An ideal quantum dot for use in TVs does not contain any cadmium due to concerns about disposal. One potential material for this purpose is InP, which adopts the zinc blende (ZnS) structure (face-centered cubic). The unit cell edge length is \(586.9 \mathrm{pm} .(\mathbf{a})\) If the quantum dot is shaped like a cube, how many of each type of atom are there in a cubic crystal with an edge length of \(3.00 \mathrm{nm} ? 5.00 \mathrm{nm} ?(\mathbf{b})\) If one of the nanoparticles in part (a) emits blue light and the other emits orange light, which color is emitted by the crystal with the \(3.00-\mathrm{nm}\) edge length? With the \(5.00-\mathrm{nm}\) edge length?

We begin by calculating the number of unit cells along each edge (n) of the crystal: \(n=\frac{\text{Edge length of crystal}}{\text{Edge length of unit cell}}\) For the 3.00 nm and 5.00 nm edge length crystals: \(n_{3.00\mathrm{nm}}=\frac{3.00\mathrm{nm}}{586.9\mathrm{pm}}=\frac{3000\mathrm{pm}}{586.9\mathrm{pm}}\approx 5.11\) \(n_{5.00\mathrm{nm}}=\frac{5.00\mathrm{nm}}{586.9\mathrm{pm}}=\frac{5000\mathrm{pm}}{586.9\mathrm{pm}}\approx 8.52\) Since we can't have a partial unit cell, we round up the number of unit cells to the nearest integer for both cases: \(n_{3.00\mathrm{nm}}=6\) \(n_{5.00\mathrm{nm}}=9\) Now we have 6 unit cells along each edge for the 3.00 nm crystal and 9 unit cells along each edge for the 5.00 nm crystal.

As we know the zinc blende unit cell has 4 atoms (2 In and 2 P) and we have the number of unit cells along each edge of the cubic crystal, we can calculate the total number of unit cells and then the number of atoms of each type in the crystal. Total number of unit cells in the crystal = \(n^3\) For the 3.00 nm and 5.00 nm edge length crystals: Total number of unit cells in the 3.00 nm crystal = \(6^3=216\) Total number of unit cells in the 5.00 nm crystal = \(9^3=729\) Now we know that each unit cell contains 2 In atoms and 2 P atoms. We can multiply the total number of unit cells by the number of In and P atoms to find the total number of atoms of each type in the crystal: For the 3.00 nm crystal: In atoms: \(216 \times 2 = 432\) P atoms: \(216 \times 2 = 432\) For the 5.00 nm crystal: In atoms: \(729 \times 2 = 1458\) P atoms: \(729 \times 2 = 1458\)

Considering that one of the nanoparticles in part (a) emits blue light and the other emits orange light, the crystal's emitted color will be determined by the size of the nanoparticles inside the crystal. It is known that as the size of the quantum dot increases, its emitted wavelength of light decreases (shifts to a longer wavelength or lower frequency) from blue to orange. Therefore, we can conclude that: In the case of the 3.00 nm crystal, with fewer and smaller nanoparticles, it emits blue light. In the case of the 5.00 nm crystal, with more and larger nanoparticles, it emits orange light. To summarize: A crystal of 3.00 nm edge length contains 432 In atoms and 432 P atoms and emits blue light. A crystal of 5.00 nm edge length contains 1458 In atoms and 1458 P atoms and emits orange light.