Question

LED is manufactured using zinc selenide then it emits

• A. infrared radiations
• B. yellow light
• C. blue light
• D. green light

Hint:

Here is a quick reference guide for common LED materials and their colors to memorize for entrance exams:
GaAs (Gallium Arsenide): Infrared
GaP (Gallium Phosphide): Red / Green
ZnSe (Zinc Selenide): Blue
InGaN (Indium Gallium Nitride): Blue / Violet / Ultraviolet

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the characteristic color of light emitted by a Light Emitting Diode (LED) manufactured using the compound semiconductor material

Zinc Selenide (ZnSe).

Step 2: Key Formula or Approach:
The color of light emitted by an LED is determined by the energy bandgap ($E_g$) of the semiconductor material used in its construction. The relationship between the bandgap energy and the wavelength ($\lambda$) of the emitted photon is given by: $$E_g = \frac{hc}{\lambda}$$ Different semiconductor alloys produce different bandgap energy levels, which correspond to specific colors across the visible and infrared spectrum.

Step 3: Detailed Explanation:
Zinc Selenide (ZnSe) is a wide-bandgap compound semiconductor material with a direct bandgap energy of approximately $2.7\text{ eV}$ at room temperature.
Let's find the approximate wavelength of light corresponding to this energy level: $$\lambda = \frac{hc}{E_g} \approx \frac{1240\text{ eV}\cdot\text{nm}}{2.7\text{ eV}} \approx 460\text{ nm}$$ A emission wavelength around $460\text{ nm}$ falls within the $450\text{--}495\text{ nm}$ band of the visible spectrum, which corresponds to

blue light. Therefore, Zinc Selenide LEDs are used to generate blue light emissions, matching option (C).

Step 4: Final Answer:
An LED manufactured using zinc selenide emits blue light, corresponding to option (C).