Question

Silicon for computer chips is grown in large cylinders called "boules" that are \(300 \mathrm{~mm}\) in diameter and \(2 \mathrm{~m}\) in length, as shown. The density of silicon is \(2.33 \mathrm{~g} / \mathrm{cm}^{3}\). Silicon wafers for making integrated circuits are sliced from a \(2.0 \mathrm{~m}\) boule and are typically \(0.75 \mathrm{~mm}\) thick and \(300 \mathrm{~mm}\) in diameter. (a) How many wafers can be cut from a single boule? (b) What is the mass of a silicon wafer? (The volume of a cylinder is given by \(\pi r^{2} h\), where \(r\) is the radius and \(h\) is its height.)

Short answer

(a) 2666 wafers can be cut from a single boule. (b) The mass of an individual silicon wafer is approximately 123.57 grams.

Step-by-step solution

Calculate the total number of wafers that can be cut from a single boule

To calculate the total number of wafers that can be cut from a single boule, we first need to find the total thickness of all wafers. We are given that each wafer is 0.75 mm thick, and the boule is 2 m long. We need to convert the length of the boule into millimeters and then divide it by the thickness of an individual wafer to see how many wafers can be cut from the boule. The length of the boule in millimeters is: \(2 \mathrm{~m} \times \frac{1000 \mathrm{~mm}}{1 \mathrm{~m}} = 2000 \mathrm{~mm}\) Now, we can divide the length of the boule by the thickness of a single wafer to find the number of wafers: \(\frac{2000 \mathrm{~mm}}{0.75 \mathrm{~mm}} \approx 2666.67 \) However, we can only cut whole wafers from the boule. So the total number of wafers that can be cut from a single boule is 2666 wafers.

Calculate the mass of an individual silicon wafer

To find the mass of an individual silicon wafer, we will first need to calculate the volume of the wafer using the given formula of a cylinder: \(V = \pi r^{2}h\), where \(r\) is the radius and \(h\) is the height of the cylinder. In this case, the height is equal to the thickness of the wafer. We are given that each wafer is 300 mm in diameter and 0.75 mm thick. So, the radius is half of the diameter: \(r = \frac{300 \mathrm{~mm}}{2} = 150 \mathrm{~mm}\) Now we can calculate the volume of a wafer: \(V = \pi (150 \mathrm{~mm})^{2}(0.75 \mathrm{~mm}) \approx 53,034.1125 \mathrm{~mm}^3\) Now, we will convert the volume to cubic centimeters: \(53,034.1125 \mathrm{~mm}^3 \times \frac{1 \mathrm{~cm}^3}{1000 \mathrm{~mm}^3} = 53.0341125 \mathrm{~cm}^3\) Next, we will find the mass of the wafer using its volume and the density of silicon. The given density is \(2.33 \mathrm{~g} / \mathrm{cm}^{3}\). Mass = Volume × Density = \(53.0341125 \mathrm{~cm}^{3} \times 2.33 \mathrm{~g} / \mathrm{cm}^{3}\) Mass \(= 123.569582125 \mathrm{~g}\) The mass of an individual silicon wafer is approximately 123.57 grams.

Key concepts

Density Calculation

Density is a fundamental material property that measures the mass per unit volume of a substance. It plays a pivotal role in various scientific and engineering problems, including the determination of the mass of an object if its volume is known. The formula for calculating density (\rho) is simple: \[ \rho = \frac{m}{V} \]where \(m\) represents mass and \(V\) is volume. When calculating the mass of an object like a silicon wafer, it's essential to know the density of the material (in this case, silicon) and the volume of the object.

Specific to our example involving silicon wafers, we use the given density of silicon, 2.33 g/cm³, to calculate the mass after determining the wafer's volume. As materials have unique densities, knowing the density of silicon is crucial to solving the exercise. Understanding density is not just about plugging numbers into a formula; it's about comprehending how much material fits within a certain space, which directly impacts the object's mass.

Cylinder Volume

The volume of a cylinder can be found using the formula \[ V = \pi r^2 h \]where \(V\) is the volume, \(\pi\) is a mathematical constant approximately equal to 3.14159, \(r\) is the radius of the base circle, and \(h\) is the height of the cylinder. It's basically the area of the base times the height, encapsulating the idea of 'stacking' the base shape up to a certain height to fill a 3D space.

For our silicon wafers, we consider each wafer as a thin cylinder, with its thickness (0.75 mm in the original problem) being the 'height' (\(h\) in the formula). Therefore, understanding the volume calculation for cylinders is fundamental to advancing towards determining the number of wafers that can be cut from the boule and the mass of each wafer.

Material Properties of Silicon

Silicon is a versatile material widely used in electronics, particularly in the fabrication of integrated circuits and solar cells. Some critical material properties of silicon include its semi-conductive nature, high melting point, and a crystalline structure that's suitable for manufacturing precision devices. Beyond its electrical properties, for physical calculations, we look at properties like density, mentioned earlier, measuring 2.33 g/cm³ for silicon. This specific density value is essential for our calculations and for understanding the characteristics and applications of silicon.

By digging into material properties like density, we can foresee its handling and performance in various applications, which in this scenario aids in determining the mass of a silicon wafer.

Unit Conversion

Unit conversion is a fundamental skill that allows us to translate different units of measurement into a common standard, enabling us to solve problems coherently. As seen in the exercise, the boule's length is provided in meters, but the wafer's thickness is in millimeters. To align these units and compute the number of wafers, we must convert meters to millimeters, knowing that \(1 m = 1000 mm\).

Additionally, for calculating the wafer's mass, the volume in cubic millimeters is converted to cubic centimeters because the density of silicon is given in grams per cubic centimeter (\(g/cm^3\)). Understanding and correctly applying unit conversions ensures accuracy in calculations and the integration of different dimensions of measurement within the same problem.