Physics, asked by singhgyan0612, 6 months ago

Explain with help of circuit diagram, the action of a forward biased p-n
junction diode which emits spontaneous radiation. State the least band gap
energy of this diode to have emission in visible region.

Answers

Answered by gopalkaluskar11
56

Answer:

Explanation:

Hey There!

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.

Answered by janvikaushik48
37

least band gap :- 1.8ev

Attachments:
Similar questions