Reef Central Online Community - View Single Post - P.Photon questions with trouble calc energy of photon and PPFD per joule

chubba hubba · 03/31/2009, 01:06 PM

Hi Sanjay, et al,

I have read your 'Facts of Light' series and I must say it is excellent! But I do have a few questions for you or anyone who can help me:

I am comparing different 'horticultural lamps' via the system wattage (lamp + ballast) and the manufactures SPD curve. But I worry I might be comparing apples to oranges in that the SPD curves were not created with the same environment or equipment (ie. ballast, spectrometer sphere, etc). What kind of error margin should I expect? Is the comparison even valid?

I intend to use my uni's spectrometer to get my own SPDs for the bulbs I am testing. I will also test/compare reflectors in terms of their foot-prints and efficiency using a PPFD meter. To do so I want to measure the PPFD one inch from the bulb and then again at 18" (using an avg of many PPFD plots). I will than find the PPFD difference and use that as a comparison for the true reflectance of a reflector. (Does that make sense?) I will use the most effective reflector to test the bulbs. Yet I will not be able to carry out any testing for a few months at least, hence my questions about using the manufacturer's SPD curve.
When weighting photosynthetic photons to match the absorption spectrum of my target plant I will use the “Inada” curve, formulated with the “Quantum Yield” (QY) of my target plant. Using this method when comparing PPFD (and PAR) of different lamps should be a valid comparison without comparing the SPDs too, correct?
Is it valid to use the SPD curve Spectral Irradiance (watt/m^2/nm) to calculate PPFD of each nm if I do not know the lamp distance used when creating the SPD curve? It seems the converted PPFD values would not be of much use because I do not know the lamp testing distance.
Is there a standard distance lamp manufacturers use when creating SPD curves?
Is the term "Action Spectra of Photosynthesis" (ASP) a valid expression for 400-500 nm U 600-700 nm? (U = union)

ARITHMETIC QUESTIONS (conversion from watt to photon):

I want to convert all the SPD wavelengths from watt/m^2/sec to PPFD. In order to do so I assume I need to be able to accomplish the following...?

J = joule
h = Plank's constant
nm = nano-meter (wavelength)
E = the energy held in a photon
c = speed of light in meters/second reduced using an exponent of 8
N = the number of photons needed to fill one joule, as in photons per joule

A.) When I use the equation provided ("Facts of Light: Part I") to find the energy of a single photon at 500 nm I continually get a different solution than Sanjay. I am using a 'TI-83 Plus' Texas Instruments calculator:

Equation: E = hc/nm*10^-9

E = ((6.626*10^-34)(3*10^8))/(500*10^-9)

My solution: E = 3.9756*10^-19 J

Sanjay solution: E = 0.039756*10^-17 J

What am I doing wrong?

B.) I am also having trouble finding the number of photons per joule at 500 nm. I continually get a different solution than Sanjay:

Equation: N = nm/hc

N = 500/((6.626*10^-34)(3*10^8))

My solution: 2.515343596*10^27 photons

Sanjay solution: 25.15*10^17 photons

What am I doing wrong?

Thanks for your GREAT work!

03/31/2009, 01:06 PM	#1
chubba hubba Registered Member Join Date: Mar 2009 Posts: 2	P.Photon questions with trouble calc energy of photon and PPFD per joule Hi Sanjay, et al, I have read your 'Facts of Light' series and I must say it is excellent! But I do have a few questions for you or anyone who can help me: I am comparing different 'horticultural lamps' via the system wattage (lamp + ballast) and the manufactures SPD curve. But I worry I might be comparing apples to oranges in that the SPD curves were not created with the same environment or equipment (ie. ballast, spectrometer sphere, etc). What kind of error margin should I expect? Is the comparison even valid? I intend to use my uni's spectrometer to get my own SPDs for the bulbs I am testing. I will also test/compare reflectors in terms of their foot-prints and efficiency using a PPFD meter. To do so I want to measure the PPFD one inch from the bulb and then again at 18" (using an avg of many PPFD plots). I will than find the PPFD difference and use that as a comparison for the true reflectance of a reflector. (Does that make sense?) I will use the most effective reflector to test the bulbs. Yet I will not be able to carry out any testing for a few months at least, hence my questions about using the manufacturer's SPD curve. When weighting photosynthetic photons to match the absorption spectrum of my target plant I will use the “Inada” curve, formulated with the “Quantum Yield” (QY) of my target plant. Using this method when comparing PPFD (and PAR) of different lamps should be a valid comparison without comparing the SPDs too, correct? Is it valid to use the SPD curve Spectral Irradiance (watt/m^2/nm) to calculate PPFD of each nm if I do not know the lamp distance used when creating the SPD curve? It seems the converted PPFD values would not be of much use because I do not know the lamp testing distance. Is there a standard distance lamp manufacturers use when creating SPD curves? Is the term "Action Spectra of Photosynthesis" (ASP) a valid expression for 400-500 nm U 600-700 nm? (U = union) ARITHMETIC QUESTIONS (conversion from watt to photon): I want to convert all the SPD wavelengths from watt/m^2/sec to PPFD. In order to do so I assume I need to be able to accomplish the following...? J = joule h = Plank's constant nm = nano-meter (wavelength) E = the energy held in a photon c = speed of light in meters/second reduced using an exponent of 8 N = the number of photons needed to fill one joule, as in photons per joule A.) When I use the equation provided ("Facts of Light: Part I") to find the energy of a single photon at 500 nm I continually get a different solution than Sanjay. I am using a 'TI-83 Plus' Texas Instruments calculator: Equation: E = hc/nm10^-9 E = ((6.62610^-34)(310^8))/(50010^-9) My solution: E = 3.975610^-19 J Sanjay solution: E = 0.03975610^-17 J What am I doing wrong? B.) I am also having trouble finding the number of photons per joule at 500 nm. I continually get a different solution than Sanjay: Equation: N = nm/hc N = 500/((6.62610^-34)(310^8)) My solution: 2.51534359610^27 photons Sanjay solution: 25.1510^17 photons What am I doing wrong? Thanks for your GREAT work!