Determining Wavelength of Light From LED's

[4lec]

im doing physics practical work at A level.
in my project i need to work out the wavelength of diff colour light from LED's. im using the equation - lambda = hc upon eV - sorry for having to type it on here. my problem is that all my results seem to be around the 5000nm mark, when according to the company the LED's were purchased from i should be around 500nm. does anybody know what would cause this massive discrepancy and/or do you think it is within error bands? as im only using basic equipment.
any idea's greatly appreciated
alec

[Buenchico]

Well, it's definitely not within acceptable error bands, that's for sure! You should be getting results around about 430nm for blue LEDs up to about 660nm for red LEDs.

If you provide some sample data from one of your experiments, someone here might be able to spot where you're going wrong. (It's probably as simple as measuring something in cm when it should be in mm. It's easily done!).

Post again, describing your experiment, together with some sample data, and we might be in a better position to identify the problem.

Chris

[4lec]

thanks chris
my experiment involves connecting up the LED to a d.c power supply and measuring the voltage and current at random points from 0V to approx 2V. after drawing a graph of these results, i have taken the approx point at which the 'curve' of best fit crosses the x - axis. which should, i think, be where the LED begins to conduct. this is then taken as the V value in the equation above.
for a green LED, this V value is 0.24V, which gives a value for Lambda as 5179.68nm.


if i try the same equation with an approx value for when light begins to be emitted, when current increases rapidly, i get something around 776.9nm for lambda. that seems more sensible


thanks again chris

[Buenchico]

Now I'm geting confused!

It's a long time since I studied physics but I would have expected the denominator in your equation to represent the energy gap, measured in electron volts rather than having a direct relationship with the p.d. across the LED. Looking at that formula, in the way which you've represented it, it would seem that doubling V should result in lambda being halved. This would mean that a single LED should be capable of displaying all frequencies in the visible light range, just by varying the voltage, whereas most LEDs I've come across don't exhibit this property. I can't be sure, but I've got a feeling that you need to think further about that formula.

Sorry I can't be of more help,

Chris

[4lec]

hiya again

its no problem, i think what im doing wrong is takin the voltage as it the LED begins to conduct current, when i should be taking it when light is finally emitted. the denominator is in electron volts (1.6x10^-19) multiplied by the p.d

thanks for your help chris, i think just explaining it to someone has helped me, coz i thought about it so much that i confused myself

cheers

alec