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 determine the wavelength of light emitted from the LED, we need to understand the relationship between the energy of a photon and its wavelength. The energy of a photon is given by the equation:

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

Where:

• \(E\) is the energy of the photon, measured in electron volts (eV).
• \(h\) is Planck's constant, approximately \(4.1357 \times 10^{-15} \, \text{eV} \cdot \text{s}\).
• \(c\) is 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 of GaAs (Gallium Arsenide) is given as \(1.42 \, \text{eV}\). This energy corresponds to the energy of the photons emitted from the LED.

We will rearrange the equation to solve for wavelength \(\lambda\):

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

Substitute the known values into the equation:

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

Calculate the wavelength:

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

Convert meters to nanometers (1 m = 10⁹ nm):

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

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

This calculation matches the correct answer, which is \(875 \, \text{nm}\).

Therefore, the correct answer is 875 nm.

Answer 2

The wavelength of light emitted from an LED is related to its energy gap by the formula:

\( \lambda = \frac{1240}{E_g} \),

where:

• \( \lambda \) is the wavelength in nanometers (nm),
• \( E_g \) is the energy gap in electron volts (eV).

Substitute \( E_g = 1.42 \, \text{eV} \):

\( \lambda = \frac{1240}{1.42} \).

Perform the calculation: \( \lambda = 875 \, \text{nm} \, \text{(approximately)}. \)
Final Answer: 875 nm.