Absorption of Solar Radiation by Formaldehyde

Problem 3.4 from "Atmospheric Change by Graedel and Crutzen", p53:

The Table below lists solar flux intensities (I) at an altitude of 1km and north latitude 40o at the summer solstice (June 21) together with absorption cross sections for the formaldehyde molecule. If the concentration of formaldehyde at 1km on Jun 11 at that location is 5 ppbv, calculate the total photon flux deposited into a cubic centrimetre of air at midday as a consequence of formaldehyde absorption of solar radiation.

Wavelength (nm) 295-305 305-315 315-325 325-335 335-345 345-355 355-365
I (photons/cm2 / sec/ (10nm) ) 2.66. 1013 4.20. 1014 1.04. 1015 1.77. 1015 1.89. 1015 2.09. 1015 2.15. 1015
sigma (10-20 cm2/molecule) 2.62 2.45 1.85 1.76 1.18 0.42 0.06


Hint:


p=1e5;                   % Air pressure (neglecting decrease due to altitute)
                         % See p46-47 in the book)
T=273;
V=1e-6;                  % Convert cm^3  to m^3
R=8.31451;               % "Molar gas constant" in J/mol/K
n=p*V/R/T ;               % number of moles

NA=6.022E23;             % Avogadros Number
N=NA*n;                  % Number of molecules in V
Nf=5E-9*N;	         % Number of formaldehyde molecules (5 ppbv or N)										


I  =  [2.66E13  4.20E14  1.04E15  1.77e15  1.89e15  2.09e15  2.15e15 ];
sigma=[2.62e-20 2.45e-20 1.85e-20 1.76e-20 1.18e-20 0.42e-20 0.06e-20];
lamda=[300:10:360]*1E-9; 

Is=I.*sigma*Nf ;
plot (lamda,Is)          % Plot the absorption as a function of wavelength
hold on 
plot(lamda,Is,'o')

abs=sum(Is)              % Total absorption by formaldehyde (photons/ cm^3/ s)

abs2=Nf*I*sigma'         % Alternative evaluation using matrix and transpose (')
 

Ann-Marie Pendrill, 11 Nov 1996
http://fy.chalmers.se/~f3aamp/matlab/absorb.html