the outer sphere and the volume of the inner sphere (hollow core).

$V_{shell} = V_{outer} - V_{inner} = \frac{4}{3}\pi R_{outer}^3 - \frac{4}{3}\pi R_{inner}^3$

$V_{shell} = \frac{4}{3}\pi (5 \text{ nm})^3 - \frac{4}{3}\pi (2 \text{ nm})^3$

$V_{shell} = \frac{4}{3}\pi (125 \text{ nm}^3) - \frac{4}{3}\pi (8 \text{ nm}^3)$

$V_{shell} = \frac{4}{3}\pi (125 - 8) \text{ nm}^3 = \frac{4}{3}\pi (117) \text{ nm}^3$

Calculating the Ratio

The ratio of the volume of the shell to the volume of the hollow core is:

Ratio $= \frac{V_{shell}}{V_{core}} = \frac{\frac{4}{3}\pi (117) \text{ nm}^3}{\frac{32}{3}\pi \text{ nm}^3}$

The terms $\frac{4}{3}\pi$ and $\text{nm}^3$ cancel out:

Ratio $= \frac{117}{\frac{32}{3}} = \frac{117 \times 3}{32}$ This step is incorrect in the calculation above. Let's re-evaluate the cancellation.

Ratio $= \frac{\frac{4}{3}\pi (117)}{\frac{4}{3}\pi (8)} = \frac{117}{8}$

Ratio $= 14.625$

Rounding Off

Rounding the ratio to one decimal place:

Ratio $\approx 14.6$