Question

If the energy gap of a semiconductor used for the fabrication of an LED is nearly 1.9 eV, then the color of the light emitted by the LED is

• A. White
• B. Red
• C. Green
• D. Blue

Hint:

For LED color problems, remember that energy and wavelength are inversely proportional. Higher energy gaps (like for blue or violet LEDs) correspond to shorter wavelengths, while lower energy gaps (like for red or infrared LEDs) correspond to longer wavelengths.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In a Light Emitting Diode (LED), when an electron recombines with a hole, it transitions from a higher energy level (conduction band) to a lower energy level (valence band). This process releases energy in the form of a photon. The energy of the emitted photon is approximately equal to the band gap energy (\(E_g\)) of the semiconductor material. The color of the emitted light is determined by the wavelength of these photons.
Step 2: Key Formula or Approach:
The energy of a photon (\(E\)) is related to its wavelength (\(\lambda\)) by the formula:
\[ E = \frac{hc}{\lambda} \] where \(h\) is Planck's constant, and \(c\) is the speed of light. For an LED, \(E \approx E_g\).
A very useful shortcut for calculations, when energy is in electron-volts (eV), is:
\[ \lambda (\text{in nanometers}) = \frac{1240}{E (\text{in eV})} \] Step 3: Detailed Explanation:
We are given the energy gap, \(E_g = 1.9 \text{ eV}\).
We can calculate the wavelength of the emitted light using the shortcut formula:
\[ \lambda = \frac{1240}{1.9} \text{ nm} \] \[ \lambda \approx 652.6 \text{ nm} \] Now, we need to match this wavelength to the corresponding color in the visible spectrum. The approximate wavelength ranges for visible colors are:
- Violet: 400 - 450 nm
- Blue: 450 - 495 nm
- Green: 495 - 570 nm
- Yellow: 570 - 590 nm
- Orange: 590 - 620 nm
- Red: 620 - 750 nm
The calculated wavelength of approximately 653 nm falls squarely within the red region of the spectrum.
Step 4: Final Answer:
The color of the light emitted by the LED is Red. Therefore, option (B) is correct.