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: What is asked.
An LED is a small lamp made from a semiconductor. We are told it is made from zinc selenide (ZnSe) and we must say what colour of light it gives.
Step 2: What sets the colour.
The colour an LED gives depends on a property of its material called the energy gap, written $E_g$. A bigger energy gap means each light packet carries more energy and the light is bluer. A smaller gap gives redder or even invisible infrared light.
Step 3: Link energy to wavelength.
The energy gap and the light's wavelength $\lambda$ are tied together by \[ E_g = \frac{hc}{\lambda} \] A large $E_g$ gives a short wavelength, and short wavelengths look blue.
Step 4: Value for zinc selenide.
Zinc selenide is a wide gap material. Its energy gap is about $2.7$ eV, which is fairly large.
Step 5: Find the wavelength.
Using $\lambda = hc / E_g$ with the handy value $hc \approx 1240$ eV nm: \[ \lambda \approx \frac{1240}{2.7} \approx 460\ \text{nm} \]
Step 6: Name the colour.
A wavelength near $460$ nm sits in the blue part of the visible band. So a zinc selenide LED glows blue. This is option (3). \[ \boxed{\text{Blue light}} \]