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Chapter 23: Problem 51

In \(\mathrm{ZnO}\), the band gap between the valence and conduction bands is \(290 \mathrm{kJmol}^{-1}\), and in \(\mathrm{CdS}\) it is \(250 \mathrm{kJmol}^{-1} .\) Show that CdS absorbs some visible light but ZnO does not. Explain the observed colors: \(\mathrm{ZnO}\) is white and \(\mathrm{CdS}\) is yellow.

Short Answer

Expert verified
CdS absorbs blue light (479 nm) and appears yellow as it reflects other colors, whereas ZnO absorbs violet light (413 nm) but appears white as it reflects all other visible light.

Step by step solution

01

Calculate the Energy of the Incident Light

The energy (E) of a photon can be calculated using the formula\[E = hc/{\lambda}\] where h is Planck’s constant, c is the speed of light, and \({\lambda}\) is the wavelength of light. For visible light, the wavelength range lies between 400 nm to 700 nm.
02

Convert the Band Gap to Energy in electron volts (eV)

1 eV is equal to 96.485 kJ/mol. Therefore, the band gaps in eV of ZnO and CdS are\[ZnO = 290/96.485 = 3.01 eV;\] \[CdS = 250/96.485 = 2.59 eV.\]
03

Calculate the Wavelength that Corresponds to the Band Gap

Use the formula for energy from step 1 to calculate the wavelength that corresponds to the band gap for ZnO and CdS. The wavelength of light absorbed by a material is given by\[{\lambda} = hc/E\]Here, h is Planck’s constant, c is the speed of light, and E is energy. Substituting the values for ZnO, we find that\[\lambda_{ZnO} = (6.63 x 10^-34 Js x 3.0 x 10^8 ms^-1)/(3.01 x 1.6 x 10^-19 J) = 413 nm.\]Similarly, for CdS,\[\lambda_{CdS} = (6.63 x 10^-34 Js x 3.0 x 10^8 ms^-1)/(2.59 x 1.6 x 10^-19 J) = 479 nm.\]
04

Relate the Absorbed Wavelength to the Observed Colors

Substances appear colored as they absorb certain wavelengths of visible light and transmit or reflect the rest. For ZnO, we calculated that it absorbs light with a wavelength of 413 nm which falls in the violet range. As a result, it would reflect all the other colors, including white light, making it appear white. For CdS, we calculated that it absorbs light with a wavelength of 479 nm which falls in the blue range. Consequently, its reflected light is subtractive of blue, tending towards yellow, hence its yellow appearance.

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