Arrange the below mentioned III-V materials to produce LEDs in order of their increasing emission wavelengths : (a) In AS (b) Si C (c) Ga P (d) Ga AS Codes :
(b) (c) (d) (a)
To arrange the given III-V materials in order of their increasing emission wavelengths for producing LEDs, we need to consider the bandgap energies of these materials. The emission wavelength is inversely proportional to the bandgap energy. Mathematically, this relationship is given by:
\(\lambda = \frac{hc}{E_g}\)
where \(\lambda\) is the emission wavelength, \(h\) is Planck's constant, \(c\) is the speed of light, and \(E_g\) is the bandgap energy.
Materials with larger bandgap energies emit light at shorter wavelengths (higher energy), and materials with smaller bandgap energies emit light at longer wavelengths (lower energy).
Let's examine the bandgap energies for the materials in question:
- InAs: Bandgap energy = 0.354 eV
- SiC: Bandgap energy = 2.36 eV
- GaP: Bandgap energy = 2.24 eV
- GaAs: Bandgap energy = 1.42 eV
Now, we arrange them in order of increasing emission wavelengths, which is equivalent to decreasing bandgap energy:
- SiC (emits shortest wavelength due to highest bandgap energy)
- GaP
- GaAs
- InAs (emits longest wavelength due to lowest bandgap energy)
Therefore, the correct arrangement in order of increasing emission wavelengths is:
Correct Answer: (b) (c) (d) (a)
This order implies that SiC has the shortest wavelength emission, followed by GaP, then GaAs, and finally InAs with the longest wavelength emission.
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