Question

A light emitting diode (LED) is fabricated using GaAs semiconducting material whose band gap is 1.42 eV. The wavelength of light emitted from the LED is:

• A. 650 nm
• B. 1243 nm
• C. 875 nm
• D. 1400 nm

Answer 1

The Correct Option is C

To ascertain the wavelength of light emitted by the LED, the relationship between photon energy and wavelength must be established. The energy of a photon is defined by the equation:

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

Where:

• \(E\) represents the photon energy in electron volts (eV).
• \(h\) is Planck's constant, approximately \(4.1357 \times 10^{-15} \, \text{eV} \cdot \text{s}\).
• \(c\) denotes the speed of light in vacuum, approximately \(3 \times 10^{8} \, \text{m/s}\).
• \(\lambda\) is the wavelength of the emitted light, measured in meters.

The band gap energy for Gallium Arsenide (GaAs) is provided as \(1.42 \, \text{eV}\). This energy is equivalent to the energy of photons emitted by the LED.

The equation will be rearranged to solve for wavelength \(\lambda\):

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

The known values are substituted into the equation:

\(\lambda = \frac{4.1357 \times 10^{-15} \, \text{eV} \cdot \text{s} \times 3 \times 10^{8} \, \text{m/s}}{1.42 \, \text{eV}}\)

The wavelength is calculated as follows:

\(\lambda = \frac{12.4071 \times 10^{-7} \, \text{m} \, \text{eV/s}}{1.42 \, \text{eV}} \approx 8.7408 \times 10^{-7} \, \text{m}\)

Meters are converted to nanometers (1 m = 10⁹ nm):

\(\lambda \approx 874.08 \, \text{nm}\)

The approximate wavelength of light emitted from the LED is \(875 \, \text{nm}\).

This computed value aligns with the correct answer of \(875 \, \text{nm}\).

Consequently, the correct answer is 875 nm.