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

The energy band gap of a semiconductor determines the range of light frequencies it can emit. For visible light emission, which falls in the approximate wavelength range of \( 400 \, \text{nm} \) to \( 700 \, \text{nm} \), the energy must coincide with the visible spectrum's energy range.

The relation between the energy of light \((E)\) in electron volts and the wavelength \((\lambda)\) in nanometers is given by:

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

Where:

• \( h \) is Planck's constant, \(\approx 4.135667 \times 10^{-15} \, \text{eV} \cdot \text{s} \)
• \( c \) is the speed of light, \(\approx 3 \times 10^8 \, \text{m/s} \)

Considering the visible light spectrum, compute the energy for the minimum wavelength (400 nm) to determine the minimum band gap for visible light emission:

\( E = \frac{(4.135667 \times 10^{-15} \, \text{eV} \cdot \text{s})(3 \times 10^8 \, \text{m/s})}{400 \times 10^{-9} \, \text{m}} \approx 3.1 \, \text{eV} \)

Considering the maximum wavelength (700 nm):

\( E = \frac{(4.135667 \times 10^{-15} \, \text{eV} \cdot \text{s})(3 \times 10^8 \, \text{m/s})}{700 \times 10^{-9} \, \text{m}} \approx 1.77 \, \text{eV} \)

Thus, a semiconductor must have at least a band gap of approximately \( 1.77 \, \text{eV} \) to emit visible light. Therefore, among the options given, the semiconductor must have at least a band gap of \( 1.8 \, \text{eV} \) to fabricate visible LEDs.

Answer 2

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 \( 1.8 \, {eV} \).