Problem 89
Question
Very small crystals composed of 1000 to 100,000 atoms, called quantum dots, are being investigated for use in electronic devices. (a) A quantum dot was made of solid silicon in the shape of a sphere, with a diameter of \(4 \mathrm{nm} .\) Calculate the mass of the quantum dot, using the density of silicon \(\left(2.3 \mathrm{~g} / \mathrm{cm}^{3}\right)\) (b) How many silicon atoms are in the quantum dot? (c) The density of germanium is \(5.325 \mathrm{~g} / \mathrm{cm}^{3}\). If you made a 4-nm quantum dot of germanium, how many Ge atoms would it contain? Assume the dot is spherical.
Step-by-Step Solution
Verified Answer
(a) The mass of the silicon quantum dot is approximately \(7.707 \times 10^{-23} \mathrm{~g}\).
(b) It contains approximately \(1.65 \times 10^{5}\) silicon atoms.
(c) A 4-nm quantum dot of germanium would contain approximately \(1.48 \times 10^5\) germanium atoms.
1Step 1: Calculate the Volume of the Quantum Dot
To find the volume of the spherical quantum dot, we can use the formula for the volume of a sphere, which is \(V = \frac{4}{3}\pi r^{3}\). Given the diameter is 4 nm, the radius, r is 2 nm or 0.000000002 cm. Let's find the volume V:
\(V = \frac{4}{3}\pi (0.000000002)^{3}\)
Calculating the volume:
\(V \approx 3.351 \times 10^{-23} \mathrm{cm}^3\)
2Step 2: Calculate the Mass of the Silicon Quantum Dot
Now that we have the volume, we can use the density of silicon to find the mass of the quantum dot. The density is given as \(2.3 \mathrm{~g} /\mathrm{cm}^3\). We can use the formula:
\(mass = density \times volume\)
Plugging in the values:
\(mass = 2.3 \times 3.351 \times 10^{-23}\)
Calculating the mass:
\(mass \approx 7.707 \times 10^{-23} \mathrm{~g}\)
3Step 3: Find the Number of Silicon Atoms in the Quantum Dot
To find the number of silicon atoms in the quantum dot, we'll first calculate the mass of one silicon atom. The molar mass of silicon is approximately 28.09 g/mol. Since there are \(6.022\times 10^{23}\) atoms in one mole, we can find the mass of one silicon atom:
\(mass_{Si} = \frac{28.09 \mathrm{~g/mol}}{6.022\times 10^{23}\mathrm{atoms/mol}} \approx 4.663 \times 10^{-23}\ \mathrm{g}\)
Now we'll divide the mass of the whole quantum dot by the mass of one silicon atom to find the number of atoms in the dot:
\(N_{Si} = \frac{7.707 \times 10^{-23} \mathrm{~g}}{4.663 \times 10^{-23} \mathrm{~g}}\)
Calculating the number of atoms:
\(N_{Si} \approx 1.65 \times 10^{5}\) silicon atoms
4Step 4: Calculate the Number of Germanium Atoms in the 4-nm Quantum Dot
To find the number of germanium atoms in a 4-nm quantum dot, we can use the same process as for silicon. First, we need to find the mass of the quantum dot using the density of germanium.
\(mass_{Ge} = density_{Ge} \times volume\)
\(mass_{Ge} = 5.325 \times 3.351 \times 10^{-23}\)
Calculating the mass of the germanium quantum dot:
\(mass_{Ge} \approx 1.783 \times 10^{-22} \mathrm{~g}\)
Next, we'll find the mass of one germanium atom. The molar mass of germanium is approximately 72.63 g/mol. So the mass of one germanium atom is:
\(mass_{Ge-atom} = \frac{72.63 \mathrm{~g/mol}}{6.022\times 10^{23}\mathrm{atoms/mol}} = 1.206 \times 10^{-22}\ \mathrm{g}\)
Finally, we'll divide the mass of the whole germanium quantum dot by the mass of one germanium atom to find the number of atoms in the dot:
\(N_{Ge} = \frac{1.783 \times 10^{-22} \mathrm{~g}}{1.206 \times 10^{-22} \mathrm{~g}}\)
Calculating the number of germanium atoms:
\(N_{Ge} \approx 1.48 \times 10^5\) germanium atoms
Key Concepts
Nanoscale MaterialsSilicon Atom Mass CalculationSpherical Volume Calculation
Nanoscale Materials
Nanoscale materials, as the name suggests, refer to materials that exist at the nanometer scale, typically between 1 to 100 nanometers (nm). To visualize, a nanometer is one billionth of a meter, which is about 10,000 times smaller than the diameter of a human hair. At this scale, materials exhibit unique physical and chemical properties that are not present in their bulk counterparts, leading to innovative applications in fields like electronics, medicine, and energy.
Quantum dots, the topic of our exercise, are an example of nanoscale materials. These tiny semiconductor particles have discrete energy levels and can emit light at specific wavelengths, making them useful for displays, solar cells, and bio-imaging. The manipulation of matter at this level allows for tailoring material properties, opening up a new realm of possibilities for technological advancement.
Quantum dots, the topic of our exercise, are an example of nanoscale materials. These tiny semiconductor particles have discrete energy levels and can emit light at specific wavelengths, making them useful for displays, solar cells, and bio-imaging. The manipulation of matter at this level allows for tailoring material properties, opening up a new realm of possibilities for technological advancement.
Silicon Atom Mass Calculation
Calculating the mass of a single silicon atom is a fundamental step in understanding the composition of nanoscale materials like quantum dots. To perform this calculation, we use the molar mass of silicon, which is 28.09 grams per mole. Since a mole contains Avogadro's number of entities, which is approximately \(6.022 \times 10^{23}\) atoms in this case, we can determine the mass of a single atom.
The calculation involves dividing the molar mass by Avogadro's number, giving us a very small value, reflecting the minuscule size of an individual atom. This precise calculation is crucial in nanotechnology where the properties depend directly on the number of atoms involved. It's the fundamental step in correlating the material's macroscopic properties, like mass or volume, to its atomic-scale characteristics.
The calculation involves dividing the molar mass by Avogadro's number, giving us a very small value, reflecting the minuscule size of an individual atom. This precise calculation is crucial in nanotechnology where the properties depend directly on the number of atoms involved. It's the fundamental step in correlating the material's macroscopic properties, like mass or volume, to its atomic-scale characteristics.
Spherical Volume Calculation
The spherical volume calculation is an essential part of determining the size and, subsequently, the mass of a quantum dot. The formula for calculating the volume of a sphere is given by \(V = \frac{4}{3} \pi r^{3}\), where \(V\) is the volume and \(r\) is the radius of the sphere. For a quantum dot with a diameter of 4 nm, the diameter must first be halved to obtain the radius, since the radius is half the diameter.
Understanding the volume calculation is critical when working with properties that depend on the size of the material, such as density. For quantum dots, and other nanoscale materials, even a slight change in volume can significantly impact their optical and electronic properties. This relationship underlines the importance of precision in the synthesis and characterization of nanomaterials for their intended applications.
Understanding the volume calculation is critical when working with properties that depend on the size of the material, such as density. For quantum dots, and other nanoscale materials, even a slight change in volume can significantly impact their optical and electronic properties. This relationship underlines the importance of precision in the synthesis and characterization of nanomaterials for their intended applications.
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