1. Using a direct band gap semiconductor, a green LED was fabricated. if the wavelength of the emitted light is 520 nm. Find the energy gap of the semiconductor?
Wavelength \(\lambda\) = 520 nm = 520× 10-9m
The bandgap is given by
\[ E_g = \frac{hc}{\lambda} = \frac{6.625 \times 10^{-34} \times 3 \times 10^8}{520 \times 10^{-9} \times 1.6 \times 10^{-19}} \text{eV} = 2.39 \text{eV} \]2. Find the wavelength of an LED whose energy gap is 3 eV.
Energy gap: Eg = 3 eV = 3×1.6×10-19
Wavelength is given by
\[ \lambda = \frac{hc}{E_g} = \frac{6.625 \times 10^{-34} \times 3 \times 10^8}{3 \times 1.6 \times 10^{-19}} = 4.14 \times 10^{-7} \text{m} \]For a comprehensive understanding, refer to a article on LED.