Red light-emitting diodes are made from GaAs and GaP solid solutions, GaP \(_{x} A s_{1-x}(\) see Exercise 12.79\() .\) The original red LEDs emitted light with a wavelength of 660 nm. If we assume that the band gap varies linearly with composition between \(x=0\) and \(x=1,\) estimate the composition (the value of \(x\) ) that is used in these LEDs.

Using Planck's equation, we can find the photon energy related to the wavelength of the emitted light: \[E = h \nu\] where: \(E\) is the energy of the photon, \(h\) is the Planck constant, which is approximately \(6.63 \times 10^{-34}\, J s\), \(\nu\) is the frequency of the light. Since the frequency \(\nu\) can also be expressed as \(\nu = \frac{c}{\lambda}\), where \(c\) is the speed of light (approximately \(3 \times 10^8\, m/s\)), and \(\lambda\) is the wavelength of the emitted light, 660 nm in this case. Now we can rewrite the equation as: \[E = h \frac{c}{\lambda}\] Plug in the given values: \[E = (6.63 \times 10^{-34} \, J s) \times \frac{3 \times 10^8 \, m/s}{660 \times 10^{-9} \, m}\]

As we have the equation \(E = h \frac{c}{\lambda}\) and the values of \(h\), \(c\), and \(\lambda\), we can now calculate the photon energy: \[E = (6.63 \times 10^{-34} \, J s) \times \frac{3 \times 10^8 \, m/s}{660 \times 10^{-9} \, m} = 3.02 \times 10^{-19} \, J\] So, the energy of the emitted photons is approximately \(3.02 \times 10^{-19}\, J\).

We are given that the band gap energy for GaP\(_x\)As\(_{1-x}\) varies linearly with the composition. This can be expressed as: \[E = E_{0} + x(E_{1} - E_{0})\] where: \(E\) is the band gap energy of GaP\(_x\)As$_{1-x}\), \(E_{0}\) is the band gap energy of GaAs (at \(x=0\)), \(E_{1}\) is the band gap energy of GaP (at \(x=1\)), \(x\) is the composition we want to find. From the literature, the values of \(E_{0}\) and \(E_{1}\) are approximately \(1.43\, eV\) and \(2.26\, eV\), respectively. We must convert the calculated photon energy from Step 2 to electron-volts (eV). To do this, we divide the energy calculated in Step 2 by the elementary charge (\(1.6 \times 10^{-19}\, C\)): \[E = 3.02 \times 10^{-19} \, J \times \frac{1 \, eV}{1.6 \times 10^{-19} \, C} = 1.89\,eV\] Now we can insert the known values into the linear equation: \[1.89\,eV = 1.43\,eV + x(2.26\,eV - 1.43\,eV)\]

Now we have only one unknown value, \(x\), we can rearrange the equation and solve for \(x\). \[x = \frac{1.89\,eV - 1.43\,eV}{2.26\,eV - 1.43\,eV} = \frac{0.46\,eV}{0.83\,eV}\] \[x \approx 0.55\] So, the composition used in the red LEDs, GaP\(_x\)As\(_{1-x}\), is approximately \(x \approx 0.55\).