Question

The semiconductor used for fabrication of visible LEDs must at least have a band gap of:

• A. \( 0.6 \, {eV} \)
• B. \( 1.2 \, {eV} \)
• C. \( 1.8 \, {eV} \)
• D. \( 0.9 \, {eV} \)

Hint:

For visible light emission from semiconductors, the band gap energy should be in the range of \( 1.8 \, {eV} \) to \( 3.0 \, {eV} \), which allows the semiconductor to produce light in the visible spectrum.

Answer 1

The Correct Option is C

To fabricate visible LEDs (Light Emitting Diodes), the semiconductor material used must have a band gap that corresponds to the energy required for the emission of visible light.
The visible spectrum of light typically falls in the wavelength range of approximately \( 400 \, {nm} \) (violet) to \( 700 \, {nm} \) (red). The energy of a photon is related to its wavelength \( \lambda \) by the equation: \[ E = \frac{hc}{\lambda} \] where \( E \) is the energy, \( h \) is Planck’s constant (\( 6.626 \times 10^{-34} \, {J·s} \)), and \( c \) is the speed of light (\( 3 \times 10^8 \, {m/s} \)). For a wavelength of \( 600 \, {nm} \), which is near the middle of the visible spectrum, the energy \( E \) is calculated as: \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{600 \times 10^{-9}} \approx 3.3 \times 10^{-19} \, {J}. \] Converting this energy into electron volts: \[ E = \frac{3.3 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 2.06 \, {eV}. \] Thus, for the emission of visible light, the band gap of the semiconductor must be at least \( 1.8 \, {eV} \), which corresponds to the energy required to produce photons in the visible range. Therefore, the correct answer is \( 1.8 \, {eV} \).