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Homework Help: Coloured light. plancks constant.

  1. Dec 8, 2008 #1
    My physics essay is on coloured light.

    after investigating i found that the higher level frequency light red yellow transfered more energy

    this is due to E=hf.

    now i want to prove my results coincide with plancks constant.
    i want to plot E against f, the gradient should be h.
    i have I,V, time, temperature rise and wavelength.
    what equation do i need to get Energy.
  2. jcsd
  3. Dec 8, 2008 #2


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    That all depends on what the context of I, V, time, temperature rise and wavelength is. Care to elaborate?
  4. Dec 10, 2008 #3
    ok this was my pratical.

    i connected a powerpack to a lamp and placed a coloured filter on top, with ammeters and voltmeters attached.

    i put a thermometer 5 cm away fom the lamp and measured the temperature rise. the

    red light heated the thermometer up the most followed by yelow green and blue.

    this was because E=hf and red has the highest frequency. i have the results but how do i find the Eneregy transfered so i can plot it on a E versus f graph?
  5. Dec 10, 2008 #4
    i measeured the temp rise for 10 minutes each
  6. Dec 10, 2008 #5


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    There are a number of things to clarify here.

    1. Red light has the lowest frequency, and longest wavelength, in the visible spectrum.

    2. If this is an ordinary incandescent lamp, it produces more power at red wavelengths than at blue. This simple fact explains the different temperatures you measured.

    3. The filters you used could be transmitting different amounts of incident light. For example, the red filter might transmit wavelengths between 630 and 700 nm, while the blue filter might transmit between 470 and 480 nm. This would lead to more heating with the red filter, simply because a larger range (70 nm) of wavelengths are transmitted than for the blue filter.

    To do what you are trying, you would need an experiment that allows a fixed number of photons, no matter what the wavelength, to be absorbed by the thermometer. I don't know just how one would accomplish this, but it would be a very complicated and expensive experiment to do.
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