Question: Yesterday in lab, we were asked why all of our green LED had large forward voltage than expected. We expected for the forward voltage to be at around 2.1V as according to hf=Egap=eV. Yet, we experienced forward voltages of 2.6-2.7 for our LEDs in the lab. Professor said that there is a reason, but I cannot seem to figure it out. Can I get help?

Could it be that the Green LED had some different composition as expected, allowing it to emit green light while having high forward voltage?

I searched through the internet, and a lot of sources say that Green LED forward voltage is 2.1V whereas other sources say that Green LED forward voltage is like 3.0V. Why the discrepancy?

Answer: 
There's green and then there's other green. This is why in engineering we use numbers instead of vague words. It is best to specify the actual wavelength.

Most common "green" LEDs are on the yellow end of what most people consider green. These do have about 2.1 V drop at reasonable currents. This type is the cheap and common "green" variety. This wavelength of green corresponds well to the peak sensitivity of our eyes.

There are other "green" LEDs that use different semiconductor substances and are more toward the bluish end of what most people would call green. These of course have a higher forward drop due to the physics you seem to understand correctly. These are much less common and typically more expensive. Their lumen specs also don't look as good, in part because this wavelength is a bit shorter than the peak sensitivity of human eyes.

My guess is that your professor deliberately gave you the unusual green type of LED to make you think a bit, learn not to take things for granted, and impress on you the need for real numbers as opposed to vague words like "green".
Wikipedia [lists](https://en.wikipedia.org/wiki/Light-emitting_diode#Colors_and_materials) green LEDs with a forward voltage between 1.9 and 4 V, so the excessive voltage drop you observed is related to technological imperfections in the LEDs you have tested. A few factors come to mind:

* electrical resistance. Any conductor introduces some voltage drop proportional to the current. You could have eliminated this factor by testing your LEDs at two different currents, and extrapolating the V/A curve to 0A.
* [crystallographic defects](https://en.wikipedia.org/wiki/Crystallographic_defect) in the die would cause electrons to lose extra energy during recombination. In the ideal lattice electrons and holes would have the same [momentum](https://en.wikipedia.org/wiki/Crystal_momentum), so all recombination energy would go to the photon. Defects result in a difference in momentum, and since photons cannot carry it, part of the energy is lost to [crystal lattice vibration](https://en.wikipedia.org/wiki/Phonon). To get rid of this effect, you'd need to test a high-grade LED with as little lattice defects as possible (I guess you got consumer-grade devices).
* there's also a potential energy loss to the [Fermi level](https://en.wikipedia.org/wiki/Fermi_level), which is defined as "the work required to add an electron to [a body], or equally the work obtained by removing an electron". The difference you observe (2.6 - 2.1 = 0.5eV) seem like a typical value. Though my knowledge of physics is not sufficient to tell whenever this phenomenon applies to electron-hole recombination in LEDs. Wikipedia [seem to suggest](https://upload.wikimedia.org/wikipedia/commons/d/d7/PnJunction-LED-E.svg) it does.