Question

In the manufacturing of computer chips, cylinders of silicon are cut into thin wafers that are \(3.00\) inches in diameter and have a mass of \(1.50 \mathrm{~g}\) of silicon. How thick \((\mathrm{mm})\) is each wafer if silicon has a density of \(2.33 \mathrm{~g} / \mathrm{cm}^{3} ?\) (The volume of a cylinder is \(\left.V=\pi r^{2} h .\right)\)

Short answer

Thickness is approximately \(0.141 \text{ mm} \).

Step-by-step solution

Find the radius

The diameter of the wafer is given as 3.00 inches. To find the radius, divide the diameter by 2. Formula: \[ r = \frac{3.00}{2} \text{ inches} \] Calculation: \[ r = 1.50 \text{ inches} \] Convert the radius to centimeters since the density unit is in \( \text{cm}^3 \): \[ 1 \text{ inch} = 2.54 \text{ cm} \] \[ r = 1.50 \text{ inches} \times 2.54 \text{ cm/inch} = 3.81 \text{ cm} \]

Find the volume of the wafer

The mass (m) of the wafer is given as 1.50 g. Using the density (d) of silicon, we can find the volume (V) using the density formula: \[ d = \frac{m}{V} \] Rearrange the formula to solve for volume: \[ V = \frac{m}{d} \] Given: \[ m = 1.50 \text{ g} \] \[ d = 2.33 \text{ g/cm}^3 \] Calculate the volume: \[ V = \frac{1.50 \text{ g}}{2.33 \text{ g/cm}^3} = 0.6444 \text{ cm}^3 \]

Set up the volume equation for the cylinder

The volume of a cylinder is given by: \[ V = \pi r^{2} h \] We know the volume (V) and radius (r). Substitute these values into the equation to solve for height (h) which represents the thickness of the wafer: \[ 0.6444 \text{ cm}^3 = \pi (3.81 \text{ cm})^2 h \]

Solve for the thickness

Rearrange the equation to solve for thickness (h): \[ h = \frac{V}{\pi r^{2}} = \frac{0.6444 \text{ cm}^3}{\pi (3.81 \text{ cm})^2} \] Calculate the value: \[ h = \frac{0.6444 \text{ cm}^3}{\pi (3.81 \text{ cm})^2} = \frac{0.6444}{45.6309} \approx 0.0141 \text{ cm} \] Convert the height to millimeters: \[ 1 \text{ cm} = 10 \text{ mm} \] \[ h \approx 0.0141 \text{ cm} \times 10 = 0.141 \text{ mm} \]

Key concepts

Volume of a Cylinder

Understanding the volume of a cylinder is crucial in many scientific and engineering tasks. The formula for the volume of a cylinder is given as: \[ V = \pi r^{2} h \] where:

• \( V \) is the volume
• \( \pi \) (Pi) is a constant, approximately 3.14159
• \( r \) is the radius of the cylinder's base
• \( h \) is the height or thickness of the cylinder

For our silicon wafer problem, we use this formula to relate its volume to its radius and thickness. By rearranging the formula to solve for the thickness (\( h \)), we can find the required dimensions of the silicon wafer given its volume and radius. Applying this math in practical scenarios helps develop a deeper understanding of geometric calculations.

Density Calculations

Density is a fundamental property in chemistry and materials science. It is defined as mass per unit volume and is calculated using the formula: \[ d = \frac{m}{V} \] where:

• \( d \) is density
• \( m \) is mass
• \( V \) is volume

In our example, we used the silicon density (\( d = 2.33 \text{g/cm}^3 \)), to find the volume of a wafer given its mass (\( m = 1.50 \text{g} \)). By rearranging the density formula to \( V = \frac{m}{d} \), we computed the volume of the silicon wafer in cubic centimeters. This step is essential for establishing the relationship between the mass, density, and volume of substances.

Unit Conversions

Many scientific problems require unit conversions. Understanding these conversions ensures the correct application of formulas. In our example, we needed to convert inches to centimeters for volume calculation coherence, because the density was given in g/cm³. The conversion factor is:

• 1 inch = 2.54 cm

Converting 1.50 inches to centimeters: \[ 1.50 \text{ inches} \times 2.54 \text{ cm/inch} = 3.81 \text{ cm} \] Additionally, we converted the thickness from centimeters to millimeters by using the factor:

• 1 cm = 10 mm

This step ensured our results were in the appropriate units for practical application and interpretation. Accurate unit conversion is vital for consistency and correctness in scientific calculations.

Problem Solving in Chemistry

Effective problem solving in chemistry often involves multiple steps and the integration of various concepts. To determine the thickness of a silicon wafer, we:

• Converted the diameter to radius
• Made unit conversions from inches to centimeters
• Used the density formula to find volume
• Substituted known values into the cylinder volume formula to isolate and solve for thickness
• Made additional unit conversion to millimeters

Each of these steps required applying fundamental principles accurately. Developing a logical sequence, making precise calculations, and verifying units ensures comprehensive and correct solutions to complex problems.