How Does Voltage Affect LED Color Through Band Gap Adjustments?

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Increasing voltage in an LED influences its color by affecting the band gap through electron and hole dynamics in the quantum wells. Higher voltage attracts electrons and repels holes, leading to a wider band gap and higher energy light emission, resulting in a different color. Conversely, applying negative voltage has the opposite effect, decreasing the band gap. The discussion seeks practical examples of this phenomenon in real-life applications. Understanding these principles is crucial for advancements in LED technology.
Janusz Bednarski
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Homework Statement
How can one use a quantum well to change the colour of emitted light from an LED without changing the semiconductor material?
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Hi!

This question is translated so that's why it might sound a bit weird. I know that the answer should mention the band diagram of the LED, and I think that the answer is to introduce an electrical contact connected to a voltage source. Then when you increase the voltage, it should attract the electrons from the quantum well, and repel the holes from the other well, thus increasing the band gap and increasing the emitted light's energy (different colour). And vice versa for a negative voltage.

Am I correct?

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I am no longer "cutting edge" for LEDs but I am unaware of practical examples of this technology. Does anyone have a real-life example of this idea?
 
Thread 'Correct statement about size of wire to produce larger extension'

Thread 'Correct statement about size of wire to produce larger extension'

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
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