I am working in photo catalytic hydrogen production by water splitting with small amount (5vol%) of glycerol as scavenger under direct solar irradiation.
I have performed solar experiment in 150 ml quartz reactor
50 ml of aqueous glycerol solution, produced H2 is 20,000 micromole/h/g.
How much Q E or else apparent quantum efficiency for above conditions?
please tell me clear information about it?
It is simple equation to calculate QE
QE = Converted number of molecule / total absorbed photons
Respected Yurii V Geletii, Youngsoo Kim and Mario J. Muñoz-Batista
Thanking very much for your reply and suggestions
Is there any simple calculation or else experiment is available to measure how many photons are absorbed ?
Start from learning Beer–Lambert law, e.g. http://en.wikipedia.org/wiki/Beer%E2%80%93Lambert_law
Meijia Li Usually when you do photocatalysis reaction, you don't need a light simulator to act as sun.
Apparent quantum yield (%) = [number of reacted electrons or holes]/[number of incident photons] · 100 = [number of H2 molecules evolved · 2]/[number of incident photons] · 100.
AQYH2 =[ 2mole of hydrogen]/[mole of incident photon]
You are right Neelu Chouhan
In detail, Follow this simple equation
A.Q.Y (%) = 2 × M × NA × h × c/ P × S × t × λinc × 100
Where
M = Amount of hydrogen produced( in mol/ time) = (e.g. 2000 μmol/h)
NA = Avogadro Number = 6.022 × 1023
c = Speed of light = 3 × 108 ms ―2
h = Planck′s constant = 6.626 × 10 ―34 Js
P = Power density of the incident monochromatic light = (e.g. 200 mW/cm2 )
S = Irradiated area to produce M (mol) of hydrogen = (e.g.15 cm2)
t = Irradiated time = 1 h = 3600 s
λinc = wavelength of the incident monochromatic light =500 nm = 500 × 10 ―9 m
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