Question

A nanowire is a (nearly) one-dimensional structure with a diameter on the order of a few nanometers. Suppose a \(100.0-\mathrm{nm}\) long nanowire made of pure silicon (density of \(\mathrm{Si}=2.33 \mathrm{~g} / \mathrm{cm}_{3}\) ) has a diameter of \(5.0 \mathrm{nm}\). This nanowire is attached at the top and hanging down vertically due to the force of gravity. a) What is the tension at the top? b) What is the tension in the middle? (Hint: Treat the nanowire as a cylinder of diameter \(5.0 \mathrm{nm}\) leneth 100 made of silican

Short answer

Please show your step-by-step solution and calculations.

Step-by-step solution

Find the volume of nanowire

We are given that the nanowire is cylindrical in shape, with a diameter of \(5.0 \,\text{nm}\) and a length of \(100.0 \,\text{nm}\). To find its volume \(V\), we can use the formula for the volume of a cylinder, \(V = \pi r^2 h\), where \(r\) is the radius and \(h\) is the height. In this case, \(h = 100.0 \,\text{nm}\). The diameter is \(5.0 \,\text{nm}\), so the radius \(r\) is half of that, which is \(r = 2.5 \,\text{nm}\): $$V=\pi(2.5\,\text{nm})^2 (100.0\,\text{nm})$$ Now, calculate the volume.

Convert the volume to cubic centimeters

Before we can use the density of silicon to find the mass of the nanowire, we need to convert the volume from cubic nanometers to cubic centimeters so the units are compatible. There are \(1\,\text{cm}=10^7\,\text{nm}\), so: $$V=\pi(2.5\,\text{nm})^2 (100.0\,\text{nm})\cdot(\frac{1\,\text{cm}}{10^7\,\text{nm}})^3$$ Calculate the value of \(V\) in cubic centimeters.

Calculate the mass of the nanowire

Now that we have the volume of the nanowire in cubic centimeters, we can use the density (\(\rho=2.33\,\text{g/cm}^3\)) of silicon to find the mass \(m\) of the nanowire using the formula \(m = \rho V\): $$m=2.33\,\text{g/cm}^3\cdot V$$ Calculate the mass.

Calculate the weight of the nanowire

Now we need to find the weight of the nanowire, which is the force acting on the nanowire due to gravity. The weight \(W\) is given by the formula \(W = mg\), where \(m\) is the mass we calculated in step 3, and \(g\) is the acceleration due to gravity (approximately \(9.81\,\text{m/s}^2\) or \(9.81\times10^4\,\text{cm/s}^2\)): $$W = mg$$ Calculate the weight.

Find the tension at the top

The tension at the top of the nanowire is simply the weight of the entire nanowire, as it must support the weight of the whole nanowire. Since we have already calculated the weight \(W\), the tension at the top of the nanowire is: $$T_\text{top} = W$$

Find the tension in the middle

To find the tension in the middle of the nanowire, we need to consider the weight of half the nanowire because the tension at this point has to support only the bottom half's weight. The mass of the bottom half of the nanowire can be found by dividing the total mass by 2, and calculating the corresponding weight (using \(W_\text{half} = \frac{m}{2}g\)): $$T_\text{middle} = W_\text{half}$$ Calculate the tension in the middle. Now, we have the tension at both the top and the middle of the nanowire.

Key concepts

Cylinder Volume Calculation

Understanding the volume of a nanowire is essential in nanowire physics. As given in the problem, a nanowire has a diameter of 5.0 nm and a length of 100.0 nm. You can think of it as a tiny cylinder. To calculate the volume of the cylinder, the formula used is: \[ V = \pi r^2 h \] where \( r \) is the radius, and \( h \) is the height (or length in this case) of the cylinder. For the diameter provided, the radius would be half, so \( r = 2.5 \text{ nm} \). Substituting these values into the formula gives you: \[ V = \pi (2.5 \,\text{nm})^2 (100.0\, \text{nm}) \] This formula gives the volume in cubic nanometers. It involves basic geometry calculations, however, nanoparticles like nanowires pose unique challenges due to their minuscule size. Therefore, these calculations enhance understanding of the dimensions and properties of materials at the nanoscale.Knowing how to calculate volume, specifically for a small scale object, helps predict other properties, such as mass and tension, when additional variables, like density, are known.

Tension Calculation

Calculating tension in a nanowire involves understanding the concept of forces acting upon it. Because the nanowire is hanging vertically, gravity exerts a force throughout its length. The tension at any point on the nanowire is the force required to support the weight of the wire below that point.1. **Top Tension:** - The tension at the top of the nanowire is equal to the total weight of the nanowire since it supports the entire wire. - If you know the weight (which is mass times gravity), the tension at the top T is calculated as: \[ T_{\text{top}} = W = mg \] 2. **Middle Tension:** - In the middle of the nanowire, the tension is due to the weight of only half of the wire. - Divide the total mass by two to find the mass of the bottom half. This gives: \[ T_{\text{middle}} = \frac{W}{2} \] Both these tension calculations help in evaluating how the nanowire would behave under the force of gravity, which is crucial for applications using nanoscale materials.

Silicon Density

Density is a key factor in determining the mass of a material and is defined as mass per unit volume. In the case of silicon nanowires, we are given that the silicon density is \(2.33\,\text{g/cm}^3\). This constant helps calculate the mass of the nanowire when its volume is known.Using the formula for density:\[ \text{Density} (\rho) = \frac{m}{V} \] where \( m \) is mass and \( V \) is the volume. Rearranging the formula allows you to find the mass:\[ m = \rho \times V \] The density value implies how many grams of silicon exist in a cubic centimeter. Therefore, after converting the nanowire's volume into cubic centimeters through appropriate unit conversion, its mass is easily calculated. Understanding silicon density becomes vital when customizing materials for specific weights and properties, significantly influencing the material's performance in applications such as semiconductors.

Nanometer to Centimeter Conversion

Converting between nanometers and centimeters is pivotal when dealing with nanowire dimensions versus macroscopic measurements. The conversion factor between these units is based on their relation to the meter, with 1 cm equaling \(10^7\) nm.To convert from nanometers to centimeters, use:\[ 1\,\text{cm} = 10^7\,\text{nm} \] For calculations concerning volume, you will cube the conversion factor to account for all three spatial dimensions:\[ \left( \frac{1\,\text{cm}}{10^7\,\text{nm}} \right)^3 \] This cubic conversion is crucial in ensuring that when multiplying volume in nanometers by density in grams per cubic centimeter, you maintain consistent units throughout the computation. Understanding these conversions allows an accurate translation of physical phenomena from the nanoscale world into units compatible with macroscopic scientific applications.