Quantum mechanics and an LED lab help

[venndicator]

so i have to do this lab for ib physics and i am a noob at physics and i am having some trouble with the lab that i have to do.

Homework Statement

the research question is "how does potential difference needed to operate an LED depend on the frequency of the light emitted?
it is set up in a circuit with a voltmeter to 6.0v, and a 220ohm resistor. there were 6 colors that i had to use, purple, blue, green, yellow, red and infrared.

i now have to potential for each of the 6 colors, and the teacher used his spectrometer to give us the wavelength and intensity of the light. the wavelength and intensity are just lots of data points, but when i graphed them, i noticed peaks following the colors of the rainbow but reversed. now my question to you guys is how do i find the frequency of the light wave?

Homework Equations

the only equation he gave was e=hf, with e being energy, h is plank's constant, and f is frequency. but i still don't know what to do here. i need to find the frequency somehow.

The Attempt at a Solution

cant find the frequency :(

 

[collinsmark]

Hello venndicator,

Welcome to Physics Forums!

so i have to do this lab for ib physics and i am a noob at physics and i am having some trouble with the lab that i have to do.

Homework Statement

the research question is "how does potential difference needed to operate an LED depend on the frequency of the light emitted?
it is set up in a circuit with a voltmeter to 6.0v, and a 220ohm resistor. there were 6 colors that i had to use, purple, blue, green, yellow, red and infrared.

i now have to potential for each of the 6 colors, and the teacher used his spectrometer to give us the wavelength and intensity of the light. the wavelength and intensity are just lots of data points, but when i graphed them, i noticed peaks following the colors of the rainbow but reversed. now my question to you guys is how do i find the frequency of the light wave?

Homework Equations

the only equation he gave was e=hf, with e being energy, h is plank's constant, and f is frequency. but i still don't know what to do here. i need to find the frequency somehow.

The Attempt at a Solution

cant find the frequency :(

Wow. What an excellent lab exercise! It's so simple, and so to the point. You're going to love this!

Okay, I'll get you started. The velocity of a wave is its frequency multiplied times its wavelength. This is true for any wave (sound waves, water waves, any type of waves [assuming the waves are periodic]). It also applies to light waves. How fast does light travel?

 

[venndicator]

thanks collinsmark,
c=3x10^8

i asked my friend, he said that the peak on the graphs, like the point with the highest point on the graph is the intensity. like the peak for intensity for purple is 0.7, and the wavelength at 0.7 is ~412nm. does that mean that the wavelength is 412nm? he then said to put it into f=c/lambda. is this the right way to do this?

and also, he told me he is stuck, he asked me "are we supposed to end up at plank's constant?" and asked me what the resistor is for.

edit: i looked it up online, i am so much derp, the wavelengths are pretty identical to my graph >.>
i just have to use f=c/lambda

 

[collinsmark]

thanks collinsmark,
c=3x10^8

i asked my friend, he said that the peak on the graphs, like the point with the highest point on the graph is the intensity. like the peak for intensity for purple is 0.7, and the wavelength at 0.7 is ~412nm. does that mean that the wavelength is 412nm? he then said to put it into f=c/lambda. is this the right way to do this?

If I'm understanding you correctly, that sounds reasonable.

and also, he told me he is stuck, he asked me "are we supposed to end up at plank's constant?"

Planck's constant is an important part of all this. You'll end up using it one way or another.

and asked me what the resistor is for.

If you were to connect the LED directly across a 6.0 V source (forward biased), the LED would burn up and would never work again. The resistor needs to be there to limit the current.

Somehow you'll end up needing voltage drop that was across each LED, when you did the experiment. Perhaps you measured the voltage across the LED with the voltmeter? Or if you measured the current through the circuit, you could use Ohms law to find the voltage drop across the resistor, and subtract that from the 6.0 V to find the voltage across the LED.

edit: i looked it up online, i am so much derp, the wavelengths are pretty identical to my graph >.>
i just have to use f=c/lambda

Okay, that's a good start.

 

[venndicator]

i just graphed the frequency vs potential, it looks linear. but i still don't know how Planck's is used here

edit: i some person on FB told me that "energy of light is E=hf. The energy loss of electrons is E=qV as they pass the voltage difference in the LED. Combine the equations and use the graph to get a value for h. compare with actual value." i think this is legit, as he is doing the same lab i am. but what i do not know is how to find the q, charge, is it the elementary charge?

 

[collinsmark]

Elementary charge of an electron: you can look this up on google. Other than that, you are on the right track. Good job!

 

[venndicator]

many thanks, collinsmark, you have helped me a lot on this. i think i should have used this a long time ago. you are so helpful, maybe i can use this site for more labs next time :)

 

[SammyS]

...

Wow. What an excellent lab exercise! It's so simple, and so to the point. You're going to love this!
...

Yes. It does look like a great lab exercise, assuming I understand what's going on here.

 

[collinsmark]

many thanks, collinsmark, you have helped me a lot on this. i think i should have used this a long time ago. you are so helpful, maybe i can use this site for more labs next time :)

As a follow up, yes, the charge you are looking for is the elementary charge of an electron. The theory goes that electrons are "pumped" up from a lower energy state to a higher energy state by applying a voltage across the LED's PN junction (and this particular voltage is determined by properties specific to the chemical makeup/structure of the semiconductor -- we don't need to concern ourselves with the particulars of this makeup/structure though for this exercise. Measuring the voltage across the terminals will suffice).

As an electron "falls" back down from the higher energy state to the lower energy state, a photon is released.

You don't need to just trust somebody on Facebook [Edit: or even me for that matter] that the relationship between the frequency of a photon and its energy is E = hf. This relationship was established by Albert Einstein 1905, in his paper "Concerning an Heuristic Point of View Toward the Emission and Transformation of Light", A. Einstein, Annelen der Physik, Mar. 18, 1905. Although this paper was published before the term "photon" was used (so you won't find the term "photon" in the paper). And of course, your instructor gave you this equation as part of the lab exercise too. [Edit: and if I'm not mistaken, you've now just proved this to yourself in your own experiment!]

And yes, the energy it takes to move an electron across a potential difference V is eV where e is the charge of an electron. The only assumption being made here is that each electron involved moves from one particular lower energy state to a particular higher energy state, and there is exactly one photon per energy state transition of each electron, as it falls back. With that, it is correct to equate the two energy equations giving eV = hf (or eV = hc/λ).

You've just shown a relationship for the minimum potential difference across the terminals of an LED (for it to produce light), and the wavelength of the LED's light (which is essentially the same thing as the light's color). This relationship involves:

• The speed of light
• Planck's constant
• The elementary charge of an electron,

all in one simple experiment. Neat, huh!