To solve this problem, we need to arrange the given semiconductors in ascending order of their wavelength based on their bandgap energy, \(E_g\). The relationship between the bandgap energy and the wavelength of light is given by:

\(E_g = \frac{hc}{\lambda}\)

where:

• \(E_g\) = Bandgap energy (in electron volts, eV)
• \(h\) = Planck's constant = \(4.135667696 \times 10^{-15}\) eV·s
• \(c\) = Speed of light = \(3 \times 10^8\) m/s
• \(\lambda\) = Wavelength (in meters)

From this equation, we can derive the wavelength as:

\(\lambda = \frac{hc}{E_g}\)

Thus, the wavelength is inversely proportional to the bandgap energy, meaning a larger \(E_g\) corresponds to a shorter wavelength. Consequently, to arrange in ascending order of wavelength, we should arrange in descending order of \(E_g\).

Let's calculate \(E_g\) for the given options:

1. A: GaAs with \(E_g = 1.4 \text{ eV}\)
2. B: \(Al_{0.2}Ga_{0.8}As\) with \(E_g = 1.62 \text{ eV}\)
3. C: \(Al_{0.4}Ga_{0.6}As\) with \(E_g = 1.92 \text{ eV}\)
4. D: \(GaAs_{0.4}P_{0.6}\) with \(E_g = 2.2 \text{ eV}\)
5. E: \(GaAs_{0.2}P_{0.8}\) with \(E_g = 2.5 \text{ eV}\)

Arranging these in descending order of \(E_g\):

1. E: \(E_g = 2.5 \text{ eV}\)
2. D: \(E_g = 2.2 \text{ eV}\)
3. C: \(E_g = 1.92 \text{ eV}\)
4. B: \(E_g = 1.62 \text{ eV}\)
5. A: \(E_g = 1.4 \text{ eV}\)

Thus, the order in ascending wavelength is:

1. E
2. D
3. C
4. B
5. A

The correct answer is therefore: E, D, C, B, A.