Question

How do we choose the semiconductor, to be used in those diodes, if the emitted radiations, is to be in the visible region?​

Answer 1

Answer:

The p-n junction diode, which emits spontaneous radiation when forward biased, is known as the light emitting diode or LED. The visible light is from 0.45 μm to 0.7 μm and corresponding energy is between 2.8eV to 1.8 eV .therefore, the least band gap of the semiconductor to be used in LED, in order to have the emitted radiation to be in the visible region, should be 1.8 eV. Phosphorus doped gallium arsenide and gallium phosphide are two such semiconductors.

Explanation:

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