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Chapter 21: Problem 14

Zinc selenide has a band gap of \(2.58 \mathrm{eV}\). Over what range of wavelengths of visible light is it transparent?

Short Answer

Expert verified
Answer: The range of wavelengths for which zinc selenide is transparent is between 481 nm and 750 nm.

Step by step solution

01

Convert the given band gap energy into units of Joules

First, we need to convert the given band gap energy of \(2.58\mathrm{eV}\) into Joules using the conversion factor \(1\mathrm{eV} = 1.6 \times 10^{-19}\mathrm{J}\). \(E = 2.58 \mathrm{eV} \times \frac{1.6 \times 10^{-19}\mathrm{J}}{1\mathrm{eV}} = 4.128 \times 10^{-19} \mathrm{J}\)
02

Find the relationship between energy and wavelength using Planck's equation

Planck's equation relates the energy of a photon to its wavelength: \(E = \frac{hc}{\lambda}\), where \(E\) is the energy of a photon, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength. Planck's constant \(h = 6.626 \times 10^{-34} \mathrm{J s}\) and the speed of light \(c = 3.0 \times 10^8 \mathrm{m/s}\).
03

Calculate the maximum wavelength for transparency

Using Planck's equation, we can find the maximum wavelength at which zinc selenide is transparent: \(\lambda_\mathrm{max} = \frac{hc}{E} = \frac{(6.626 \times 10^{-34} \mathrm{J s})(3.0 \times 10^8 \mathrm{m/s})}{4.128 \times 10^{-19} \mathrm{J}} = 4.81 \times 10^{-7} \mathrm{m}\) Convert the wavelength to nanometers: \(\lambda_\mathrm{max} = 4.81 \times 10^{-7} \mathrm{m} \times \frac{10^9 \mathrm{nm}}{1 \mathrm{m}} = 481 \mathrm{nm}\)
04

Determine the range of wavelengths for transparency

Zinc selenide will be transparent for wavelengths in the visible spectrum longer than the maximum wavelength we calculated in the previous step. The visible light spectrum ranges from \(380\mathrm{nm}\) to \(750\mathrm{nm}\), so the range of wavelengths for transparency will be: \(481\mathrm{nm} \le \lambda \le 750\mathrm{nm}\) This means that zinc selenide is transparent to wavelengths of visible light between \(481\mathrm{nm}\) and \(750\mathrm{nm}\).

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