Experiment to find Plank's Constant

[Jimmy87]

Homework Statement

Determine Plank's constant by measuring threshold voltage for a range of different colored LED's. The experiment involves increasing the voltage across a diode until a current starts to flow. You then measure the voltage across the LED at this time (Vo - threshold voltage). The theory goes that the threshold voltage is the voltage needed to give the electrons the same energy as a photon that is emitted. This is the bit I don't quite get.

Homework Equations

E = hc/lambda = eVo (E=energy of photon, h = plank's constant, c = speed of light, lambda = wavelength of light, e = charge of electron, Vo = threshold voltage)

The Attempt at a Solution

I have done this so that the threshold voltage is recorded when a current of 0.01mA flows. However, at this point, despite a current flowing, the LED does not emit photons of enough intensity to visually see them. If you increase the voltage, after the current increases by a few mA, the LED's shine. I don't understand why only the threshold voltage is equal to the same amount of energy as the photons for that LED? When you increase the voltage and it shines brightly surely the voltage is still transferred directly to the photons? What is the significance of the threshold voltage and is it supposed to be the voltage at which the led first shines or the instant a current starts to flow (i.e. 0.01A)?

Thanks for any guidance given.

 

[DEvens]

The idea is, photons come as little clumps of energy. That is, a photon of a particular color also has a particular wavelength, and also has a particular energy.

On the LED side, the notion is that there is a gap of some kind. And when an electron just manages to hop that gap the result is it gives up some energy in the form of a photon. So to get the electron through this gap you have to push hard enough. And when you do you put energy into the electron. When it gets through the gap it gives up this energy. The characteristic energy for the gap is fixed for a given LED.

If you push harder then the electron only gives up the characteristic energy of the gap. So the voltage across the LED will rise with increasing applied voltage. But it will always have the deficit due to the energy given up at the gap.

The reason the color of the LED is pure is because the gap is finely controlled to be a single energy. As well, it is controlled to be in a single layer, meaning a given electron only gives up energy in one gap not several. That is part of why monochrome LEDs have that spiffy sparkly look. It's basically that you are seeing a facet of this layer. And the reason they are relatively cool is because very little energy is lost to other things like resistance heating.

So pushing harder means more electrons go through, giving more current. But they all have this step in energy at the gap, each producing a photon of the same color. So the voltage you measure across the diode behaves as you describe.

As a metaphor (possibly somewhat misleading) think of two level planes separated by a ramp. You roll marbles at the ramp from the bottom. If you roll them too slowly they just deflect off and don't make it up. If you roll them hard enough they make it up, losing a bit of energy. If you roll them harder still they still make it up, losing the same amount of energy. In an LED that energy is lost as a photon. Faster moving electrons hit the gap (the ramp) and lose the same amount of energy.

 

[Jimmy87]

The idea is, photons come as little clumps of energy. That is, a photon of a particular color also has a particular wavelength, and also has a particular energy.

On the LED side, the notion is that there is a gap of some kind. And when an electron just manages to hop that gap the result is it gives up some energy in the form of a photon. So to get the electron through this gap you have to push hard enough. And when you do you put energy into the electron. When it gets through the gap it gives up this energy. The characteristic energy for the gap is fixed for a given LED.

If you push harder then the electron only gives up the characteristic energy of the gap. So the voltage across the LED will rise with increasing applied voltage. But it will always have the deficit due to the energy given up at the gap.

The reason the color of the LED is pure is because the gap is finely controlled to be a single energy. As well, it is controlled to be in a single layer, meaning a given electron only gives up energy in one gap not several. That is part of why monochrome LEDs have that spiffy sparkly look. It's basically that you are seeing a facet of this layer. And the reason they are relatively cool is because very little energy is lost to other things like resistance heating.

So pushing harder means more electrons go through, giving more current. But they all have this step in energy at the gap, each producing a photon of the same color. So the voltage you measure across the diode behaves as you describe.

As a metaphor (possibly somewhat misleading) think of two level planes separated by a ramp. You roll marbles at the ramp from the bottom. If you roll them too slowly they just deflect off and don't make it up. If you roll them hard enough they make it up, losing a bit of energy. If you roll them harder still they still make it up, losing the same amount of energy. In an LED that energy is lost as a photon. Faster moving electrons hit the gap (the ramp) and lose the same amount of energy.

Thank you for your detailed answer, it has helped a lot. What I still don't get though is why you use the threshold voltage. The theory in my book says that the threshold voltage is:

"the voltage needed to give the electrons the same energy as a photon emitted by the LED."

You can therefore set the energy of a photon equal to the charge of an electron multiplied by the threshold voltage. I don't get why you need to measure the threshold voltage though? Let's say you increase the voltage across the LED until a current just begins to flow (0.1mA) and you measure the threshold voltage to be 1.6V. When you increase the voltage further to say 2.3V the LED starts to shine and more current flows through it (3mA). If you use 2.3V instead of 1.6V for a certain color you get a different answer for plank's constant. Why must you use the threshold voltage - if you increase the voltage surely the energy of the electrons is still all going to the emitted photon?

 

[haruspex]

When you increase the voltage further to say 2.3V the LED starts to shine and more current flows through it (3mA).

I think the answer is implied here:

If you push harder then the electron only gives up the characteristic energy of the gap.

That is, the electron will retain any spare energy. Since you can't measure that, you have to stick to the threshold voltage so that you can assume it is zero.
That leaves the question of why increasing the voltage increases the brightness. Presumably that has to do with the rate of electron flow (current), not the energy per electron.

 

[Jimmy87]

I think the answer is implied here:
That is, the electron will retain any spare energy. Since you can't measure that, you have to stick to the threshold voltage so that you can assume it is zero.
That leaves the question of why increasing the voltage increases the brightness. Presumably that has to do with the rate of electron flow (current), not the energy per electron.

Thank you!